Say I have two independent random variables $$X \sim N(u_1, \sigma_1)$$ and $$Y \sim N(u_2, \sigma_2)$$. I want to get the conditional distribution of X given whether X is bigger than Y or not.

$$P(X|X = ... and

$$P(X|X>Y)$$ = ...

I am thinking solving this in this way:

\begin{align} P(X|XX|X)P(X)}{P(Y>X)} \\ &= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})}\\ P(X|X>Y) &= \frac{P(Y

My questions are:

(1) Whether above solution is correct

(2) How to get the mean and sd for $$P(X|X and $$P(X|X>Y)$$ if they are still normal?

Say I have two independent random variables $$X \sim N(u_1, \sigma_1)$$ and $$Y \sim N(u_2, \sigma_2)$$. I want to get the conditional distribution of X given whether X is bigger than Y or not.

$$P(X|X = ... and

$$P(X|X>Y)$$ = ...

I am thinking solvesolving this in this way:

\begin{align} P(X|XX|X)P(X)}{P(Y>X)} \\ &= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})} \end{align}\begin{align} P(X|XX|X)P(X)}{P(Y>X)} \\ &= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})}\\ P(X|X>Y) &= \frac{P(Y

My questions are:

(1) Whether above solution is correct

(2) How to get the mean and sd for $$P(X|X and $$P(X|X>Y)$$

Say I have two independent random variables $$X \sim N(u_1, \sigma_1)$$ and $$Y \sim N(u_2, \sigma_2)$$. I want to get the conditional distribution of X given whether X is bigger than Y or not.

$$P(X|Y-X>0)$$$$P(X|X = ... and

$$P(X|Y-X<0)$$$$P(X|X>Y)$$ = ...

I am not sure whether a closed-formthinking solve this in this way:

\begin{align} P(X|XX|X)P(X)}{P(Y>X)} \\ &= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})} \end{align}

My questions are:

(1) Whether above solution existsis correct

(2) How to get the mean and sd for this problem.$$P(X|X