Exercise 7.1. Show that that the exponential distribution obeys

which is referred to as memoryless, or Markov property.

Solution:

Exercise 7.2. Find the th moment of an exponentially distributed random variable with parameter .

Solution:

Using the L'Hospital's rule it is easy to prove that the value of the first part is and thus

Taking into account the recursion one can easily see that

Exercise 7.3. Let us assume that two independent activities start. Their durations are denoted by , and are supposed to be exponentially distributed random variables with parameters , respectively. Let , , .

Find

1. The distribution and mean of ,

2. that is completes first,

3. The distribution and mean of , that is the distribution of the time between the first and second events,

4. The probability that at an arbitrary time

, ,

5.

6.

7.

8. .

Solution:

Distribution of the first event

that is, V is exponentially distributed with parameter consequently ,

completes first

distribution of the time between the first and second event

given represents the residual time of and similar argument is valid for .

Due to the memoryless property of ,

Consequently

has been completed, but is still running

the first event has been completed, but the second is still running

both events have been completed

distribution of the sum of and

Since W , V are dependent random variables their convolution cannot be applied. However, it is easy to see that

distribution of X given

distribution of X given

that is, it follows an exponential distribution with parameter ,

distribution of X given

Exercise 7.4. Find the probability that , supposing that , and are independent.

Solution:

By the law of total probability

Exercise 7.5. Find the distribution and mean of a series system consisting of independent and exponentially distributed components.

Solution:

In the case of series system

Exercise 7.6. Find the distribution, mean and variance of a parallel system consisting of independent and exponentially distributed components with the same failure rate, that is .

Solution:

Apply the following useful relation. If then

Substitute then

Due to the memoryless property of the exponential distribution the time between the consecutive failures are also exponentially distributed and are independent of each other. It is easy to see that the parameter of the time between th and th failure is , . This fact can be used to calculate the mean and variance of the time to the th failure.

Hence

In particular, the variance of the life time of a parallel system is the variance of the last failure, that is

Exercise 7.7. Let , , be independent random variables.

Show that

Solution:

hence

thus

from which the statement follows.

Similarly

thus

from which the statement follows.

Exercise 7.8. Prove that the distribution function of the Erlang distribution with parameters is

Solution:

Using integration by parts, where we get

Consequently

Exercise 7.9. Let and and independent random variables.

Find their convolution.

Solution:

Exercise 7.10. Find the mean of the previous convolution by using the density function.

Solution:

Obviously thus we could check the correctness of the density function.

Exercise 7.11. Derive the density function of the Erlang distribution with parameters from the 2-phase hypoexponential distribution.

Solution:

As we have seen

Taking the limit as we get the desired result, that is

therefore we apply the L'Hospital's rule.

Thus we obtain what is the density function we needed.

Exercise 7.12. Find the distribution function of the 2-phase hypoexponential distribution.

Solution:

To check its correctness let us take the limit as . Applying the L'Hospital's rule we have

which is exactly the distribution function of the Erlang distribution with parameters .

Exercise 7.13. Let , and independent random variables. Find the conditional density function .

Solution:

Specially, if , then using the L'Hospital's rule and taking substitution we get

that is we have the uniform distribution.

If at the beginning we assume that then

since follows the Erlang distribution with parameters.

Exercise 7.14. Find the coefficient of variation of the Erlang distribution with parameters

Solution:

Exercise 7.15. Verify the density function of the hyperexponential distribution.

Solution:

It is easy to see that it is nonnegative, furthermore

Exercise 7.16. Show that the squared coefficient of variation of the hyperexponential distribution is always at least 1

Solution:

To prove it, we need

which follows from the Cauchy-Bunyakovszkij-Schwartz inequality with substitutions

Exercise 7.17. Let , , independent random variables.

Show that

Solution:

It is well-known that

from which our statement follows.

In particular, if we obtain the relations valid for the exponential distribution.