Question

(Statistics)

Answer 1

Are they really asking for E(Z) in a)? It seems too obvious for this question lol

Answer 2

Cause \(Z\sim N(0,1)\),
then \(E(Z)=0\)
\(E(Z^2)=Var(Z)+[E(Z)]^2=1+0^2=1\)

Answer 3

yeah, it does ask for E(Z), most likely for completeness sake

Answer 4

Well the answer is as above. There's nothing to it since you know Z is a standard normal.

Answer 5

For b):
Recall that a Chi square distribution is also a Gamma distribution through the relation below:
\[\chi^2(\nu)=\text{Gamma}\left(\frac{\nu}{2},2\right)\]

And recall the general form of a gamma distribution, \(\text{Gamma}(\alpha,\beta)\), has pdf:
\[\large f_X(x)= \frac{1}{\Gamma(\alpha)\beta^{\alpha}}x^{\alpha-1}e^{-x/\beta},\,\,\,\alpha>0,\beta>0,x>0\]

Hence in your case,
\[\large f_X(x)=\frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\nu/2}}x^{\frac{\nu}{2}-1}e^{-x/2}\]

\[\large\begin{aligned} E(X^a)&=\int_o^{\infty}x^a\cdot \frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\nu/2}}x^{\frac{\nu}{2}-1}e^{-x/2}\,dx\\
&=\frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\nu/2}}\int_0^{\infty}x^{\frac{\nu}{2}+a-1}e^{-x/2} \,dx\end{aligned}\]
Now, let
\(t=x/2\implies x=2t\)
\(dt=1/2 \,\,dx\implies 2\,\,dt=dx\)
The bounds of integration stay the same from this substitution.

\[\large\begin{aligned} E(X^a)&=\frac{1}{\Gamma\left(\frac{\nu}{2}\right)2^{\nu/2}}\int_0^{\infty}(2t)^{\frac{\nu}{2}+a-1}e^{-t}\cdot 2 \,dt \\
&=\frac{2^{\frac{\nu}{2}+a}}{\Gamma\left(\frac{\nu}{2}\right)2^{\nu/2}}\underbrace{\int_0^{\infty}t^{\frac{\nu}{2}+a-1}e^{-t} \,dt}_{\Gamma(\nu/2+a)} \\
&=\frac{2^a}{\Gamma\left(\frac{\nu}{2} \right)}\Gamma\left(\frac{\nu}{2}+a \right) \end{aligned}\]

Answer 6

Oh I see why they ask for the result in a) now lol. We use it in c)

Answer 7

\[E(T)=E\left( \frac{Z}{\sqrt{Y/\nu}}\right)\\
=E(Z)E\left(\frac{1}{\sqrt{Y/\nu}} \right),\text{by indepence of }X,Y\\
=0,\,\,\text{since }E(Z)=0\]

\[Var(T)=Var\left( \frac{Z}{\sqrt{Y/\nu}}\right)\\
=E\left[\left( \frac{Z}{\sqrt{Y/\nu}}\right)^2\right]-\left[E\left( \frac{Z}{\sqrt{Y/\nu}}\right)\right]^2,\text{by definition of variance}\\
=E\left[\left( \frac{Z}{\sqrt{Y/\nu}}\right)^2\right]-0, \,\,\text{since }E(T)=0 \text{ from above} \\
=E\left( \frac{Z^2}{Y/\nu}\right)\\
=\nu E\left(\frac{Z^2}{Y}\right)\\
=\nu E(Z^2)E(Y^{-1}),\text{by indepence of }X,Y\\
=\nu (1)E(Y^{-1})\text{ use the result in b) on }Y\text{ since } Y \text{ is chi-square, and }a=-1 \\
=\nu\cdot 2^{-1}\frac{\Gamma\left(\frac{\nu}{2}-1 \right)}{\Gamma\left(\frac{\nu}{2}\right) }\\
=\nu\frac{1}{2} \frac{\cancel{\Gamma\left(\frac{\nu}{2}-1 \right)}}{\left(\frac{\nu}{2}-1\right)\cancel{\Gamma\left(\frac{\nu}{2}-1\right)} }\\
=\frac{\nu}{\nu-2} \]

Answer 8

I only noticed now.. the independence is from Z and Y (not X and Y)