Exercise 6.1. Show that if then , and !

Solution:

Exercise 6.2. Show that if , then , and !

Solution:

Exercise 6.3. Show that if if , then , and !

Solution:

We used the fact that in the case of absolute convergent series the summation and derivative are interchangeable.

Thus

In the following we can show how these results can be obtained by using the property of the geometric distribution.

So . Similarly

Thus

Hence

Exercise 6.4. Find the mean and variance of a modified geometric distribution with success parameter .

Solution:

As we know the modified geometric distribution is and , where . Hence

Exercise 6.5. Show that that the geometric distribution yields

that is the so-called memoryless property holds.

Solution:

Exercise 6.6. Let , and independent random variables. Find that .

Solution:

Since and are independent then the convolution of , is also binomial so we have

that is we obtain the hypergeometric distribution.

Exercise 6.7. Let and independent random variables. Find that .

Solution:

Exercise 6.8. Let us consider a supermarket at which customers arrive according to a Poisson distribution with parameter and choose the th cashier with probability . Find the distribution of the number of customers at cashier .

Solution:

Let us perform a random experiment with independent and identical trials. Let describe the number of th outcome. As the joint distribution of is a multinomial distribution with parameters and we have

Since .

It follows that , and are independent random variables.