Let $X, Y$ be independent random variables such that $X,Y \sim N(\mu,\sigma^2)$, show that $X+Y$ and $X-Y$ are independent using the moment generating function.
I know that the moment generating function of a sum of independent random variables is the product of the MGF.
So, I´m trying to solve that but i don´t know if my process is correct
$M_{X+Y}(t_1,t_2)=M_X(t_1)M_Y(t_2)=M^2_{N(\mu,\sigma^2)}(t)$ ?
Recall that for an $\mathcal{N}(\mu, \sigma^2)$ random variable, the moment generating function of it is $$M(t) = \exp\left(\mu t + \frac{1}{2}\sigma^2t^2\right). \tag{1}$$
By condition, $X + Y \sim \mathcal{N}(2\mu, 2\sigma^2)$ and $X - Y \sim \mathcal{N}(0, 2\sigma^2)$. Therefore by $(1)$, we have: $$M_{X + Y}(t) = \exp\left(2\mu t + \sigma^2 t^2\right), \; M_{X - Y}(t) = \exp\left(\sigma^2 t^2\right).$$
On the other hand, as a bivariate random vector $(X + Y, X - Y)$, its MGF can be computed by definition as follows: \begin{align} & M_{(X + Y, X - Y)}(t_1, t_2) \\ = & E[\exp(t_1(X + Y) + t_2(X - Y))] \\ = & E\left\{\exp[(t_1 + t_2)X] \times \exp[(t_1 - t_2)Y]\right\} \\ = & E\left\{\exp[(t_1 + t_2)X] \right\}\times E\left\{\exp[(t_1 - t_2)Y]\right\} \quad \text{by independence of $X$ and $Y$.}\\ = & M_X(t_1 + t_2) M_Y(t_1 - t_2) \\ = & \exp\left(\mu(t_1 + t_2) + \frac{1}{2}\sigma^2(t_1 + t_2)^2\right)\exp\left(\mu(t_1 - t_2) + \frac{1}{2}\sigma^2(t_1 - t_2)^2\right) \\ = & \exp\left(2\mu t_1 + \sigma^2 t_1^2\right)\exp\left(\sigma^2 t_2^2\right) \\ = & M_{X + Y}(t_1) M_{X - Y}(t_2). \end{align}
Hence $X + Y$ and $X - Y$ are independent.
An answer using moment generating functions has already been given, but please also consider the following simpler approach. Note that two jointly normal random variables are independent if and only if they are uncorrelated. Since $X,Y$ are independent normals, the pair $(X,Y)$ is normal. Any linear transformation of a normal is also a normal, so $(X+Y, X-Y)$ is normal, i.e. $X+Y$ and $X-Y$ are jointly normal. Then $E[(X+Y)(X-Y)] = E[X^2 - Y^2] = \sigma^2 - \sigma^2 = 0 = (2 \mu) \cdot 0 = E[X+Y]E[X-Y]$. Thus $X+Y$ and $X-Y$ are independent.
For the general case that $X$ and $Y$ have a bivaraite normal distribution and have the same variances, it can be shown that $X+Y$ and $X-Y$ are independent. It follows from the three facts:
1- $X+Y$ and $X-Y$ are uncorrelated, i.e., their covariance is zero.
2- $X+Y$ and $X-Y$ also jointly follow a bivariate normal distribution.
3- For $(W,U)$ with a normal bivariate distribution, $W$ and $U$ are independent if and only if $W$ and $U$ are uncorrelated.
