If $X$ and $Y$ are normally distributed random variables with the same variance, what kind of distribution their substraction follows?

Question

Let $X \sim N(\mu_1, \sigma^2)$ and $Y \sim N(\mu_2, \sigma^2)$. I find the distribution of $X - Y$. I suspect that $X - Y \sim N(\mu_1 - \mu_2, \sigma^2)$. Indeed,

$\mu_{x-y} = \mu_x - \mu_y = \mu_1 - \mu_2$.

but I don't know how to prove that $\sigma^2_{X-Y} = \sigma^2$. I think the covariance must be zero, but the exercise doesn't say that the variables are independent. Any help? Thanks.

Answer 1

You need some info on the joint distribution of $X,Y$ to describe the distribution of $X-Y$. You cannot generally conclude $X-Y$ is normally distributed (see here for counterexample). On the other hand, if $X,Y$ are jointly bivariate normal, then any linear combination of $X,Y$ is normal.

Without info on the joint distribution, we can say that for constants $a,b\in \mathbb{R}$, and $X\sim N(\mu_1,\sigma^2_1)$,$Y\sim N(\mu_2,\sigma^2_2),$

$$E[aX+bY]=a\mu_1+b\mu_2,\\ \text{Var}(aX+bY)=a^2\sigma^2_1+b^2\sigma^2_2+2ab\text{Cov}(X,Y).$$