WEBVTT
Kind: captions
Language: en
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This is the Stochastic processes, model 1;
probability theory refresher, lecture 3; problems
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in random variables and distributions.
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Let as a first problem; let X be a random
variable
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having geometric distribution
with the parameter p. Our interest is to find;
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our interest is to prove that the probability
of X=n+k, given X takes the value greater
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than n that is as same as the probability
that X takes the value k for every integers,
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n and k.
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You can prove this result by starting from
the left hand side that is probability of
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X takes the value, n + k, given X greater
than n by definition this is same as probability
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of X = n + k intersection, X greater than
n, divided by probability of X greater than
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n, that is same as; that is same as the numerator;
X = greater than n means, all possible values,
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n=n + k that means that the intersection is
going to be probability of X takes the value,
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n + k.
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Whereas the denominator is the probability
of X is greater than n, that is same as since
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X is the geometric distribution with the parameter
p, the probability of X = n + k, that is nothing
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but 1- p power n+k-1 into p. Whereas the denominator,
probability of X is greater than n, that means
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summation i=n+1 to infinity 1-p power i-1
multiplied by p, that is same as numerator
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can keep it as it is.
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Whereas the denominator since the summation
i = n + k to infinity, you can take p times
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1- p power n, common outside, the remaining
terms are 1+1- p, the third term will be 1-p
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whole square and so on. Therefore, you can
still simplify you will get 1- p power n+k-1
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divided by 1- p power n, keep it as it is,
this series you will have the value 1-(1-p).
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therefore if you further simplify you will
get a 1- p power k-1 multiply by p.
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That is nothing but probability of X= k. So
this results are the probability of X= n +
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k given X is greater than n, that is same
as probability of X=k for all n and k. This
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is the important property of geometric distribution
and this property is called a memory less
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property. We will move into the next problem.
Let X be a random variable
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having gamma distribution
with the parameter n, you assume that n is
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a positive integer, the other parameter is
lambda.
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Then the cumulative distribution function
CDF of x is given by F(x) for the random variable
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X that is 1-summation i=0 to n-1 lambda x
power i, e power –lambda times x divided
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by i factorial. So, whenever X is a gamma
distribution with the parameters n and lambda,
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then the CDF can be written in this way. We
know that the probability density function
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of the gamma distribution is lambda power
n, x power n - 1 e power –lambda x divided
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by gamma of n.
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Since n is a positive integer, gamma of n
is a n-1 factorial. Now we can find out the
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CDF of this random variable that is nothing
but –infinity to x, the probability density
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function that is same as since the f(x) is;
this is valid for a x is greater than 0 and
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lambda is greater than 0. So this integration
is valid from 0 to x lambda power n t power
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n-1 e power – lambda times t divided by
gamma of n dt.
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So now you have to integrate this one and
get an expression for this CDF of the random
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variable x. So what we can do, make a substitution
lambda times t, that is same as; we make it
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as some mu. Therefore, this integration becomes
the integration from 0 to lambda times x,
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mu power n-1 e power – mu, divided by gamma
of n * d (mu). That is same as 1- integration
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goes from lambda x to infinity mu power n-1
e power – mu divided by gamma of n d mu.
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Since n is a positive integer gamma of n is
n-1 factorial so you can take it outside.
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You can do this integration by parts so you
will get a; mu power n-1 e power - mu divided
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by -1 between the limits lambda x to infinity
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minus integration from lambda x to infinity
n-1 times mu of n-2 e power –mu divided
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by -1 d mu. So the whole thing is multiplied
by n-1 factorial.
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Now you can integrate the second term again
by integration by parts and when you substitute
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the limits for mu is infinity and as well
as mu = lambda x and subsequently if you do
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the integration by parts, you will get a 1-
1 divided by (n-1) facorial lambda x power
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n-1 e power – lambda x. Then the next term
will be –lambda x power n-2 e power - lambda
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x by n-2 factorial. Similarly, the other terms.
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The last term will be by doing integration
by parts again and again, the last term you
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will get minus of lambda x power 0 e power
–lambda x by 0 factorial. These we can write
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it in the form 1-summation i=0 to n-1 lambda
x power i e power –lambda x by i factorial.
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So here we are finding this CDF of the gamma
distribution, when one of the parameters is
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a positive integer.
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This result will be useful in finding the
total time spending the queuing system, that
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will be discussed in the later models.