# Normal distribution of $X_1$ and $X_2$

Let $X_1$ and $X_2$ be independent with normal distributions $N(6,1)$ and $N(7,1)$ respectively. Find $P(X_1 > X_2)$. Hint: Write $P(X_1 > X_2) = P(X_1 - X_2 > 0)$ and determine the distribution of $X_1 - X_2$.

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Since $X_1\sim N(6,1),\;\; X_2\sim N(7,1)$, we have $X_1-X_2\sim N(-1,2)$. Hence $P(X_1>X_2) = P\bigg(\frac{X_1-X_2+1}{\sqrt{2}}>\frac{1}{\sqrt{2}}\bigg) =1-\Phi(1/\sqrt{2})=0.24$

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why is it N(-1,2)? I get the -1 part but why is it 2? –  user48495 Nov 14 '12 at 4:39
Because If X and Y are uncorrelated random variables, then $Var(X \pm Y) = Var(X) + Var(Y)$. Try proving it yourself. –  Gautam Shenoy Nov 14 '12 at 4:57