2
$\begingroup$

Are the two concepts distinct (is one stronger than the other) or equivalent?

In other words given two random variables $X$ and $Y$ defined on a sigma algebra $\mathcal{F}$ with sub sigma algebra $\mathcal{G} \subset \mathcal{F}$, is $E[X|\mathcal{G}] \perp E[Y|\mathcal{G}]$ (independence of the conditional expectations) equivalent to $X$ and $Y$ being conditionally independent given $\mathcal{G}$ (for an example, let $\mathcal{G} = \sigma(Z)$ for some random variable $Z$)?

What if $X$ and $Y$ were independent? Would that guarantee conditional independence (no!, I have counterexamples $X$ first coin flip, $Y$ second coin flip, $Z = X+Y$) or independence of conditional expectations?

$\endgroup$
1
$\begingroup$

Consider the following very simple example.

Let $\pi$ $\sim U[0,1]$ or something like that -- a random number between $0$ and $1$. And let $$X=\begin{cases}1&\text{ with probability }\pi\\0&\text{ with probability }1-\pi&\end{cases}$$ and let $$Y=\begin{cases}1&\text{ with probability }1-\pi\\0&\text{ with probability }\pi\end{cases}$$ and make sure that if $\pi$ is given then $X$ and $Y$ realize independently.

That is $$P(X=1\mid \pi\ )=P(Y=0\mid\pi\ )=\pi,\ P(Y=1\mid \pi\ )=P(X=0\mid \pi\ )=1-\pi$$

and $X$ and $Y$ are conditionally independent with respect to $\pi$:

$$P(X=1\cap Y=1\mid \pi\ )=P(X=0\cap Y=0\mid \pi\ )=\pi(1-\pi)$$

$$P(X=0\cap Y=1\mid \pi\ )=(1-\pi)^2$$ $$P(X=1\cap Y=0\mid \pi\ )=\pi^2.$$


Now, compute the conditional expectations:

$$E[\ X\mid \pi\ ]=\pi, \ \text{ and } E[\ Y\mid \pi\ ]=1-\pi.$$

$\pi$ and $1-\pi$ are not independent.


EDIT

Considering independent $X$ and $Y$.

Let $\Omega=\{1,2,3,4,5,6,7,8,9\}$, $\mathscr F=2^{\Omega}$, $P(I)=\frac19$ and let's visualize $\Omega$ lie this:

[![enter image description here][1]][1]

Furthermore let $$X(\omega)=\begin{cases}1&\text{ if }\omega \in\{1,2,3\}\\ 2&\text{ if }\omega \in\{4,5,6\}\\ 3&\text{ if }\omega \in\{7,8,9\} \end{cases}$$ and let $$Y(\omega)=\begin{cases}1&\text{ if }\omega \in\{1,4,7\}\\ 2&\text{ if }\omega \in\{2,5,8\}\\ 3&\text{ if }\omega \in\{3,6,9\}. \end{cases}$$ Check that $X$ and $Y$ are independent.

Now, define $\mathscr G$ as $$\{\Omega,\emptyset, G_1=\{1,2,4,5,7\},G_2=\{3,6,8,9\}\}.$$

In order to get the conditional expectations $E[X\mid \mathscr G\ ]$ , we need to calculate the following conditional probabilities:

$$P(X=i\mid G_1\ )=\frac{P(X=i \cap G_1 )}{P(G_1)}=\begin{cases}\frac35&\text{ for }&i=1\\\frac25&\text{ for }&i=2\\0&\text{ for }& i=3,\end{cases}$$ $$P(X=i\mid G_2\ )=\frac{P(X=i \cap G_2 )}{P(G_2)}=\begin{cases}0&\text{ for }&i=1\\\frac14&\text{ for }&i=2\\\frac34&\text{ if }&i=3.\end{cases}$$ So

$$E[\ X\mid G_1\ ]= 1\times P(X=1\mid G_1\ )+2\times P(X=2\mid G_1)+3\times P(X=3\mid G_1\ )=$$ $$=\frac35+2\times \frac25=\frac75$$ and $$E[\ X\mid G_2\ ]= 1\times P(X=1\mid G_2\ )+2\times P(X=2\mid G_2)+3\times P(X=3\mid G_2\ )=$$ $$=2\times \frac14+3\times \frac34=\frac{11}4$$ then

$$E[X\mid \mathscr G\ ](\omega)=\begin{cases}\frac75&\text{ if }&\omega\in G_1\\\frac{11}4&\text{ if }&\omega\in G_2\end{cases}.$$

In order to get the conditional expectations $E[Y\mid \mathscr G\ ]$ , we need to calculate the following conditional probabilities:

$$P(Y=i\mid G_1\ )=\frac{P(Y=i \cap G_1 )}{P(G_1)}=\begin{cases}\frac25&\text{ for }&i=1\\\frac25&\text{ for }&i=2\\\frac15&\text{ for }& i=3,\end{cases}$$ $$P(Y=i\mid G_2\ )=\frac{P(Y=i \cap G_1 )}{P(G_2)}=\begin{cases}\frac14&\text{ for }&i=1\\\frac14&\text{ for }&i=2\\\frac12&\text{ if }&i=3.\end{cases}$$ So

$$E[\ Y\mid G_1\ ]= 1\times P(Y=1\mid G_1\ )+2\times P(Y=2\mid G_1)+3\times P(Y=3\mid G_1\ )=$$ $$=\frac25+2\times \frac25+3\times \frac15=\frac95$$ and $$E[\ Y\mid G_2\ ]= 1\times P(Y=1\mid G_2\ )+2\times P(Y=2\mid G_2)+3\times P(Y=3\mid G_2\ )=$$ $$=1\times \frac14+2\times \frac14+3\times \frac12=\frac{9}4$$ then

$$E[Y\mid \mathscr G\ ](\omega)=\begin{cases}\frac95&\text{ if }&\omega\in G_1\\\frac{9}4&\text{ if }&\omega\in G_2\end{cases}.$$


As far as the independence of $E[X\mid\mathscr G\ ]$ and $E[Y\mid\mathscr G\ ]$. Consider the following example

$$P\left(E[\ X\mid \mathscr G\ ]=\frac75\right)=P(G_1)=\frac59,\ \ P\left(E[\ Y\mid \mathscr G\ ]=\frac95\right)=P(G_1)=\frac59$$ and $$P\left(E[\ X\mid \mathscr G\ ]=\frac75 \cap E[\ Y\mid \mathscr G\ ]\right)=P(G_1)=\frac59$$ and not $\frac{25}{81}$.

$\endgroup$
  • $\begingroup$ Thanks. So, if $X$ and $Y$ were independent would we have independence of conditional expectations $E[X|\mathcal{G}]$ and $E[Y|\mathcal{G}]$? Or would we need $\mathcal{G}$ to satisfy some properties? $\endgroup$ – user423750 Feb 21 '18 at 16:01
  • $\begingroup$ I am going to edit my answer: I will include an example with independent $X$ and $Y$. $\endgroup$ – zoli Feb 21 '18 at 17:37
  • $\begingroup$ please check my calculations carefully because I was quite tired when I did al that.. $\endgroup$ – zoli Feb 21 '18 at 21:55

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.