1. ## [SOLVED] Please check what i did wrong

Hi, can someone please check what i did wrong..
So I know that
$\displaystyle V(x) = M''(0)-(M'(0))^2$
My $\displaystyle M(t)=e^{\mu t + \frac{\sigma^2 t^2}{2}}$
$\displaystyle M'(t) = \mu e^{\mu t} + e^{\sigma^2 t}$
$\displaystyle M'(0) = \mu$
$\displaystyle M''(t) = \mu^2 e^{\mu t} + e^{\mu t} + \sigma^2 e^{\sigma^2 t}$
$\displaystyle M''(0) = \mu^2 + 1 + \sigma^2$

Then
$\displaystyle V(x) = \mu^2 + 1 + \sigma^2 - \mu^2$
$\displaystyle V(x) = 1 + \sigma^2$

But the answer is supposed to be $\displaystyle V(x) = \sigma^2$

Can someone please check what i did wrong

2. Don't worry, I know what i did wrong lol

3. I found

$\displaystyle M'(t) = (\mu+\sigma^2 t) e^{\mu t +\frac{\sigma^2 t^2}{2}} \neq \mu e^{\mu t} + e^{\sigma^2 t}$

4. Originally Posted by pickslides
I found

$\displaystyle M'(t) = (\mu+\sigma^2 t) e^{\mu t +\frac{\sigma^2 t^2}{2}} \neq \mu e^{\mu t} + e^{\sigma^2 t}$
haha thanks, i know, i just realised that right after i posted it