# Statistics: If $X_1$ and $X_2$ are both normally distributed then explain why $X_1 - X_2$ can be standardized with mean 0 and standard deviation of 1

I am currently studying hypothesis testing for two populations and I would like a math major or someone experienced to explain to me why this particular statistic has a mean of 0 and a standard deviation of 1:

$\displaystyle z_{\bar{X_1}-\bar{X_2}} = \frac{\bar{X_1}-\bar{X_2} - \left(\mu_1 - \mu_2\right)}{\sigma_{\bar{X_1}-\bar{X_2}}}$

The course that I'm taking is under the political science department. I see a lot of theoretical questions here and I would like it if someone can explain the fundamentals. How do you know that if $X_1$ and $X_2$ are both normally distributed then $X_1 - X_2$ is also normal? Why is it that when you standardize $X_1 - X_2$ you get the same mean and standard deviation like when $X_1$ or $X_2$ is standardized?

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A linear combination of independent normal RVs is normal. The mean of $X=\sum_i a_iX_i$ is $\mu =\sum_i a_i \mu_i$, and the variance is $\sigma^2=\sum_i a_i^2\sigma_i^2$. So you have just standardized a normal RV $X \sim N(\mu,\sigma^2)$, and a standardized normal RV has distribution $N(0,1)$.
For every random variable $X$ with mean $\mu_X$ and standard deviation $\sigma_X$, the random variable $Z=(X-\mu_X)/\sigma_X$ has mean $0$ and standard deviation $1$. Hence all your questions are explained if, forgetting the normal distribution and nearly everything else, you can show that the mean of $\bar X_1-\bar X_2$ is $\mu_1-\mu_2$. Can you?
Yes, in fact, you have completely answered the OP's question. The OP probably should have been wondering instead why the distribution was normal, not why it had mean $0$ and variance $1$--much more difficult in my mind. +1 for you. – MPW Feb 15 '14 at 23:48