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