What is the expected value of $X$ given $XWhat is $\mathbb{E}[X|X<Y]$ if $X,Y\overset{iid}{\sim}\mathcal{N}(\mu,\sigma^2)$?
I have found that $\mathbb{E}[X|X<Y]=\int_{-\infty}^{\infty} -\log(\Phi(\frac{x-\mu}{\sigma}))x\phi(\frac{x-\mu}{\sigma})dx$ but from there couldn't procede.
 A: Basically you want to find $E[X_{(1)}]$ given the sample size is 2, where is $X_{(1)}$ the first order statistic. So, for now assume $X \sim N(0,\sigma^2)$ we are just shifting the location , later in the end we will add $\mu$ to our answer.(This is just to make the calculations easier). Let $f(x)$ be the pdf of $N(0,\sigma^2)$ and $f_{(1)}(x)$ be the pdf of $X_{(1)}$. Now I hope you know how to show 
$f_{(1)}(x) = 2[1-F(x)]f(x)$ 
so, $E[X_{(1)}] = \int_{-\infty}^{\infty}{2x(1-F(x))f(x)} = 2 \int_{-\infty}^{\infty}{xf(x)} -  2\int_{-\infty}^{\infty}{xF(x)f(x)} = -  2\int_{-\infty}^{\infty}{xF(x)f(x)}$
Now use by parts to integrate $ \int_{-\infty}^{\infty}{xF(x)f(x)}$ so,
$ \int_{-\infty}^{\infty}{F(x)f(x)} = F(x)\int_{-\infty}^{\infty}{xf(x)} - \int_{-\infty}^{\infty}{f(x)\int{xf(x)}} = - \int_{-\infty}^{\infty}{f(x)\int{xf(x)}}$ 
Now use the result that $\int{xf(x)} = -\sigma^2f(x)$ (you can prove this)
And substitute the result back and solve the integrals to obtain your answer. Your "final" answer should come as $-\frac{\sigma}{\sqrt{\pi}} + \mu$, this is for accounting the fact that we shifted the location to the origin,so after shifting it back this should be your result
A: By $\mathbb{E}(X | X < Y)$, you probably mean $\mathbb{E}_{X \sim X | X < Y}(X)$. I prefer this notation since it clearly splits the information in two parts: 
1- in subscript, the random variable and the distribution that are used to compute the expectation (in this case, $X$ and $P(X|X < Y)$, resp.); 
2 - between parenthesis, the function whose expectation we are computing (in this case, the identity function $f(X) = X$).
Hints to solve the problem:
1 - From the fact that $X$ and $Y$ are independent, you have that $P(X,Y) = P(X)P(Y)$.
2 - Using $P(X,Y)$, derive an expression for the function $f(Y) = P(X < Y)$.
3 - Use $f(Y)$ and the definition of conditional probability to obtain the pdf $P(X | X < Y)$.
4 - Use $P(X | X < Y)$ and the definition of expectation to compute your desired $\mathbb{E}_{X \sim X | X < Y}(X)$.
