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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.

$P(X|X<Y)$ = ... and

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

I am thinking solving this in this way:

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

My questions are:

(1) Whether above solution is correct

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

$P(X|X<Y)$ = ... and

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

I am thinking solving this in this way:

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

My questions are:

(1) Whether above solution is correct

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

$P(X|X<Y)$ = ... and

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

I am thinking solving this in this way:

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

My questions are:

(1) Whether above solution is correct

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

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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.

$P(X|X<Y)$ = ... and

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

I am thinking solvesolving this in this way:

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

My questions are:

(1) Whether above solution is correct

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

$P(X|X<Y)$ = ... and

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

I am thinking solve this in this way:

\begin{align} P(X|X<Y) &= \frac{P(Y>X|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 is correct

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

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.

$P(X|X<Y)$ = ... and

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

I am thinking solving this in this way:

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

My questions are:

(1) Whether above solution is correct

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

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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<Y)$ = ... 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|X<Y) &= \frac{P(Y>X|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.$P(X|X<Y)$

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.

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

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

I am not sure whether a closed-form solution exists for this problem.

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.

$P(X|X<Y)$ = ... and

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

I am thinking solve this in this way:

\begin{align} P(X|X<Y) &= \frac{P(Y>X|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 is correct

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

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