$X$ and $Y$ be i.i.d $U(0,1)$ random variables. Then $E(X|X>Y)$ [duplicate]

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?

• FYK, I did the same procedure as yours Jan 19, 2021 at 13:02
• $$\mathbb E[X\mid X>Y]=\frac{\mathbb E[X\boldsymbol 1_{\{X>Y\}}]}{\mathbb P\{X>Y\}}.$$
– Surb
Jan 19, 2021 at 13:03
• 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?
– mag
Jan 19, 2021 at 13:12
• I think you are missing a factor 2. Jan 19, 2021 at 13:49
• @Thomas yes I understood thanks Jan 20, 2021 at 7:29

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}

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 $$\mathbb E[X1_{\{X>Y\}}] = \mathbb E[XU(X)] = \int_0^1xU(x)u(x)\,\mathrm dx = \int_0^1x^2\,\mathrm dx = \frac{1}{3}$$