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So I have this problem that says $W,X,Y$, and $Z$ are all independent random variables and all have standard normal distribution. Find $P(W+X>Y+Z+1)$. Now I know that $W+X-Y-Z$ will also be independent and standard normal distribution, however what I'm confused about is this:

I know that if I make $W$ and $X$ standard normal I could say to let $U=W+X$ and then $U\sim N(\mu_W +\mu_X , \sigma^2_W +\sigma^2_X)$.

Now lets say I want to find $P(W+X-Y-Z>1)$ .

How would I normalize that (is that the correct verbage)? Would I say to let $V=W+X-Y-Z$ and then $V\sim N\left(\mu_W +\mu_X-\mu_Y-\mu_z , \sigma^2_W +\sigma^2_X-\sigma^2_Y-\sigma^2_Z\right)$?

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Kyle, this is minor but when you say, "Now I know that $W+X-Y-Z$ will also be independent", I'm not sure what you are trying to say here but statements about independence must involve the comparison of at least two random variables. –  Patrick Dec 2 '12 at 6:20

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up vote 3 down vote accepted

You are fairly close, but the random variable $W+X-Y-Z$ has variance $\sigma_W^2 +\sigma_X^2+\sigma_Y^2+\sigma_Z^2$ (plus signs only).

In general if $X_1,X_2,\dots,X_n$ are independent normal with means $\mu_i$ and variances $\sigma_i^2$, and $a_1,a_2,\dots, a_n$ are any constants, then $\sum_{i=1}^{n} X_i$ is normally distributed, with mean $\sum_1^n a_i\mu_i$ and variance $\sum_1^n a_i^2 \sigma_i^2$.

Note the squaring of the $a_i$. In your case, the $a_i$ are always $1$ or $-1$, so their squares are $1$.

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How come with the variance were adding them all vs what I had? –  TheHopefulActuary Dec 1 '12 at 21:26
4  
@Kyle: Well, one could prove it, by going back to the formula for variance. But informally, think of two random variables $X$ and $Y$, and their difference $X-Y$. Both the wiggliness of $X$ and the wiggliness of $Y$ contribute to the wiggliness of $X-Y$. Take for example the case where $X$ and $Y$ have equal variance. Could $X-Y$ have variance $0$? That would mean totally predictable, constant!. And if the variance of $Y$ is bigger than the variance of $X$, could the variance of $X-Y$ be negative? Impossible, variance is never negative. –  André Nicolas Dec 1 '12 at 21:33

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