(a) Consider

But since $U ⊥ V$ then the above reduces to

Hence

|---------------------------------------| |U ˜ χ2 with (n − 1) degrees of freedom | ----------------------------------------

(b)

Now use the moment generation function to ﬁnd the expectations of $U$ and $V.$

Need to ﬁnd $M ′X (t)$ where $m- MX (t) = (1 − 2t)− 2$ where $m$ is the degree of freedom

Hence

M ′(t) = m (1 − 2t)− m2 −1

(1)

at $t = 0$ the above becomes

|------------| M-X′(0)-=-m---

For $U$ , we found that $m = (n − 1)$ , hence

|----------------| -E-(U-) =-(n −-1)|

and for $V$ we are told its degree of freedom is $m = 1$ hence

|----------| E-(V-)-=-1--

Therefore

E (W ) = (n − 1 ) + 1

Hence

|-----------| -E-(W-)-=-n-|

Now

But $U ⊥ V$ so the above becomes

|-------------(---)-----(---)---------------------| |V ar(W ) = E U 2 + E V 2 + 2E (U )E (V ) − n2| --------------------------------------------------

Lets ﬁnd $2 E (Z )$ for a $Z$ chi square random variable of degree of freedom $m$ . We already found $′ M (t)$ above in (1)

At $t = 0$

( ) E (Z2 ) = − 2m − m-− 1 2

Hence

( 2) E Z = m (m + 2)

(2)

Hence using (2) above, we now can ﬁnd $E (U 2)$ and $E (V 2)$

For $U$ it has degree of freedom $m = (n − 1)$ , hence

For $V$ it has degree of freedom $m = 1$ , hence

Hence

|--------------| V-ar(W--) =-2n-|

λxe−λ- X (x) ˜ x!

Moment generating function for a Poisson r.v. or parameter $λ$ is (from page 144)

|-----------------| | −λ etλ | -MX--(t) =-e--e---

Now

Hence

( ) M ′(t) = E √xe √xt Y

and

Therefore

M ′′(0) = E (x) Y

But $\text{[math]}$

Hence

M ′′(0 ) = λ Y

But

′′ ( 2) M Y (0) = E Y

then

|--(---)----| |E Y 2 = λ| -------------

Now to ﬁnd $V ar(Y )$

( ) 2 V ar(Y ) = E Y 2 − [E (Y )]

Where

So we need to ﬁnd $√ -- E ( x )$ to complete the solution.

Hence

|-------------(--------------)| |V ar(Y ) ≃ λ 1 − λe−0.58578λ | ------------------------------

(a)

Now

E (X ) = E (E (X |Y ))

But $E (X |Y )$ is expectation of a Poisson r.v. with parameter $2 Y$ . But we know that mean of a poisson r.v. with parameter $λ$ is $λ$ . Hence $2 E (X |Y ) = Y$ since we are told $2 Y$ is the parameter.

Hence

( ) E (X ) = E Y 2

But the moment generating function for Gamma is $(-λ-) α MY (t) = λ− t$ (book page 145 second edition).

Hence $E (Y 2) = M ′Y′(0) = α(αλ+21)$ (page 145)

Hence

|-------------------| | α (α + 1) | |E (X ) = -----2--- | --------------λ-----

(b)

( 2) 2 Var (X ) = E X − [E (X )]

(1)

But

E (X2 ) = E (E (X2 |Y ))

But $E (X2 |Y)$ is $E (X2 )$ of a poisson r.v. with parameter $Y 2$ . But we know that $E (X2 )$ of a poisson r.v. with parameter $λ$ is $λ2 + λ$ (book page 144 example A). Hence since we are told $Y 2$ is the parameter, then