(a)
Problem review:
is a random variable and is a random variable, where and
and can be thought of as the failure rate for each respective component. is the lifetime of component . Hence means to ask for the probability of the first component to have a lifetime of given that the failure rate of this kind of components is
solution:
Now we know that
Looking at the following diagram to help determine the region to integrate:
Hence
But since then the joint density is the product of the marginal densites.
Hence
Therefore
We take since we expect the lifetime to go to zero eventually. Also this is a requirment for the integrals to not diverge.
Hence the above becomes
Hence
(b)
Hence
Hence
(c)Need to find which is the same as , hence this is the same as part(a) but replace by as show in the following diagram
Hence
Hence
Then
Problem review: Poisson probability density is a discrete probability function (We normally call it the probability mass function ). This means the random variable is a discrete random variable.
The random varible in this case is the number of success in trials where the probability of success in each one trial is and the trials are independent from each others. The difference between Poisson and Binomial is that in Poisson we are looking at the problem as becomes very large and becomes very small in such a way that the product goes to a fixed value which is called , the Poisson parameter. And then we write where The following diagram illustrates this problem, showing the three r.v. we need to analyze and the time line.
But what is "trials" in this problem? If we divide the time line itself into very small time intervals then the number of time intervals is the number of trials, and we assume that at most one event will occure in this time interval (since it is too small). The probability of event occuring in this is the same in the interval and in the interval . Now let us find for and and based on this. Since where is the number of trials, then for we have where we divided the time interval by the time width to obtain the number of time slots for . We do the same for and obtain that
Similary, , hence
Let us refer to the random variable as and the r.v. as and the r.v. as
The problem is then asking to find and to identify
To help in the solution, we first draw a diagram to make it more clear.
We take to the same for the random variables .
But is the same as hence
Now r.v. , since the number of events in is indepenent from the number of events that could occur in .
Given this, we can now write the joint probability of as the product of the marginal probabilitites. Hence the numerator in the above can be rewritten and we obtain
| (1) |
Now since each of the above is a poisson process, then
Hence (1) becomes
| (2) |
Now we simplify this further and try to idensity the resulting distribution First we note
Hence (2) becomes
Let then the above becomes
We see that the parameter will occure in the numerator and denomerator with the same powers, hence we can factor it out and cancel it. Hence we obtain
Hence
But we found that , hence the exponential term above vanish and we get
Let , then hence the last line above can be written as
But this is a Binomial with parameters , hence