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Let $X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$ equals?

The way I interpreted it is as follows:

$E(X|X>Y)$ is $E(X)$ realizing $X>Y$

so $E(X)=\int_{0}^{1}\int_{y}^{1}xdxdy=\int_{0}^{1}(\frac{x^2}{2})_{y}^{1}dy=\frac{1}{2}\int_{0}^{1}(1-y^2)dy=\frac{1}{2}(y-\frac{y^3}{3})_{0}^{1}=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$

What can be a better alternative to solve this problem? I tried so hard but couldn't get the answer using the proper definition of expectation, can anyone help me?And can we use law of total expectation here?

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    $\begingroup$ FYK, I did the same procedure as yours $\endgroup$
    – tommik
    Jan 19, 2021 at 13:02
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    $\begingroup$ $$\mathbb E[X\mid X>Y]=\frac{\mathbb E[X\boldsymbol 1_{\{X>Y\}}]}{\mathbb P\{X>Y\}}.$$ $\endgroup$
    – Surb
    Jan 19, 2021 at 13:03
  • $\begingroup$ Your interpretation is correct, however the first equal sign $E(X)=\int_0^1\cdots$ is not correct. Do I understand it right, that you now want to know how to use the proper definition of conditional expectation to justify your interpretation? $\endgroup$
    – mag
    Jan 19, 2021 at 13:12
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    $\begingroup$ I think you are missing a factor 2. $\endgroup$
    – Thomas
    Jan 19, 2021 at 13:49
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    $\begingroup$ @Thomas yes I understood thanks $\endgroup$
    – Daman
    Jan 20, 2021 at 7:29

2 Answers 2

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This is an answer how to use strictly the proper definition of conditional expectation to justify the interpretation in the question:

The conditional expectation conditioned by the event $\{X>Y\}$ is the conditional expectation conditioned by the $\sigma$-algebra generated by the event: $$\mathcal M:=\sigma(\{X>Y\})=\{\emptyset,\Omega,\{X>Y\},\{X>Y\}^\complement\}.$$ This $E(X|\mathcal M)$ is a $\mathcal M$-measureable random variable satisfying $$\int_MX\mathrm dP=\int_ME(X|\mathcal M)\mathrm dP $$ for all $M\in\mathcal M$. So you have to find out which $\mathcal M$-measureable RV satisfies $$\int_{\{X>Y\}}X\mathrm dP\color{blue}{=}\int_{\{X>Y\}}E(X|\mathcal M)\mathrm dP,\\ \int_{\{X\leq Y\}}X\mathrm dP\color{blue}{=}\int_{\{X\leq Y\}}E(X|\mathcal M)\mathrm dP,\\ \int_\Omega X\mathrm dP\color{blue}{=}\int_\Omega E(X|\mathcal M)\mathrm dP. $$ As being $\mathcal M$-measureable it should be constant on the sets $\{X>Y\}$ and $\{X>Y\}^\complement$, i.e. $E(X|\mathcal M)(\omega)=c_1$ for $\omega\in\{X>Y\}$ and $E(X|\mathcal M)(\omega)=c_2$ for $\omega\in\{X\leq Y\}$. So you want to have (calculated for this special case; everything written above holds general) \begin{align} \frac{1}{3}=\int_{\{X>Y\}}X\mathrm dP&\color{blue}{=}\int_{\{X>Y\}}c_1\mathrm dP=c_1P(\{X>Y\})=c_1/2,\\ \frac{1}{6}=\int_{\{X\leq Y\}}X\mathrm dP&\color{blue}{=}\int_{\{X\leq Y\}}c_2\mathrm dP=c_2P(\{X\leq Y\})=c_2/2,\\ \frac{1}{2}=E(X)=\int_\Omega X\mathrm dP&\color{blue}{=}\int_\Omega (c_1 \cdot1_{\{X>Y\}}+c_2\cdot1_{\{X\leq Y\}})\mathrm dP\\&=c_1P(\{X>Y\})+c_2P(\{X\leq Y\})=(c_1+c_2)/2. \end{align} Usually when you determine this random variable $E(X|\mathcal M)$ you are just interested in the values for $\omega\in\{X>Y\}$, since this is the inital intention of $E(X|X>Y)$. But to write it completely \begin{align} E(X|\mathcal M)(\omega)=\begin{cases}c_1=\frac{2}{3}, &\text{for }\omega\in\{X>Y\}\\ c_2=\frac{2}{6}=\frac{1}{3}, &\text{for }\omega\in\{X\leq Y\}. \end{cases} \end{align}

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The cdf of $\mathcal U(0,1)$ is $U(x) = x$, and the the pdf $u(x) = 1$.

Note that $\mathbb E[X|X>Y] = 2\mathbb E[X1_{\{X>Y\}}]$ since $$\mathbb E[X|X>Y] = \frac{\mathbb E[X1_{\{X>Y\}}]}{\mathbb P(X>Y)},$$ and $\mathbb P(X>Y) = \frac{1}{2}$. Moreover, $$\mathbb E[X1_{\{X>Y\}}] = \mathbb E[\mathbb E[X1_{\{X>Y\}}]|X] = \mathbb E[X\mathbb P(X>Y|X)]$$ using the law of iterated expectations. Since $\mathbb P(X>Y|X) =U(X)$, we obtain $$\mathbb E[X1_{\{X>Y\}}] = \mathbb E[XU(X)] = \int_0^1xU(x)u(x)\,\mathrm dx = \int_0^1x^2\,\mathrm dx = \frac{1}{3}$$

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