Question

Derive Moment Generating Function for Normal Distribution:

Answer 1

So the correct answer is \[e^{\mu t + \frac{\sigma^{2}t^{2}}{2}}\]

But when I try to derive it:
\[E(e^{tx}) = \int\limits_{-\infty}^{\infty}e^{tx}\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(x-\mu)^{2}}{2\sigma^{2}}}dx\]
\[= \int\limits\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(x-\mu)^{2}+2\sigma^{2} tx}{2\sigma^{2}}}dx\]
\[= e^{\frac{-\mu^{2}}{2\sigma^{2}}}e^{(\mu +\sigma^{2}t)^{2}} \int\limits\limits\limits_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{ -x^{2}+(2)(\mu +\sigma^{2}t)x - (\mu +\sigma^{2}t)^{2}}{2\sigma^{2}}}dx\]
\[= e^{\frac{-\mu^{2}}{2\sigma^{2}} + (\mu+\sigma^{2}t)^{2}} \]

and I'm just lost from here.

Answer 2

Oof I don't know this I wish I could help

Answer 3

But maybe a video can help ....... I'm sorry

https://www.youtube.com/watch?v=L8NS33q3YYc

Answer 4

The first few seconds of that video literally gives the entire answer out.

Answer 5

Huhhh so it helped

Answer 6

Yay! Good luck