# if X and Y are independent Normal random variables, what would be the probability that X is less than Y?

How can I find this probability $P(X<Y)$ ? knowing that X and Y are independent Normal random variables.

$X$~$N(m_1 ,v_1)$ (with mean m1 and variance v1)

$Y$~$N(m_2 ,v_2)$

I know that $$\Pr \left[ X < Y \right] = \int F_X \left( y \right) f_Y \left( y \right) \mathrm{d} y$$ but I cannot solve it for normal random variable.

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The key theorem is the following: Let $X$ and $Y$ be independent normal, with means $\mu$ and $\nu$, and variances $\sigma^2$ and $\tau^2$ respectively. Let $W=X-Y$. Then $W$ is normally distributed, with mean $\mu-\nu$, and variance $\sigma^2+\tau^2$.

Now it is easy to compute $\Pr(W\lt 0)$.

If $\mu=\nu$, we don't need the above machinery. For then the random variable $W$ has distribution which is symmetric about $0$, so the probability is $1/2$.

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From symmetry, we have $\Pr[X< Y] = \Pr[Y<X]$. Since $\Pr[X< Y] + \Pr[Y<X] = 1$, we get $$\Pr[X< Y] = 1/2.$$

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Their summation is one but this does not mean that Pr[X<Y]=1/2. –  May Jan 27 '13 at 1:27
so Pr[X<Y]=Pr[Y<X] is not correct –  May Jan 27 '13 at 1:29
@May, note that $\Pr[X<Y] = \Pr[Y<X]$ (since $X$ and $Y$ have the same distribution). –  Yury Jan 27 '13 at 1:29
yes they have normal distribution but their means and variances are different. –  May Jan 27 '13 at 1:31
I see. I thought they were identically distributed. –  Yury Jan 27 '13 at 1:34