Problem source

Let $X$ and $Y$ be independent standard normal random
variables: the probability density function, $\f$, of each
is therefore given by 
\[
\f(x)=\left(2\pi\right)^{-\frac{1}{2}}\e^{-\frac{1}{2}x^{2}}.
\]

\begin{questionparts}
\item  Find the moment generating function $\mathrm{E}(\e^{\theta X})$
of $X$. 
\item  Find the moment generating function of $aX+bY$ and hence
obtain the condition on $a$ and $b$ which ensures that $aX+bY$
has the same distribution as $X$ and $Y$. 
\item  Let $Z=\e^{\mu+\sigma X}$. Show that 
\[
\mathrm{E}(Z^{\theta})=\e^{\mu\theta+\frac{1}{2}\sigma^{2}\theta^{2}},
\]
and hence find the expectation and variance of $Z$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$ \begin{align*}
&& \E[e^{\theta X}] &= \int_{-\infty}^{\infty} e^{\theta x} \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2 } \d x\\
&&&= \int_{-\infty}^{\infty}  \frac{1}{\sqrt{2\pi}} e^{-\frac12 x^2+\theta x} \d x\\
&&&= \int_{-\infty}^{\infty}  \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x^2-2\theta x)} \d x\\
&&&= \int_{-\infty}^{\infty}  \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2+\frac12\theta^2  } \d x\\
&&&= e^{\frac12\theta^2 }\int_{-\infty}^{\infty}  \frac{1}{\sqrt{2\pi}} e^{-\frac12 (x-\theta )^2  } \d x\\
&&&=e^{\frac12\theta^2 }
\end{align*}

\item \begin{align*}
&& M_{aX+bY} (\theta) &= \mathbb{E}[e^{\theta (aX+bY)}] \\
&&&= e^{\frac12(a\theta)^2} \cdot  e^{\frac12(b\theta)^2} \\
&&&= e^{\frac12(a^2+b^2)\theta^2}
\end{align*}

Therefore we need $a^2+b^2 = 1$

\item $\,$ \begin{align*}
&& \E[Z^\theta] &= \E[e^{\mu \theta + \sigma \theta X}] \\
&&&= e^{\mu \theta}e^{\frac12 \sigma^2 \theta^2} \\
&&&=e^{\mu \theta + \frac12 \sigma^2 \theta^2} \\
\end{align*}
\begin{align*}
\mathbb{E}(Z) &= \mathbb{E}[Z^1] \\
&= e^{\mu + \frac12 \sigma^2} \\
\var[Z] &= \E[Z^2] - \left ( \E[Z] \right)^2 \\
&= e^{2 \mu+ 2\sigma^2} - e^{2\mu + \sigma^2} \\
&= e^{2\mu+\sigma^2} \left (e^{\sigma^2}-1 \right)
\end{align*}

[NB: This is the lognormal distribution]
\end{questionparts}