# Conditional Expectation

I am learning conditional expectation for fun. I cannot solve the following problem, I think it is just my vocabulary is limited. This problem came from a book on Stochastic Processes:

Let $\Omega = [0,1]\times [0,1]$ with the usual Lebesgue measure and $\sigma$-algebra of Borel sets. Let $X$ and $Y$ be random variables on $\Omega$ with joint density $$f_{X,Y}(x,y) = \left\{ \begin{array}{ccc} x+y & \text{ if }& x,y\in [0,1] \\ 0 & \text{ if }& \text{ otherwise} \end{array} \right.$$ Show that $\displaystyle E(X|Y) = \frac{2 + 3Y}{3 + 6Y}$.

The part that I do not understand is in the solutions, the author begins by writing, $$Y^{-1} (B) = [0,1] \times B$$ where $B$ is a Borel set in $[0,1]$.

How can be possibly say that? There is no reason at all why $X$ and $Y$ are defined by different coordinates. Unless I completely misunderstand what it means to be of "joint density".

• As explained below, this assumes implicitely that $(X,Y):(x,y)\mapsto(x,y)$ on $\Omega=[0,1]^2$. What is your source? – Did Sep 1 '15 at 14:58
• @Did "Basic Stochastic Processes" by Brezniak. How do you know that this is what it implicitly assumes? – Nicolas Bourbaki Sep 2 '15 at 18:55
• 'Cause this is by far the simplest way to define $(X,Y)$ on $\Omega=[0,1]^2$ with the desired joint distribution. – Did Sep 2 '15 at 18:59
• Note however that the identity of the probability space $(\Omega,\mathcal F,P)$ and of the random variables $(X,Y)$ is irrelevant to solve the exercise; all that is needed is the joint distribution of $(X,Y)$, characterized by the PDF $f_{X,Y}$. – Did Sep 2 '15 at 19:38

This is to establish the following for Borel set $B\subseteq [0,1]$, \begin{align} \mathsf P(Y\in B) & = \iint_{Y^{-1}(B)} f_{X,Y}(x,y) \mathrm dx\mathrm dy \\ & = \iint_{[0;1]\times B} f_{X,Y}(x,y) \mathrm dx\mathrm dy \\ & = \int_B \int_0^1 (x+y) \mathrm dx\mathrm dy \\ & =\int_B (y + \frac 1 2 ) \mathrm dy. \end{align}

Thus, the marginal PDF $f_Y(y)$ is $y+\frac 1 2$ for $y\in [0;1]$, and $0$ otherwise.

In fact, the joint density of $X$ and $Y$ tells us how $X: \Omega\rightarrow \mathbb{R}$ and $Y:\Omega\rightarrow \mathbb{R}$ are distributed with $X(x,y)=x$ and $Y(x,y)=y$. We have for any Borel set $A\subseteq [0,1]\times [0,1]$,

$$P((X,Y)\in A) = \iint_A f_{X,Y}(x,y) \mathrm dx\mathrm dy.$$ Thus, \begin{align} (x,y)\in Y^{-1}(B) & \iff Y(x,y) \in B \\ & \iff x\in [0,1] \wedge y\in B \\ & \iff (x,y)\in [0;1]\times B \end{align}

• I am sorry, this does not answer my question. I can do the computations, I do not understand why $Y^{-1}(B) = [0,1]\times B$? – Nicolas Bourbaki Aug 6 '15 at 4:34
• I have explained it at the end. – i707107 Aug 6 '15 at 4:40
• Define $f:[0,1]^2\to \mathbb{R}$ as $f(x,y) = 0$. Let $B = \{0\}$. Then $f^{-1}(B) = [0,1]^2$, it is not equal to $[0,1]\times B$. – Nicolas Bourbaki Aug 6 '15 at 4:43
• The random variable $Y$ takes $(x,y)$ to $y$. – i707107 Aug 6 '15 at 4:45
• Can you please explain why? (I think my vocabulary is limited.) I thought a random variable is any measurable function from $\Omega \to \mathbb{R}$. – Nicolas Bourbaki Aug 6 '15 at 4:46

we can write $E(X|Y)=\int_X xf(x|y)dx$

but $f(x|y)=\frac{f(x,y)}{f(y)}$

and $f(y)=\int_X f(x,y)dx=\int_{0}^{1} (x+y)dx=(x^2/2+xy |_0^1)=1/2+y$

Also

$$E(X|Y)=\int_X xf(x|y)dx=\int_{0}^{1}x \frac{x+y}{1/2+y} dx$$

$$=\int_{0}^{1} \frac{x^2+xy}{1/2+y} dx=\frac{1}{1/2+y} (x^3/3+x^2y/2|_0^1)$$ $$= \frac{(1/3-y/2)}{(1/2+y)}=\frac{2+3y}{3+6y}$$