Distribution of normal variable subtracted from another normal random variable?

$$X\sim N(a,c^2), \text{ }\text{ } \text{ }\text{ }Y\sim N(b,d^2)$$

Then $$X+Y \sim N(a+b,c^2+d^2)$$

What is the distribution of $X-Y$?

I can't figure out if it is $$N(a-b,c^2-d^2)$$ or $$N(a-b,c^2+d^2)$$

• Would your first guess make sense when $d^2$ exceeds $c^2$?
– whuber
Jan 26 '18 at 23:37

Using your notation, the distribution will be $N(a-b,c^2+d^2)$: $$X-Y \sim N(a-b,c^2+d^2)$$
Its is correct as stated in the previous answer that the difference of two independent normal random variables $$X\sim N(a,c^2)$$ and $$X\sim N(b,d^2)$$ is distributed as $$\begin{equation} X-Y \sim N(a-b,c^2 + d^2) \end{equation}$$ This intuition follows naturally from the additive property of the expected value $$E(X+Y) = E(X) + E(Y)$$ and that the variance of the difference between two random variables are given by $$\begin{equation} Var(X-Y) = Var(X) + Var(Y) - 2Cov(X,Y) \end{equation}$$ with $$Cov(X,Y) = 0$$ for two independent variables. This also imply that the formula presented in the other answer is not correct when the variables are correlated. Then, $$\begin{equation} X-Y \sim N(a-b,c^2 + d^2-2\rho_{x,y} c\cdot d) \end{equation}$$ where $$\rho_{x,y}$$ is the correlation between the two random variables.