MGF of Normal Distribution

Question

Let X $\backsim \mathcal{N}(0,\sigma^2)$ and $t>0$. I want to find $E(e^{-t x})$

$$E(e^{-t x})=\int_{-\infty}^{\infty} e^{-tx} \dfrac{1}{\sqrt{2\pi \sigma^2}}e^{\frac{-x^2}{2\sigma^2}}dx = \int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2}} e^{-tx\frac{-x^2}{2\sigma^2}}dx = \int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-2\sigma^2tx-x^2}{2\sigma^2}}dx=\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x+\sigma^2t)^2+\sigma^4t^2}{2\sigma^2}}dx=e^{\frac{\sigma^4t^2}{2\sigma^2}}\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x+\sigma^2t)^2}{2\sigma^2}}dx=e^{\frac{\sigma^2t^2}{2}}$$

Is that ok or did I missed something? For $\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x-\sigma^2t)^2}{2\sigma^2}}dx$ I just assumed that this is $1$ intuitively. Is that correct?

Answer 1

What you have done is right. For the part where you have a doubt just make the substitution $y=x-\sigma^{2}t$ and use the fact that normal density function integrates to $1$.

Answer 2

1. You have a sign error on the 4th equation since $-2\sigma^2tx-x^2=-(x+\sigma^2t)^2+\sigma^4t^2$.
2. That integral is indeed 1, as you are integrating the density of the distribution $\mathcal{N}(\sigma^2t,\sigma^2)$ (with the correction from point 1, it is now $\mathcal{N}(-\sigma^2t,\sigma^2)$) over all $\mathbb{R}$.

Answer 3

No answer but a trick that decreases the chance on mistakes considerably.

First find $M_U(t)$ where $U$ has standard normal distribution.

Then make use of: $$M_X(t)=\mathbb e^{-tX}=\mathbb e^{-t\sigma U}=M_U(\sigma t)$$

This also works more generally.

Answer 4

If we only look at the exponents, by completing the square we have

$$-\frac{x^2}{2\sigma^2}-tx=-\frac{(x+\sigma^2t)^2-\sigma^4t^2}{2\sigma^2}=-\frac{(x+\sigma^2t)^2}{2\sigma^2}+\frac{\sigma^2t^2}2.$$

As the first term is just a shift of the integration variable and the integral is on $\mathbb R$, the shift has no effect, and as the second term is a constant, which we can pull if out as the factor

$$e^{\sigma^2t^2/2}$$ and this is the requested MGF.