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{\bf Question}

Suppose that $X$ and $Y$ are independent standard normal random
variables.

\begin{description}
\item[(a)]
Prove that $Z=aX+bY$ is normally distributed where $a$ and $b$ are
given constants such that both are not equal to zero at the same
time.  Give the mean and variance of $Z$.

\item[(b)]
Find $P(X-Y>0)$ and $P(X-Y>1)$.

\item[(c)]
Derive the distribution of $Z=X^2+Y^2$.  Hence find $P(X^2+Y^2
\leq 1)$.
\end{description}



\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$M_X(t)=E(e^{tX})=e^{\frac{t^2}{2}}$

$M_Y(t)=E(e^{tY})=e^{\frac{t^2}{2}}$

\begin{eqnarray*} M_Z(t) & = & E(e^{tZ})\\ & = &
E\left\{e^{t(aX+bY)}\right\}\\ & = & E(e^{atX}e^{btY})\\ & = &
M_X(at)M_Y(bt)\ {\rm since}\ X\ {\rm and}\ Y\ {\rm are\
independent}\\ & = & e^{\frac{1}{2}(a^2+b^2)t^2}
\end{eqnarray*}

The above is the mgf of a normal r.v. with mean 0 and variance
$a^2 + b^2$.

By using the uniqueness theorem of the mgf $Z \sim N(0,a^2 +
b^2).$

\item[(b)]
$Z=X-Y \sim N(0,2)$

$P(Z>0)=\frac{1}{2}$

$P(Z>0)=1-\Phi\left(\frac{1}{\sqrt 2}\right)=1-0.76=0.24$

\item[(c)]
$Z=X^2+Y^2 \sim \chi^2$ with 2 degrees of freedom by using the mgf
technique $\star$

\begin{eqnarray*} P(Z \leq 1) & = & \ds\int_0^1 f(z) \,dz =
\ds\int_0^1
\ds\frac{1}{\Gamma\left(\frac{2}{2}\right)2^{\frac{2}{2}}}z^{\frac{2}{2}-1}e^{-\frac{1}{2}z}
\,dz\\ & = & \ds\int_0^1 \ds\frac{1}{2}e^{-\frac{1}{2}z} \,dz =
1-e^{-\frac{1}{2}} \end{eqnarray*}

$\star M_Z(t)=E\left(e^{tX^2 +
tY^2}\right)=\left(\ds\frac{1}{1-2t}\right)^{\frac{2}{2}}$ if
$t<\frac{1}{2}$
\end{description}




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