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