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

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 $$N(a-b,c^2+d^2)$$

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    $\begingroup$ Would your first guess make sense when $d^2$ exceeds $c^2$? $\endgroup$
    – whuber
    Jan 26 '18 at 23:37
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Using your notation, the distribution will be $N(a-b,c^2+d^2)$: $$X-Y \sim N(a-b,c^2+d^2)$$

Here is a detailed explanation that should help: http://mathworld.wolfram.com/NormalDifferenceDistribution.html

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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.

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