$X,Y_1,Y_2$ 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 $\mathbf{E}[X|Y_1,...,Y_n] = Z$.

  • $\begingroup$ 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

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 $(Y_1,Y_2)$?

  • $\begingroup$ 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

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