Question

1. Derive the moment generating function for which is distributed as normal with mean 6 and variance .

Answer 1

There are blanks in your question.

Answer 2

But regardless, I can still show you the derivation and you can just plug in the variance and mean in the end result:
\[\Large \begin{align} M_X(t) &=E(e^{tX})\\
&=\int_{-\infty}^{\infty}e^{tx}\cdot\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\, dx \\
&=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2+tx}\, dx \\
\end{align}\]

Let \[w=\frac{x-u}{\sigma}\implies x=\sigma w+\mu\\
\sigma \, dw= dx\]The bounds of integration remain the same with this substitution.
But regardless, I can still show you the derivation and you can just plug in the variance and mean in the end result:
\[\Large \begin{align} M_X(t)
&=\frac{1}{\sqrt{2\pi}\sigma}\int_{-\infty}^{\infty}e^{-\frac{1}{2}w^2+t(\sigma w+\mu)}\sigma\, dw \\
&=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}w^2+t\sigma w}e^{\mu t}\, dw\\
&=\frac{e^{\mu t}}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(w^2-2t\sigma w)}\,dw \\
&= \frac{e^{\mu t}}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(w^2-2t\sigma w +t^2\sigma^2 - t^2\sigma^2)}\,dw \\
&= \frac{e^{\mu t}}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}\left[ (w-t\sigma)^2 - t^2\sigma^2\right]}\,dw\\
&= \frac{e^{\mu t}}{\sqrt{2 \pi}}\int_{-\infty}^{\infty}e^{-\frac{1}{2}(w-t\sigma)^2}e^{\frac{1}{2} t^2\sigma^2}\, dw\\
&= e^{\mu t+\frac{1}{2}t^2\sigma^2}\underbrace{\int_{-\infty}^{\infty}\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2}(w-t\sigma)^2 }\, dw}_{\text{PDF of }N(t\sigma,1)\text{ integrates to 1}}\\
&=e^{\mu t+\frac{1}{2}t^2\sigma^2}
\end{align}\]

Answer 3

thank you very much... :)