(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