0
$\begingroup$

$X,Y_1,Y_2$ are random variables. Suppose $X$ and $Y_1$ are independent, and $X$ and $Y_2$ are independent. Then by definition we have: $$Pr[X \leq x, Y_1\leq y_1] = Pr[X \leq x]Pr[Y_1\leq y_1],$$ $$Pr[X \leq x, Y_2\leq y_2] = Pr[X \leq x]Pr[Y_2\leq y_2].$$ I was wondering if the following equation holds: $$Pr[X \leq x, Y_1\leq y_1, Y_2\leq y_2] = Pr[X \leq x]Pr[Y_1\leq y_1, Y_2\leq y_2].$$

Actually, what I want to show is that if $\sigma$-algebras $\sigma\{X\}$ and $\sigma\{Y_1\}$ are independent, and $\sigma\{X\}$ and $\sigma\{Y_2\}$ are independent, then $\sigma\{X\}$ and $\sigma\{Y_1,Y_2\}$ are independent (use $\pi-\lambda$ theorem).


It is useful to calculate conditional expectations of joint Gaussian random variables $\mathbf{E}[X|Y_1,...,Y_n]$. Here $X$ and $Y_1$ to $Y_n$ are joint Gaussian.

Set $\mathbf{E}[X|Y_1,...,Y_n] := Z = a + b_1Y_1+...+b_nY_n$. Set $\mathbf{E}[X-Z]=0$ and $\mathbf{E}[(X-Z)Y_i]=0$ to solve $a$ and $b_1$ to $b_n$. Then $X-Z$ and $Y_i$ are independent. Therefore(?) $\sigma\{X-Z\}$ and $\sigma\{Y_1,...,Y_n\}$ are independent. We have $\mathbf{E}[X|Y_1,...,Y_n] = Z$.

$\endgroup$
  • $\begingroup$ No. The result fails to hold, already for two-valued random variables. $\endgroup$ – Did Nov 7 '15 at 17:46
  • $\begingroup$ @Did Opps, what if $X, Y_1, Y_2$ are joint Gaussian? $\endgroup$ – Stupid_Guy Nov 7 '15 at 17:49
  • $\begingroup$ Completely different question. Then yes. $\endgroup$ – Did Nov 7 '15 at 17:56
  • $\begingroup$ @Did Opps, what makes the difference? Should I open a new question? Do you think the calculation of conditional expectation for joint Gaussian is correct? Thx so much! $\endgroup$ – Stupid_Guy Nov 7 '15 at 17:57
1
$\begingroup$

Example: Toss two fair dice. Let $Y_i$ be the number showing on die $i$, $i=1,2$. Let $X$ be $1$ or $0$ depending on whether the sum of the two dice is even or odd. Then $Y_1$ is independent of $X$, $Y_2$ is independent of $X$, but $X$ is not independent of the pair $(Y_1,Y_2)$.

For joint Gaussian: What has to be true about the covariance matrix of $(X,Y_1,Y_2)$ for $X$ to be independent of $(Y_1,Y_2)$?

$\endgroup$
  • $\begingroup$ For joint Gaussian, the co-variance matrix of $(X,Y_1,Y_2)$ is $diag(v,B)$ where $v$ is the variance of $X$ and $B$ is the co-variance matrix of $(Y_1,Y_2)$. Sorry I still can't figure it out. Could you please be more clear why it is true for joint Gaussian? $\endgroup$ – Stupid_Guy Nov 7 '15 at 18:03
  • $\begingroup$ This is assuming $X$ is independent of $Y_1$ and $X$ is independent of $Y_2$, right? What does this block-diagonal form of the covariance matrix tell you about the joint characteristic function of $(X,Y_1,Y_2)$? (Or, more tediously, about their joint density function?) $\endgroup$ – John Dawkins Nov 7 '15 at 18:07
  • $\begingroup$ Opps, got it! We could write the joint density out as the product form. Thx so much!:) $\endgroup$ – Stupid_Guy Nov 7 '15 at 18:14

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.