If $X,Y$ are uniform on $[0,1]$, what is expectation of $X$ knowing $X+Y>1$?

Question

Let $X,Y\sim U(0,1)$, I'm trying to figure out $E[Y\mid X+Y>1]$.

I get to the correct answer by following this line:

$$E[Y\mid X+Y>1] = \frac{E[Y\,1_{Y>1-X}]}{P(X+Y>1)}=2\iint y1_{y>1-x}\,dx\,dy=\frac{2}{3} $$

However I can't wrap my head around why this is false:

\begin{align} E[Y\mid X+Y>1] &= \int{E[E[Y\mid X+Y>1]\mid X=x]}f_{X}(x)\,dx \\&= \int{E[Y\mid Y>1-x]}\,dx \\&=\int\frac{\int y1_{y>1-x}\,dy}{P(Y>1-x)}\,dx \\&= \int\left(1-\frac{x}{2}\right)dx \\&=\frac{3}{4} \end{align}

I used to be familiar with this concept but this eludes me right now, any intuition would be appreciated. Thank you.

Answer 1

I spent quite a while trying to figure out what formula you were attempting to use. In the end, I just had to rephrase what the formulas seemed to say.

• $E[Y \mid X+Y > 1]$ is finding the mean $y$ coordinate in the triangle $x+y > 1, x,y < 1$
• $\int_0^1 E[Y \mid Y > 1-x]\,dx$ is finding the mean $y$ coordinate on the line $y = 1-\frac{x}{2}$ (being the mean, for each $x$, of the slice $1-x < y < 1$)

It's apparent that this won't work because the points on the line $y=1-\frac{x}{2}$ in the second case have equal weight instead of the weight depending on the width of the slice $1-x < y < 1$.

In other words, in order for this to work, we would need some sort of independence assumption, and conditioned on $X+Y > 1$, $Y$ and $X$ are definitely not independent.

Answer 2

By definition of conditioning over an event: $$\mathsf E(Y\mid Y>1-X)=\mathsf E(Y\mathbf 1_{Y>1-X})\div \mathsf E(\mathbf 1_{Y>1-X})$$

Always use this before attempting to use the Law of Total Expectation.

$$\begin{align}\mathsf E(Y\mid Y>1-X)&=\dfrac{\mathsf E(\mathsf E(Y~\mathbf 1_{Y>1-X}\mid X))}{\mathsf E(\mathsf E(\mathbf 1_{Y>1-X}\mid X))}\\[2ex]&=\dfrac{\int_0^1\int_{1-x}^1 y\,\mathrm d y\,\mathrm d x}{\int_0^1\int_{1-x}^1 1\,\mathrm d y\,\mathrm d x}&&\neq \int_0^1\dfrac{\int_{1-x}^1 y\,\mathrm d y}{\int_{1-x}^1 1\,\mathrm d y}\,\mathrm d x\end{align}$$

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Conditioning over events does not mix well with conditioning over sigma-algebrae.   They are related concepts with similar symbolic representation, but they are not compatible.