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If I have two random variables, $X_1, X_2 \sim N(\mu, \sigma^2)$ where $\mu, \sigma^2$ are unknown, what can I say about $X_1-X_2$? More specifically, I know that the difference should be a normal as well, but how would I calculate the variance? Thanks!

user123276
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    The variance calculation question is answered in many threads here (such as http://stats.stackexchange.com/questions/75010, inter alia). The conclusion about normality is incorrect and that is addressed in many other threads. – whuber Oct 04 '15 at 15:16
  • Hi, thanks for your response, do you know why the normality is incorrect? I was always under the assumption that $X_1 - X_2 \sim N(\mu_1-\mu_2, \sigma^2_1+\sigma^2_2)$, did I miss something? – user123276 Oct 04 '15 at 15:29
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    Regardless of whether $X$ and $Y$ are normal or not, it is true (whenever the various expectations exist) that \begin{align} \mu_{X-Y}&=\mu_X-\mu_Y\ \sigma_{X-Y}^2&=\sigma_{X}^2+\sigma_{Y}^2-2\operatorname{cov}(X,Y) \end{align} where $\operatorname{cov}(X,Y)=0$ whenever $X$ and $Y$ are independent or uncorrelated. The only issue is whether $X-Y$ is normal or not and the answer to this is that $X-Y$ is normal only when $X$ and $Y$ are jointly normal (including, as a special case, when $X$ and $Y$ are independent). Note: $X$ and $Y$ being just uncorrelated does not suffice. – Dilip Sarwate Oct 04 '15 at 15:42
  • Hi, do you know why they normally say that linear combinations of normally distributed random variables are normal? Are they assuming them to be jointly normal to begin with? – user123276 Oct 04 '15 at 19:15
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    @user123276 The statement is typically, or at least should be, that linear combinations of independent normal variables are normal. – Danica Oct 04 '15 at 19:50
  • "Hi, do you know why they normally say ..." Ask them. – Dilip Sarwate Oct 04 '15 at 20:16
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    If $X_1,X_2$ are bivariate normal, $X_1-X_2$ is normal. Otherwise, not so much. See here for some discussion of the sum of general correlated normals (which given that $-X_2$ is normal, applies to differences as well). The question there links to an uncorrelated case, as well. – Glen_b Oct 05 '15 at 12:27

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