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Hello and welcome to our second lecture in
this series on Random Variables and Probability
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Distributions for a course Introduction to
Data Analytics. In the previous lecture we
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saw, we made a more broad introduction to
the concept of random variables and distributions,
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we spoke about the concept of having continuous
distributions and discrete distribution, we
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presented some examples.
And we also spoke about the idea of a probability
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density function versus cumulative density
function. In todayÕs lecture, we are going
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to pick a 5 or 6 very common distributions
and discuss them one at a time. And the first
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one that and these distributions are going
to be both, some of them are going to be discrete,
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some of them are going to continuous and some
of them could be both.
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So, we start with the most common distribution,
which is the uniform distribution. The uniform
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distribution has a discrete version and a
continuous version. Now, we will start with
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the most simple one, which is the discrete
version. You already seen examples of this.
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So, for instance the example that we saw on
the last class of the six sided dice, where
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we were quantifying the probability of getting
any particular face values.
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So, the face value of the dice is essentially,
you throw the dice you get something on top,
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so you see the 1 or 2 or 3 or 4 or 5 or 6.
So, the probabilities associated with these
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6 possible outcomes, assuming a fair dice
is one sixth for each and, so that is example
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of discrete uniform distribution. We also
saw the case of the coin toss, where here
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you have only two possible outcomes and as
long as they are both equal, it is still uniform.
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So, the formula in terms of the PDF the Probability
Density Function, which we discussed in the
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last class, it is fairly straight forward,
it is just 1 by k when there are k outcomes.
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So, if this is 6 sided dice it is 1 by 6 for
each of those six possible outcomes. If it
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is coin toss, which is two possible outcomes
it is going to be 1 by 2 and the ideas that
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for each of those k possibilities, it is 1
by k and for the rest of the universe, it
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is 0.
So, the probability of getting a 7 when you
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role a 6, a single 6 sided dice is 0, so you
cannot get 7 and you cannot get minus 45 either,
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so that is what this formula says. With the
continuous version, here again you are looking
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at a uniform distribution, so the probability
is a uniform, but like we discussed the variable
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that we are quantifying is continuous. So,
the variable in the six sided dice was, what
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is the number that shows up and that number
is either 1, 2, 3, 4, 5, or 6 and that 6 discrete
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possibilities, but you might have many things
that are not discrete and this goes back all
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the way to our discussions and quantitative
variables, which can be continuous or discrete.
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So, examples of this and the truth is the
uniform distribution, while it is theoretically
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very intuitive, could be quite convenient
in some cases. There are not a lot of examples,
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real world things that tend to be uniforms,
some of them are something like number of
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seconds pass the minute. So, if you were to
randomly, if you have to have random process,
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which just looked at the clock over the course
of date during some random intervals, the
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number of the seconds pass the last minute
could be uniform and that is essentially a
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space, where you can have 0 seconds pass the
minute all the way to 60 seconds pass the
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minute.
So, your interval is from 0 to 60 and where
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you find yourself in that interval is described
potentially by uniform distribution. And again
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the idea that it is continuous, just means
that you not discretized the seconds as 1
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or 2 or 3 all the way to 60, you could be
in that in 4.5432 seconds pass the minute.
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So, it is continuous, the variable time is
being monitored continuous as a continuous
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variable.
Another example could be the exact age of
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the randomly selected person between the ages
of 50 and 60 perhaps in a certain country.
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Now, again you know you have to come up with
fairly specific examples, because something
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simple like the age of the randomly selected
person is not likely to be uniform. You are
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not likely to find as many people between
80 and 90, as you are between 70 and 80.
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And even if you stop the clock at a certain
point with typically with things like age,
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you see that the number of people or the probability
of a person in that window decreasing as sometimes
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age increases and it depends on your state
space and it depends on your countries, is
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the population increasing decreasing, so on
and so forth. But, sometimes over a short
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enough interval, even if the overall distribution
is not uniform you can create an interval
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and say that is my universe.
So, my universe is people between the age
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30 to 40, within that interval perhaps the
distribution of the exact age of people. So,
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imagine I take the universe of all people
between the ages of 30 to 40 in India and
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then, I try to create a distribution of the
exact age of a randomly selected person from
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this bucket and that could potentially be
a uniform. The uniform is also a great distribution,
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think of when you want to make absolutely
no assumptions.
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So, if I want to make the assumption that
the countryÕs population is and some of these
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need not even be assumptions, so it could
be a fact. So, something simple like the countryÕs
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population is increased and so on, then I
would be hard pressed to come up with a uniform
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distribution and I might find richer distribution
to represent my data. But, this uniform distribution
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can be thought of us like, this ground 0 I
do not make any assumptions, the probability
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of everything occurring within a certain interval
is equal and, so I could potentially use it.
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I have also come across the usage of this
a little bit in different aspects like physics
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and chemistry and so on, for instance the
probability of certain types of molecules
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be in certain locations over certain space
could be uniformly distributed and so on.
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So, we have spoken about, so the formula for
the uniform distribution, we spoke about the
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discrete uniform PDF and we said, the PDF
is essentially 1 by k. So, with the continuous,
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the PDF is essentially 1 by b minus a.
So, the idea here is that, this is essentially
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a and this is essentially b and as you can
see in the graph, the probability between
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a and b is uniform and just like in the discreet
case, if this distance b minus a, then the
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probability is 1 by b minus a. And another
way, another quiet simple way of thinking
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about it is, between this interval, over this
entire interval from the lower limit to the
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upper limit to the area under this curve should
be equal to 1 and if you look at the simple
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math of it, 1 by b minus a times b minus a,
which should be, so the height is 1 by b minus
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a, the length of this rectangle is b minus
a.
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So, 1 by b minus a times b minus a, would
give you 1 and, so you can essentially think
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of it that ways well. And; obviously, like
we discussed it has to be 0 for if you are
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less than a if you are greater than b. So,
what is this CDF of this distribution? Again,
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it is x minus a by b minus a, we will not
be exactly deriving it out. We will give you
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formulas at the end of this lecture to give
you an idea of, how to get to the CDF, how
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to get to the mean, how to get to the variance.
But, the core idea here is that, essentially
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you take some point and let us call this point
x, so this is x, there are better x. And the
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whole idea is that we know this CDF is essentially
the area under the curve to the left of x.
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So, the question could be if you knew that
this height is 1 by b minus a, how do you
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go about writing out this area, how do you
calculate this area and it can be a function
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of x. So, that general formula would be the
CDF and the idea is fairly simple. If this
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is x and you are looking at the CDF is a ratio,
it is essentially the area, we know that the
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total area is 1.
So, this total area of the blue rectangle
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is 1 that we have discussed. So, what percentage
of that rectangle have you essentially covered
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and the idea is, because it is uniform, because
this height is constant. If this is x and
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this is a and this is b, then of the 1, you
covered x minus a is this area that you have
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covered and b minus a would be the full area
that you could be possibly covers. If you
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think of it as a ratio of, how much have covered
and how much I can potentially cover, where
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this b minus a kind of represents the 1 in
some sense of the full area.
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Then, I have covered x minus a of the b minus
a that I could cover and therefore, I covered
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the percentage in some sense, where; obviously,
if x was equal to b, then I would have covered
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the 100 percent and that would be equal to
1, so that is the over all idea behind getting
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this formula. Again the formula for the mean
should be fairly intuitive, if this is a and
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you know, you have this is b, this central
point in some sense is b by you know half
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of between b and a essentially.
And again with variance we will not derive
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it, accept to tell you that the core concept
is to say how much do we deviate on average
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from the mean. So, we will talk through some
of this formula some of it is to give you
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an intuition, some of it is to actually give
you these problems in the assignment. So,
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you get a feel for actually figuring out what
the mean of variance exactly is.
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Let us move on to the next distribution. The
next distribution we are going to talk about
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is the binomial distribution. So, the binomial
distribution is also another distribution,
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where you have a lot of toy problems associated
with it, but by nature of it in the real world
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sometimes it is more useful to approximate
it with another distribution and user, but
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what exactly is the binomial distribution,
let us start there. So, we spoke about this
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class of distributions and if you did not,
then let us just do that right now, so let
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us.
So, there are these class of distributions
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where called the Bernoulli distributions and
the idea behind that distribution is that,
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it is very similar to the first example we
saw in the previous lecture, where you have
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an event and it has some probability. So,
the 30 percent chance is going to rain, that
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is what the exact example we use. And, so
therefore, the probability that it is not
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going to rain is 70 percent. What is key about
it is that, there are only two possible outcomes
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and these two possible outcomes, discrete
outcomes sum to 100 percent and that class
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of distribution is called Bernoulli distribution.
So, our standard example of tossing a coin
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and saying, what is the probability of heads
and what is the probability of tails, would
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be an example for Bernoulli distribution.
Now, if just, so that there is no confusion
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if it happened that the probability of heads
and tails are both equal, you can also think
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of it as a discrete uniform distribution and,
but out here we are just saying we saying
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something different. We saying that Bernoulli
distribution are distributions, where there
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are only two possible outcomes.
The probability of the two outcomes need not
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be the same as required by the uniform. But,
the key out here is that there are only two
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possible outcomes, two possible discrete outcomes
that sum to a 100 percent. So, great, so we
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spoken about the Bernoulli distribution, what
is it happen to do with the binomial distribution,
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is not that, what this slide is about. Well,
if you take the problem and you quantify the
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probability of heads and tails, then you are
talking about Bernoulli.
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But, instead if I rephrase that problem and
said, what is the probability of getting 5
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heads out of 10 tosses. Then, I am describing
the problem associated with a binomial distribution.
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So, an example of a binomial distribution
would be, if I said I am going to toss the
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coin 10 times, I am going to take part in
a BernoulliÕs process, n number of times
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and I asked myself the question, what is the
probability of getting k successes and success
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could be defined as getting a heads or it
raining or whatever it is. It is one of those
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two outcomes essentially; you call one of
those two possible outcomes of a Bernoulli
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process as a success.
And then, you say, what is the probability
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of getting k successes out of n possible trials?
So, if I say I am going to toss a coin 10
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times, can you tell me the probability associated
with 0 heads, 1 head, 2 heads, 3 heads, 4
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heads all the way up to 10 possible heads.
Now, note if I toss the coin 10 times, I can
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either get 0 heads or any of those numbers
in between till 10 heads, I cannot get minus
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1 heads I cannot get 11 heads. So, the probability
associated, if I toss the coin 10 times, the
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some probability that I am going to get 0
heads, the sum probability that I am going
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to get 5 heads.
But, each of these numbers each of these probabilities
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when added together should again be equal
to 100 percent or 1. So, quantifying the probability
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of getting k successes out of n trials of
a Bernoulli process is the binomial distribution
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and you can think a various real world example.
So, for instance the probability of, you know
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let say there are 10 mergers the companies
considering.
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So, in mergers and acquisitions since it is
a small enough number, what is the probability
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of getting 3 out of 10 of them or you might
say, probability of having 5 defective products
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in a batch of 20 products, what is the probability
of the 3 defective products, answering questions
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like that. And one just I noticed, you might
have noticed that I have take an example of
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small enough a numbers.
So, technically the binomial could be answering
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a question like, what is the probability that
out of you know 1 million possible toys. Let
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us say that I distribute, I am a toy maker
and I send out 1 million, what is the probability
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that I get 500, 5000 toys as broken. And technically
that would still be a binomial distribution,
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but what we will learn probably in the next
classes, why thatÉ The computation of that
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is a little messy, you get very large numbers
and you have some other distributions that
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can approximate something like this to solve
problems like that.
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But, the core concept is this. You have a
Bernoulli process, where something isÉ You
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are looking at something that is binary a
or b, 1 or 0 and you turn around and you say,
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what is the probability of getting k successes
out of n trails of this process. And, so logically
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the formula and note that I have used the
word PMF of here, that is an important distinction
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that is worth mentioning. PMF is the same
thing as PDF, PDF Probability Density Function,
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we spoke about that we spoke about probability
density functions and cumulative density functions.
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PMF is just the standard way of calling a
PDF if you are dealing with a discrete distribution.
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And since this is a discrete distribution,
because you cannot get out of 10 tosses of
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a coin you cannot get 3 and half heads. So,
you can only get either 0 heads, 1 head, 2
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head or 3 head and so. Since this is a discrete
distribution, it is technically called Probability
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Mass Function or PMF. So, the PMF, this distribution
should be fairly intuitive, it is nothing
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but, n choose k, so this symbol out here just
means you might have seen it like this.
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So, that is n choose k and that is has to
do with a combinations, something you might
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have studied in the permutations and combinations.
The idea is, how many ways are there of choosing
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n from k trails. So, here I think the words
n and k are swapped as suppose to, what I
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was mentioning earlier here. So, here for
instance I was interested in finding out,
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what is the probability of getting 5 heads,
then it would really be 10 would be the n
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c 5.
So, how many ways are there of getting 5 of
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choosing 5 out of 10, so I toss the coin 10
times, there are many ways in which, I could
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get 5 heads out of 10 tosses. It is either
that the first 5 could all be heads and the
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next 5 could all be tails or you can have
1 head 1 tail 1 head 1 tail. So, that number
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of different ways in which I can get 5 heads
out of 10 tosses is, what is being quantified
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by this number, which relates to this part
of the formula. Once I figure out the numbers
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of possible ways, this part, which is p power
k tells me, what is the probability of getting
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the 5 heads in the first place.
So, it might be 0.5 power 5, if the probability
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of head was different from a tails, let us
say we were dealing with the problem where
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60 percent chance of heads following, because
it is an un even coin, then it could be 0.,
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00:20:26,159 --> 00:20:33,660
it will be 0.6 power 5 and this part, which
is the remaining part talks about the probability
192
00:20:33,660 --> 00:20:39,810
of getting the tails, the remaining 5 as tails.
So, essentially the 3 parts of this formula
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are the different ways of getting those heads,
probability of getting so many of those heads,
194
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probability of getting the remaining number
as tails.
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So, that is essentially, what the PDF captures
and the formula for the cumulative density
196
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function again is the same logic that we were
discussing with the uniform, which is that
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for a given k, which might mean for let us
say I am interested in knowing, what is the
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00:21:07,450 --> 00:21:12,720
probability of 6 heads out of 10 tosses. You
are essentially looking at nothing but, everything
199
00:21:12,720 --> 00:21:17,320
to the left of the curve and to the left of
the curve here means, what is the probability
200
00:21:17,320 --> 00:21:22,049
of getting 0 tails I mean.
So, let us see what is the probability of
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getting 6 heads is the question out of 10
tosses and that would be nothing but, the
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00:21:26,651 --> 00:21:32,440
PDF of getting 0 heads out of 10 tosses, 1
head out of 10 tosses, 2 heads out of 10 tosses,
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all the way up to 6 heads out of 10 tosses,
that summation of those probabilities, because
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that is essentially what would be to the left
of the curve.
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And here I am doing nothing but, thinking
of a curve, where 1 out of 10 tosses. We should
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start with 0 out of 10 tosses and going all
the way to, you know 2 dot, dot, dot, dot
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6 and then, it will go 7 and it will go all
the way till 10 out of 10 tosses. So, if this
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was some kind of PDF, I am essentially looking
for this area and the curve and you know,
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the top part of the line is not uniform I
am just representing, which side I am interested
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in, that is all I am doing there and also
it would be discrete.
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So, you have discrete rectangles popping out
of the 1, 2 and 3 and you just summing them
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up and the formula also with the summation
does not really simplify too much, so you
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are dealing with the CDF as it is. As I mention
it is more useful for small values of n, when
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you get really large values of n, you took
it other approximations. The mean is nothing
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00:22:44,130 --> 00:22:48,080
but, the number of times you are looking at
times of probability.
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00:22:48,080 --> 00:22:54,669
So, if the probability of getting the heads
is 60 percent and you are going to toss the
217
00:22:54,669 --> 00:23:01,360
coin 10 times, the average number of heads
you are going to receive is nothing but, 10
218
00:23:01,360 --> 00:23:08,779
times 60 percent, which would be 6. So, on
average I should get 6 heads out of 10 tosses,
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00:23:08,779 --> 00:23:13,470
because this is 60 percent chance of getting
heads. And, so the intuition of that should
220
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be fairly obvious and again with variance,
the intuition itself might not be very obvious.
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It is really about looking at how much the
deviation, how much of a deviation there is
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from the mean.
So, how much does one out of 10 deviate from
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00:23:27,900 --> 00:23:31,529
the mean and we would look at the probability
that you would first of all see of 1 out of
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10. And again I think there the formulas for
variance might probably help you to get a
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better idea of it, great.
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00:23:38,710 --> 00:23:48,419
So, let us move to our next distribution this
is also a discreet distribution and this is
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very similar to the binominal in many ways
both of them are discrete, but essentially
228
00:23:54,179 --> 00:23:59,340
while this is discrete, which also counts
the number of possible occurrences x. So,
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it says, what is this quantifies the probability
of having getting x occurrences. So, probability
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00:24:04,850 --> 00:24:11,919
of getting one of those occurrences two of
those occurrences, but it is over a certain
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00:24:11,919 --> 00:24:16,659
period of time or space see in the binominal
we were saying, what is the probability of
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5 heads out of 10 tosses.
So, the 5 heads was discrete but, so was the
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10 tosses here you are asking the question,
what the probability of x is where x is discrete
234
00:24:29,409 --> 00:24:35,169
meanings, what is the probability of there
being 5 people. But, instead we wonÕt say
235
00:24:35,169 --> 00:24:39,240
out of 10 people we will instead say, what
is the probability of there being 5 people
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coming to this bus stop over the next 5 minutes
the key difference is 5 minutes is continuous
237
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as oppose to 10 tosses, which is discrete
the ramifications of that is; however, low
238
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the probability is of people arriving in a
given time there is still technically a probability,
239
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that you have a really large number of people
infinite number of people coming over the
240
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next 5 minutes, where as I can say with certainty
in the case of the binominal that out of 10
241
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tosses I cannot get 11 heads.
And this has nothing to do with the probability
242
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of getting a heads in the first place even
if I had a probability of 0.1 of getting a
243
00:25:24,000 --> 00:25:30,460
heads in a toss, because of such an unfair
coin I still know or 0.9 whatever you know
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whatever the probability is I still know that
its technically not possible to get 11 heads
245
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out of 10 tosses.
Here, you are looking at a discrete occurrence
246
00:25:38,620 --> 00:25:47,380
such as probability that of n number of defaults
in a given month or you know number of people,
247
00:25:47,380 --> 00:25:52,860
who are going to arrive at the bus stop in
next 5 minutes the number of defaults the
248
00:25:52,860 --> 00:25:58,570
number of people, who arrive in the bus stop
are discrete you cannot have less than 0 people
249
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I mean this in this particular case you cannot
have less than 0 and you cannot have two and
250
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half people who arrive at the bus stop, so
it is a discrete distribution.
251
00:26:06,049 --> 00:26:12,200
But, the distribution is defined over at time
or a space, which is continuous. So, it is
252
00:26:12,200 --> 00:26:18,840
not how many heads can I get out of 10 tosses
its more, how many of a certain occurrence
253
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can happen over a certain period of time and
that is the core concept there the PMF again
254
00:26:27,470 --> 00:26:33,559
note it is not the PDF, because it is a discrete
distribution its characterized as lambda power
255
00:26:33,559 --> 00:26:39,120
k and lambda here is a parameter. So, it is
essentially a number that that you have and
256
00:26:39,120 --> 00:26:44,010
it represents like the average rate.
So, 3 people are arriving per minute that
257
00:26:44,010 --> 00:26:50,309
would be more like a three out there and k
is the variable of interest where you say
258
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k is equal to 1 and you get a probability
you say k is equal to 0 you get a probability.
259
00:26:55,960 --> 00:27:00,429
And technically this k can go all the way
up to infinity like I was discussing right
260
00:27:00,429 --> 00:27:08,510
and the sum of all those possibilities discrete
possibilities is equal to 1. So, if you wanted
261
00:27:08,510 --> 00:27:13,180
to know the probability that 3 people will
arrive at the bus stand all you will do is
262
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you will say, what is the average arrival
rate.
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00:27:15,940 --> 00:27:19,940
So, may be the average arrival rate, which
is what is essentially represented by lambda.
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So, the average arrival rate could be something
like well two people arrived per minute, so
265
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that is two and I want to ask the question
what is the probability 0 people will arrive
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00:27:29,500 --> 00:27:33,570
over the next minute. So, I would say on a
average two people arrive at the bus stop
267
00:27:33,570 --> 00:27:41,960
k is that 0. So, I will put this I will substitute
k with 0 and this will this formula will spit
268
00:27:41,960 --> 00:27:45,220
out a answer, which is the exact probability
of 0 people.
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00:27:45,220 --> 00:27:50,620
And you will replace k with 1, 2, 3, 4 and
so on, all the way up to infinity and the
270
00:27:50,620 --> 00:27:57,970
sum of all those probabilities are is essentially,
what is essentially the distribution is essentially,
271
00:27:57,970 --> 00:28:07,140
what quantifying. Because, of how it is defined
the mean is lambda we defined lambda as essentially
272
00:28:07,140 --> 00:28:17,620
that rate parameter and its interesting property
of the that variance also is lambda and lambda
273
00:28:17,620 --> 00:28:22,950
has to be greater than 0. But, again just
to recap this stateÕs space for a Poisson
274
00:28:22,950 --> 00:28:30,450
distribution that is the possible value that
k can take are always greater than 0 in greater
275
00:28:30,450 --> 00:28:37,440
and less than and goes all the way up to infinity.
But, it is a discrete distribution, because
276
00:28:37,440 --> 00:28:41,120
you can never have two and a half people arriving
at a bus stop.
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00:28:41,120 --> 00:28:48,149
So, the next distribution we are going to
look at is the geometric distribution the
278
00:28:48,149 --> 00:28:55,549
geometric distribution is also an discrete
distribution, but it is a very interesting
279
00:28:55,549 --> 00:29:01,830
counterpart to the binominal distribution.
So, take the fact that we said that is a Bernoulli
280
00:29:01,830 --> 00:29:06,010
process, which means there is some probability
of event happening and some probability that
281
00:29:06,010 --> 00:29:10,740
the event will not happen and they add up
to one. So, probability of a heads, let us
282
00:29:10,740 --> 00:29:14,330
say is 60 percent.
Therefore, by definition probability of the
283
00:29:14,330 --> 00:29:19,760
tails is 40 percent that was Bernoulli and
then, we said the bi nominal is nothing but,
284
00:29:19,760 --> 00:29:28,110
the probability of getting k successes out
of n trails. So, what is the probability of
285
00:29:28,110 --> 00:29:36,260
getting 2 heads out of 10 tosses 3 heads out
of 10 tosses that space is defined by the
286
00:29:36,260 --> 00:29:44,340
binominal the geometric defines the probability
of the number of times you need to toss the
287
00:29:44,340 --> 00:29:52,799
coin before getting your first heads or you
can think of it as getting the next heads.
288
00:29:52,799 --> 00:29:57,541
So, number of attempts before an event is
what you are looking at and before you can
289
00:29:57,541 --> 00:30:03,510
think of it as in inter arrival counterpart
to your binominal distribution. So, if you
290
00:30:03,510 --> 00:30:10,730
take the coin toss case right essentially
it is more like, how many times do I need
291
00:30:10,730 --> 00:30:17,640
to toss the coin before I get my first heads
or a next heads. So, you start at some point
292
00:30:17,640 --> 00:30:22,799
and then, you say so; obviously, anything
that I have tossed before it is not influence
293
00:30:22,799 --> 00:30:28,529
my future tosses, because they are independent.
And now, I am going to start tossing the coin
294
00:30:28,529 --> 00:30:34,590
and tell me the probability that I will have
to toss the coin 0 times before I get my first
295
00:30:34,590 --> 00:30:39,740
head or 1 you can count you can say, what
is the number of tails I am going to see before
296
00:30:39,740 --> 00:30:47,769
I get my next heads. So, that could be 0,1,2,3
and technically it can be infinity meaning
297
00:30:47,769 --> 00:30:53,460
they could be this really bizarre world, where
even though there is a finite non zero probability
298
00:30:53,460 --> 00:30:59,010
of getting a heads whatever that number is
it could be 0.9 or it could be 0.1 its technically
299
00:30:59,010 --> 00:31:05,000
possible that I wind up tossing the coin and
infinite number of times I keep getting tails
300
00:31:05,000 --> 00:31:10,080
and, so am still waiting for my heads.
So, this is again a distribution, which starts
301
00:31:10,080 --> 00:31:17,470
with 0 and goes all the way to infinity depending
on how itÕs defined exactly the geometry
302
00:31:17,470 --> 00:31:23,669
is sometimes defined as the number of tails
you see before the next heads. And there the
303
00:31:23,669 --> 00:31:28,960
distribution starts at 0, because you can
see 0 tails another version of this distribution
304
00:31:28,960 --> 00:31:34,720
can could start with, how many tosses do you
need to make to see the first heads in that
305
00:31:34,720 --> 00:31:39,360
case the very you need to at least toss the
coin 1 time to see the heads.
306
00:31:39,360 --> 00:31:46,970
And, so both these distributions you might
find in text books the PMF and CDF should
307
00:31:46,970 --> 00:31:52,799
be fairly intuitive all we are doing with
the probability mass function you are saying,
308
00:31:52,799 --> 00:32:01,950
what is the probability of getting k minus
1. So, here you might say k minus 1 is nothing
309
00:32:01,950 --> 00:32:09,539
but, the number of tosses before the actual
success that you keep getting the tails. So,
310
00:32:09,539 --> 00:32:13,720
if p is the probability of getting the heads
you are saying, what is the probability of
311
00:32:13,720 --> 00:32:19,289
getting k minus 1 tails.
So, if you here we are defining the distribution
312
00:32:19,289 --> 00:32:24,580
as on the first toss I get the heads then
what is the probability that, so let us say
313
00:32:24,580 --> 00:32:28,360
I want to know the probability of getting
it taking 3 tosses before I get a heads, then
314
00:32:28,360 --> 00:32:34,870
you are saying; that means, for the first
two tosses 3 minus 1 is 2 the first 2 tosses
315
00:32:34,870 --> 00:32:38,789
I should have gotten a tails. So, what is
the probability of getting a tail like that
316
00:32:38,789 --> 00:32:43,039
and that is 1 minus p.
So, 1 minus p to the power of k minus 1 says
317
00:32:43,039 --> 00:32:47,870
before that success happens I need to say
I need to say k minus 1 failures and what
318
00:32:47,870 --> 00:32:54,191
is the probability of that and then finally,
the probability of that one success. Again
319
00:32:54,191 --> 00:32:59,970
essentially the CDF you are doing a summation,
but the summation kind of neatly simplifies
320
00:32:59,970 --> 00:33:06,669
to the formula that we have shown here and
the mean being 1 by p should be fairly intuitive
321
00:33:06,669 --> 00:33:18,950
meaning if the probability of getting a heads
is let us say 10 percent it should be intuitive
322
00:33:18,950 --> 00:33:26,080
that it on average should take 10 tosses before
I get my first heads, so 1 by 0.1 would be
323
00:33:26,080 --> 00:33:29,980
10 and so on.
And again even if it is not entirely intuitive
324
00:33:29,980 --> 00:33:35,519
to you may be working through some problems
and formulas, where we will be discussing
325
00:33:35,519 --> 00:33:42,409
those can help. Finally, the variance is also
1 minus p by p square and even that is something
326
00:33:42,409 --> 00:33:46,169
that you might have to work through a couple
of times.
327
00:33:46,169 --> 00:33:51,159
So, we come to the last distribution that
we are going to be discussing in this class
328
00:33:51,159 --> 00:33:56,929
and that distribution is the exponential distribution
in many ways the exponential distribution
329
00:33:56,929 --> 00:34:03,710
essentially you know how the geometric was
looking at the inter arrival time of a binominal
330
00:34:03,710 --> 00:34:08,780
distribution the same way the exponential
looks at the inter arrival times of the Poisson
331
00:34:08,780 --> 00:34:13,300
distribution. So, what does that mean in terms
of our examples you know how we spoke about
332
00:34:13,300 --> 00:34:18,150
the Poisson distribution as being discrete
distribution, where you said, what is the
333
00:34:18,150 --> 00:34:21,660
possibility of three people arriving at the
bus stop in the next 5 minutes what is the
334
00:34:21,660 --> 00:34:25,230
probability of two people arriving at the
bus stop in the next 5 minutes, so on.
335
00:34:25,230 --> 00:34:29,270
So, you have fixed the time 5 minutes or ten
minutes whatever is of your interest and you
336
00:34:29,270 --> 00:34:34,590
looked at the probabilities of various discrete
possible occurrences. So, you said what is
337
00:34:34,590 --> 00:34:38,940
the probability zero people arrived one person
arrives two people arrive three people arrives.
338
00:34:38,940 --> 00:34:44,770
Here with the exponential you are describing
the inter arrival that is how long should
339
00:34:44,770 --> 00:34:51,080
I wait before the next person arrives.
So, it is the exact same thing the exponential
340
00:34:51,080 --> 00:34:56,090
is the same thing to the Poisson the way the
geometric is to the binomial. The binomial
341
00:34:56,090 --> 00:35:03,130
is quantifying the number of the probabilities
of getting 3 out of 10 out of 10 tosses the
342
00:35:03,130 --> 00:35:07,030
probability of getting 3 heads, where as the
geometric saying how long should I wait before
343
00:35:07,030 --> 00:35:13,920
the next head heads arrives. The Poisson is
quantifying over some time scale or space
344
00:35:13,920 --> 00:35:18,680
scale its quantifying the probabilities of
different occurrences like, what is the probability
345
00:35:18,680 --> 00:35:23,620
of 1 occurring 2 occurring 3 occurring this
is how long should I wait before the next
346
00:35:23,620 --> 00:35:28,060
thing occurs.
So, you can think of the exponential as moving
347
00:35:28,060 --> 00:35:34,140
in time and waiting for that next occurrence
and how long should I wait, what you can think
348
00:35:34,140 --> 00:35:42,740
of it a space I keep walking and you know
I encounter occurrences over some length scale
349
00:35:42,740 --> 00:35:47,880
the poisons gives me the probability of seeing
n number of such occurrences. But, if I start
350
00:35:47,880 --> 00:35:52,590
walking on that scale how long should I walk
before I get the next occurrence and long
351
00:35:52,590 --> 00:35:56,810
can be in length or it can be in time.
So, think of it is time or space, but essentially
352
00:35:56,810 --> 00:36:01,610
the continues version of the you can think
of it as the continuous version of the geometric
353
00:36:01,610 --> 00:36:12,540
distribution. The more again the PDFs and
CDFs it is there for your reference, but the
354
00:36:12,540 --> 00:36:19,860
important thing about this distribution is
that people call it memory less meaning that
355
00:36:19,860 --> 00:36:29,061
the probability of something occurring over
a time if the same if you condition that it
356
00:36:29,061 --> 00:36:35,530
is not happened yet. So, think of this way
I am not saying for instance, so let us think
357
00:36:35,530 --> 00:36:42,240
exponential is often used to say to describe
may be the failures of a light bulb over time.
358
00:36:42,240 --> 00:36:47,660
So, how long should I wait before this light
bulb fails, but as Poisson would describe
359
00:36:47,660 --> 00:36:53,900
saying out of 100 light bulbs over a tenure
horizon how many would fail, but let us leave
360
00:36:53,900 --> 00:36:57,150
that. So, out here an exponential am dealing
with one light bulb I am saying how long should
361
00:36:57,150 --> 00:37:03,510
I wait before it fails. So, if this distribution
were uniform the probability that it would
362
00:37:03,510 --> 00:37:10,190
fail between year 1 and 2 would be the same
as probability that it would fail between
363
00:37:10,190 --> 00:37:15,740
year 5 and 6 that would be uniform what we
mean by seeing exponential is memory less
364
00:37:15,740 --> 00:37:18,750
is that.
The probability that the bulb would fail between
365
00:37:18,750 --> 00:37:26,540
the year 1 and 2 is the same as the probability
of the bulb would fail between year 5 and
366
00:37:26,540 --> 00:37:33,590
6 if I tell you at the start of year 5 that
the bulb is already not failed. So, if you
367
00:37:33,590 --> 00:37:38,490
are standing at time 0 the probability the
bulb will fail between year 1 and 2 is very
368
00:37:38,490 --> 00:37:43,210
different from the probability the bulb will
fail between the year 5 and 6.
369
00:37:43,210 --> 00:37:48,400
But, the probability that the bulb will fail
between year 1 and 2 is the same as the probability
370
00:37:48,400 --> 00:37:55,330
of the bulb will fail between 5 and 6 if I
go to year 5 and tell you that the bulb has
371
00:37:55,330 --> 00:37:59,640
not failed yet same way if I go to year one
and tell you that the bulb has not failed
372
00:37:59,640 --> 00:38:04,870
yet. So, condition on the bulb not failing
the probability is over a future time horizon
373
00:38:04,870 --> 00:38:08,550
or length horizon would be the same.
374
00:38:08,550 --> 00:38:13,420
And just for your convenience we have kind
of put together the four distributions that
375
00:38:13,420 --> 00:38:21,880
are partly related and saying that these are
discrete over the time arrivals versus how
376
00:38:21,880 --> 00:38:25,820
the distribution is counted. And I have used
the color coding to describe which are continuous
377
00:38:25,820 --> 00:38:29,070
distribution and which are discrete distribution
and you can see that clearly its only the
378
00:38:29,070 --> 00:38:33,950
exponential that is continuous the other three
are discrete distributions great.
379
00:38:33,950 --> 00:38:39,670
Now, before I sign off from this class I just
wanted to give you some formulas on how to
380
00:38:39,670 --> 00:38:44,620
calculate mean and standard deviation and
so on. But, before we do that let us start
381
00:38:44,620 --> 00:38:51,220
with the more basic thing, which is given
a PDF, how are you going to get the cumulative
382
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density function and the idea here is we have
always said it is the area to the left side
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of the curve. So, the left side of the curve
is from minus infinity, so this is the PDF
384
00:39:03,670 --> 00:39:07,020
and the left side extreme is referred to as
minus infinity.
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00:39:07,020 --> 00:39:10,660
Now, if you know that distribution starts
from another location you just essentially
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00:39:10,660 --> 00:39:14,020
want to get to the starting point all this
minus infinity means it is the starting point
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00:39:14,020 --> 00:39:18,920
of the distribution if a distribution can
technically go to minus infinity it can, but
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00:39:18,920 --> 00:39:24,290
like the uniform for instance it starts at
a. So, that would be a you would replace that
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00:39:24,290 --> 00:39:28,960
and x is you leave x as it is, because your
cumulative density function is a function
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00:39:28,960 --> 00:39:33,340
of x and this integral is what would solve
it.
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00:39:33,340 --> 00:39:38,920
Obviously, if you are dealing with a discrete
distribution you would instead have this summation
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00:39:38,920 --> 00:39:48,401
not integral, but everything else about the
formula would stay the same way if you are
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00:39:48,401 --> 00:39:53,790
going from CDF to PDF you will essentially
differentiate the CDF and there is not much
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00:39:53,790 --> 00:40:00,930
more to that. So, now, coming to mean and
variance the idea here is that the mean x
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00:40:00,930 --> 00:40:06,450
bar is the kind of mean that we have discussed,
so far. So, you have some data points you
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00:40:06,450 --> 00:40:12,590
take them up you average them divide by, but
if you wonÕt start thinking of a mean in
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00:40:12,590 --> 00:40:16,040
terms of a distribution.
So, you given instead a distribution not actual
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00:40:16,040 --> 00:40:21,950
data or you use the data and created a distribution
out of it you created an f of x out if it.
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00:40:21,950 --> 00:40:28,460
Then, how do you calculate a theoretical mean
of the distribution and that you do by essentially
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00:40:28,460 --> 00:40:32,860
this formula for the continuous distribution
and this formula for the discrete distribution
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00:40:32,860 --> 00:40:37,800
and these you can call them expected values
these typically gets called mu it is that
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00:40:37,800 --> 00:40:43,090
Greek alphabet mu, that the core idea here
is that you take each possible outcome in
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00:40:43,090 --> 00:40:47,550
the discrete case multiplied by the probability
that there outcome can take on and we all
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00:40:47,550 --> 00:40:51,950
know that these probabilities multiply all
the way to the sum of these probabilities
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00:40:51,950 --> 00:40:57,800
is going to be equal to one.
But, you are multiplying pi and xi for each
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00:40:57,800 --> 00:41:05,260
I and; that is the discrete case the equivalent
of that is nothing, but f of x is the equivalent
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00:41:05,260 --> 00:41:11,720
of pi and x is x in both cases. So, this is
the equivalent of the continue this is the
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00:41:11,720 --> 00:41:18,320
continuous version of the same problem. The
standard deviation is the same here you have
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00:41:18,320 --> 00:41:22,760
the n minus 1, because you are dealing with
x bar, but if you have a theoretical mean
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00:41:22,760 --> 00:41:31,120
that is given to you can use the n and in
the discrete case this formula, which is the
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00:41:31,120 --> 00:41:35,310
same formula that you have used to with the
exception of one by n takes place.
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00:41:35,310 --> 00:41:40,380
Now, this actually simplifies and for your
convenience on the continuous version I have
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00:41:40,380 --> 00:41:46,370
given you the simplified version this to me
this version, which is used for the discrete
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00:41:46,370 --> 00:41:53,630
is more intuitive but this formula simplifies
to a formula, which looks more like summation
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00:41:53,630 --> 00:42:00,120
of x square and so on. Separately just like
on a case of the mean discrete uses a summation
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00:42:00,120 --> 00:42:06,950
and continuous uses an integration we have
the same distinction even here, but for your
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00:42:06,950 --> 00:42:11,740
benefit on this one I have shown you the simplified
formula.
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00:42:11,740 --> 00:42:18,390
So, even here originally you have started
off with integral minus infinity to infinity
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00:42:18,390 --> 00:42:26,000
x minus mu the whole square dot f of x and
so on. But, that is not what we are doing
420
00:42:26,000 --> 00:42:31,260
here and here actually this would not be dx
this would just be probability, so it will
421
00:42:31,260 --> 00:42:41,260
be p of I out here. But, out here it would
be d x and it would be f of x because it represents
422
00:42:41,260 --> 00:42:48,310
that. So, I hope these formulas give you some
idea and we will definitely be looking to
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00:42:48,310 --> 00:42:54,920
see you apply some of these formulas on distributions
to get answers. But, thatÕs it for this class
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00:42:54,920 --> 00:42:58,500
next class we will talk about the distribution
we have not talked about, so far which is
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00:42:58,500 --> 00:43:00,220
the normal distribution.
Thank you.