Distribution of difference of two normally distributed random variables divided by square root of 2

Question

In Statistical Inference by Casella & Berger, in the proof of a theorem (theorem 5.3.1, page 220 in particular) I came across with the following proposition:

"the distribution of $\frac{X_2 - X_1}{\sqrt 2}$ is $N(0,1)$" where $X_2, X_1$ are iid normally random variables with the same mean $\mu$ and variance $\sigma^2$.

But despite of hours of trying, I couldn't prove this proposition even if it seems quite straightforward. I tried to use MGFs:

\begin{align*} M_{\frac{X_2-X_1}{\sqrt 2}}(t)&=E\left[e^{t\frac{X_2-X_1}{\sqrt 2}}\right]\\ &=E\left[e^{t\frac{X_2}{\sqrt 2}}\right]E\left[e^{t\frac{-X_1}{\sqrt 2}}\right]\\ &=\left(e^{\frac{1}{\sqrt 2}\mu t+\frac{1}{4}t^2\sigma ^2}\right)\left(e^{\frac{-1}{\sqrt 2}\mu t+\frac{1}{4}t^2\sigma ^2}\right)\\ &=e^{0t+\frac{1}{2}t^2\sigma ^2} \end{align*}

thus $\frac{X_2-X_1}{\sqrt 2} \sim N(0,\sigma ^2)$, so I found the variance not equal to $1$. What's my mistake here?

Answer 1

Your calculation is correct. What is true is $\frac {X_1-X_2} {\sqrt 2 \sigma}$ is normal with mean $0$ and variance $1$.

Answer 2

In general, one has that for independent $X_i, i=1,2\dots n$ which are independent $N(\mu_i, \sigma^2_i)$ $$\alpha_1X_1+\alpha_2X_2\dots a_nX_n \text{ is } N(\sum_{i=1}^n\alpha_i\mu_i, \sum_{i=1}^n\alpha^2_i\sigma^2_i)$$

and this is easily proven using MGFs.

In particular, taking $n=2, \alpha_1=\frac{1}{\sqrt{2}}, \alpha_2=-\frac{1}{\sqrt{2}}$ gives your answer