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What 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.

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  • $\begingroup$ You should add the 'self-study' tag and read its wiki. $\endgroup$ – StubbornAtom Mar 16 '18 at 13:51
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    $\begingroup$ As for references similar questions have been asked before, like this one: stats.stackexchange.com/q/326436/119261. $\endgroup$ – StubbornAtom Mar 16 '18 at 14:05
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    $\begingroup$ You may ignore $\mu$ and $\sigma$ (that is, work with $\mu=0$ and $\sigma=1$) because all they do is establish the units in which $X$ and $Y$ are expressed. Then, by changing variables to $X=(U+V)/\sqrt{2},Y=(U-V)/\sqrt{2}$ (so that $U$ and $V$ are also independent standard Normal variables), the problem reduces to an elementary integral. $\endgroup$ – whuber Mar 16 '18 at 14:07
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    $\begingroup$ @Martin You're not ignoring $\mu$ and $\sigma$, but simply recognizing that they play no essential role: they merely determine the units in which the values are expressed. Thus the symmetry is maintained and the variables $U$ and $V$ work just fine. You can also show this algebraically. Of course it is crucial that the parameters for the distributions of $X$ and $Y$ be the same! $\endgroup$ – whuber Mar 16 '18 at 16:33
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    $\begingroup$ @whuber ah of course the two different $\mu$ and $\sigma$ are the same. so in the representation with standard normal distribution the relation becomes something like $X-Y<a$ with equal weights for X and Y. (I was imagining a change of the slope as well, which I believe occurs only when the $\sigma$ are different, so different $\mu$ don't matter) . $\endgroup$ – Sextus Empiricus Mar 16 '18 at 22:15
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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

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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 $\mathbb{E}_{X \sim X | X < Y}(X)$.

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