Let $X_1, X_2, X_3$ be mutually independent random variables from $(\Omega,\mathcal{F},P)$ to $(\mathbb{R},\mathcal{B})$. Show that $X_3$ is independent of $\sigma(X_1,X_2)$, i.e., for all $B \in \mathcal{B}$ and $A \in \sigma(X_1,X_2)$, $P(X_3^{-1}(B) \cap A) = P[X_3^{-1}(B)]P[A]$.

My attempt:
We know from the mutual independence of $X_1,X_2,X_3$ that for all $B_1,B_2,B_3 \in \mathcal{B}$, $$ P[X_1^{-1}(B_1) \cap X_2^{-1}(B_2) \cap X_3^{-1}(B_3)] = P[X_1^{-1}(B_1)]P[X_2^{-1}(B_2)]P[X_3^{-1}(B_3)] $$

Take an arbitrary $B_3 \in \mathcal{B}$ and $A \in \sigma(X_1,X_2)$. Now, $\sigma(X_1,X_2) = \sigma(\sigma(X_1) \cup \sigma(X_2))$.

I have worked out three cases.

First case: $A \in \sigma(X_1) \cup \sigma(X_2)$.
Since $X_3$ and $X_1$ (or $X_2$) are independent, $P[X_3^{-1}(B_3) \cap A] = P[X_3^{-1}(B_3)]P[A]$, and we are done.

Second case: $A =C_1 \cap C_2$ for some $C_1 \in \sigma(X_1)$ and $C_2 \in \sigma(X_2)$.
$C_1=X_1^{-1}(B_1)$ for some $B_1 \in \mathcal{B}$ and $C_1=X_1^{-1}(B_2)$ for some $B_2 \in \mathcal{B}$. Hence, by independence of $X_1,X_2,X_3$, $$P[X_3^{-1}(B_3) \cap (C_1 \cap C_2)] = P[X_3^{-1}(B_3)]Pr[C_1]Pr[C_2] = P[X_3^{-1}(B_3)]Pr[C_1 \cap C_2]$$ and we are done.

Third case: $A =C_1 \cup C_2$ for some $C_1 \in \sigma(X_1)$ and $C_2 \in \sigma(X_2)$.
We can show that $A^c$ and $X_3^{-1}(B_3)$ are independent by case 2. Hence, $A$ and $X_3^{-1}(B_3)$ are independent. $\tag*{$\blacksquare$}$

How do we proceed for a general set $A$ in $\sigma(X_1,X_2)$?


1 Answer 1


You can prove this using the Dynkin's $\pi-\lambda$ Theorem. Consider $\{A \in \sigma (X_1,X_2): P(X_3^{-1}(B) \cap A) = P[X_3^{-1}(B)]P[A]$. By case 2) you know that this class contains $C_1\cap C_2$ whenever $C_1 \in \sigma (X_1)$ and $C_2 \in \sigma (X_2)$. Now verify that this class is a $\lambda-$ system. Next verify that $\sigma (X_1,X_2)$ is the $\sigma-$ field generated by sets of the type $C_\cap C_2$ where $C_1 \in \sigma (X_1)$ and $C_2 \in \sigma (X_2)$. These sets form a $\pi-$system. By the $\pi-\lambda$ Theorem it follows that our class must contain all sets in $\sigma (X_1,X_2)$.

  • $\begingroup$ Thank you for the answer. I couldn't prove that the set you defined (let's call it $S$) is a $\sigma$-algebra. Let $E_1, E_2 \in S$. How do we show that $E_1 \cap E_2 \in S$? $\endgroup$ Commented Aug 21, 2022 at 9:01
  • 1
    $\begingroup$ @MathewsBoban Actually, you cannot prove directly that it is a $\sigma-$algebra. I have corrected my proof by switching over to $\pi-\lambda$ Theorem: en.wikipedia.org/wiki/Dynkin_system $\endgroup$ Commented Aug 21, 2022 at 9:25

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