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".

  • $\begingroup$ As explained below, this assumes implicitely that $(X,Y):(x,y)\mapsto(x,y)$ on $\Omega=[0,1]^2$. What is your source? $\endgroup$ – Did Sep 1 '15 at 14:58
  • $\begingroup$ @Did "Basic Stochastic Processes" by Brezniak. How do you know that this is what it implicitly assumes? $\endgroup$ – Nicolas Bourbaki Sep 2 '15 at 18:55
  • $\begingroup$ 'Cause this is by far the simplest way to define $(X,Y)$ on $\Omega=[0,1]^2$ with the desired joint distribution. $\endgroup$ – Did Sep 2 '15 at 18:59
  • $\begingroup$ 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 $f_{X,Y}$. $\endgroup$ – 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, $$\begin{align} (x,y)\in Y^{-1}(B) & \iff Y(x,y) \in B \\ & \iff x\in [0,1] \wedge y\in B \\ & \iff (x,y)\in [0;1]\times B \end{align}$$

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.