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(a)

Problem review:

T1   is a random variable and T2   is a random variable, where T1 ˜αe−αt1   and T2 ˜βe−βt2.

α  and β  can be thought of as the failure rate for each respective component. Ti  is the lifetime of component i  . Hence P (T  = t )
    1    1  means to ask for the probability of the first component to have a lifetime of t1   given that the failure rate of this kind of components is α.

solution:

Now we know that

              ∫  ∫
P (T1 > T2) =      fT ,T  (t1,t2)dt2dt1
                     1 2

Looking at the following diagram to help determine the region to integrate:

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Hence

              ∫      ∫
                 t1=∞    t2=t1
P (T1 > T2) =               fT1,T2 (t1,t2)dt2 dt1
                t1=0    t2=0

But since T1 ⊥ T2,  then the joint density is the product of the marginal densites.

Hence

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Therefore

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We take α, β ≥ 0  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

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Hence

|----------------------|
|                 β    |
P (T1 > T2 ) = --------|
---------------(α-+-β)--

(b)

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Hence

             (w )    d  ( w)
fW (w ) = fT2 --  ×  ---  --
               2     dw   2

Hence

|--------------------|
|         1    ( w ) |
|fW (w) = --fT2  --  |
----------2------2---

(c)Need to find P (T1 > 2T2 )  which is the same as P (T1 > W )  , hence this is the same as part(a) but replace T2   by W  as show in the following diagram

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Hence

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Hence

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Then

|------------------------|
|                  β     |
|P (T1 > W ) = --------- |
---------------(2α-+-β-)-

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Problem review: Poisson probability density is a discrete probability function (We normally call it the probability mass function pmf  ). This means the random variable is a discrete random variable.

The random varible X  in this case is the number of success in n  trials where the probability of success in each one trial is p  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 n  becomes very large and p  becomes very small in such a way that the product np  goes to a fixed value which is called λ  , the Poisson parameter. And then we write P (X  = k) = λke− λ
             k!     where k = 0,1,2,⋅⋅⋅ The following diagram illustrates this problem, showing the three r.v. we need to analyze and the time line.

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But what is "trials" in this problem?  If we divide the time line itself into very small time intervals δt  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 p  of event occuring in this δt  is the same in the interval [t0,t1] and in the interval [t1,t2]  . Now let us find λ  for X  and Y  and Z  based on this. Since λ =  np  where n  is the number of trials, then for X  we have              (t −t)
λx =  nxp =  -1δt0-p     where we divided the time interval by the time width δt  to obtain the number of time slots for X  . We do the same for Y  and obtain that λ  =  (t2−t1)p
 y      δt

Similary,       (t2−t0)     (t2−t1)+(t1−t0)    (t2−t1)    (t1−t0)
λZ =  --δt--p =  -----δt-----p = --δt--p + --δt--P  , hence λz = λx + λy

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Let us refer to the random variable N (t1,t2)  as Y  and the r.v. N  (t0,t1)  as X  and the r.v. N (t0,t2)  as Z

The problem is then asking to find P (X  = x |Z  = n)  and to identify pmf (X |Z )

To help in the solution, we first draw a diagram to make it more clear.

We take λ  to the same for the 3  random variables X, Y,Z  .

                     P-(X-=--x,Z-=--n)-
P  (X  =  x|Z  = n) =     P (Z  = n)

But Z =  n  is the same as X +  Y = n  hence

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Now r.v. X  ⊥ Y  , since the number of events in [t0,t1]  is indepenent from the number of events that could occur in [t1,t2]  .

Given this, we can now write the joint probability of X,Y  as the product of the marginal probabilitites. Hence the numerator in the above can be rewritten and we obtain

P (X =  x|Z = n ) = P-(X--=-x)P--(Y--=-n-−-x-)
                           P (Z  = n)
(1)

Now since each of the above is a poisson process, then

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Hence (1) becomes

                    ( (λx)x    ) ( (λy )n− x    )     1
P (X  = x|Z =  n) =   -----e−λx    --------e− λy   (λz)n-----
                       x!          (n − x)!       -n!-e−λz
(2)

Now we simplify this further and try to idensity the resulting distribution.  First we note

Hence (2) becomes

                    ( (       )x           )  ( (       )n −x           )
                        (t1−t0)p     ((t1−t0))       (t2−t1)p       ( (t2−t1) )
P (X  = x|Z =  n) = ( ----δt------e−   δt  p)  |( ----δt--------e−   δt  p |) (-------)n1-(------)
                           x!                      (n −  x)!               --(t2δ−tt0)p-- − (t2−δtt0)p
                                                                              n!    e

Let p-
δt = φ  then the above becomes

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We see that the parameter φ  will occure in the numerator and denomerator with the same powers, hence we can factor it out and cancel it. Hence we obtain

                    ----n!----(t1 −-t0)x-(t2-−-t1)n−-x
P (X =  x|Z = n ) = x!(n − x)!      (t − t )n
                                      2   0

Hence

|--------------------(--)-----------------------|
|                      n   (t1-−-t0)x(t2 −-t1)n−x|
|P (X  = x|Z =  n) =            (t −  t)n       |
-----------------------x----------2----0--------|

                                                           λ
                    ----n!-----(    x −λx) (    n−x  −λy)-e-z--
P (X  = x|Z =  n) = x!(n − x )!  (λx) e      (λy)   e     (λz)n

But we found that λz =  λx + λy  , hence the exponential term above vanish and we get

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Let k = --λx-
    λx+λy   , then 1 − k = 1 − --λx- =  λx+λy−λx = --λy-
            λx+ λy     λx+λy     λx+λy   hence the last line above can be written as

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But this is a Binomial with parameters n,k  , hence

|----------------------------(----------)--|
|                                  λx      |
P (X  = x |Z  = n) ˜Binomial   n, --------  |
---------------------------------λx-+-λy----

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