Let $\Sigma _1,\Sigma _2\subset 2^\Omega$ be independent $\sigma$-algebras and suppose they both contain events for which $0<P(X)<1$.

Show that $\Sigma := \Sigma_1\cup\Sigma _2$ cannot be a $\sigma$-algebra.

Assume $\Sigma$ is a $\sigma$-algebra. Let $A_i\in \Sigma _1, j\in I$ be the events for which $0<P(A_i)<1$ and $B_i\in\Sigma _2,i\in J$ events for which $0<P(B_i)<1$.

Due to independence $0<P(A_i\cap B_j)<1$ so $A_i\cap B_j\in\{\emptyset, \Omega\}$ is impossible. Pick $(k,l)\in I\times J$, then $A_k\cap B_l\in \{A_i\}_{i\in I}$ or $A_k\cap B_l\in\{B_i\}_{i\in J}$. If $A_k\cap B_l = A_k$ (or $A_k^c$), then $$P(A_k)P(B_l) = A_k\Longrightarrow P(B_l)=1\quad\mbox{and}\quad A_k\cap B_l = A_k^c\Longrightarrow A_k\cap B_l =\emptyset, $$ which are contradictions.

What is the problem with $A_k\cap B_l = A_m$ (or $A_m^c$) ?

If we considered a simpler case e.g $$\Sigma _1 = \{\emptyset,\Omega, A,A^c\}\quad\mbox{and}\quad \Sigma _2 = \{\emptyset,\Omega, B,B^c,C,C^c,D,D^c\} $$ How would one exhibit a contradiction for $A\cap B = C$? (Assuming $\Sigma _1$ and $\Sigma _2$ are independent)

We don't, of course, have to only consider intersections, but since there is some talk about independence, then intersection seems the logical place to look for a contradiction.

In an attempt to play around with it: $$A\cap B= C\Longrightarrow A\cap (B\cap C) = C\Longrightarrow P(A)P(B\cap C) = P(C) \Longrightarrow P(B) = P(B\cap C) $$ Likewise $$ A\cap B = C\Longrightarrow A\cap B=B\cap C\Longrightarrow P(A)P(B) = P(B\cap C)$$ so we get $$P(A)P(B) = P(B) \Longrightarrow P(A)=1$$ which seems problematic. Turns out to be sufficient.

  • 1
    $\begingroup$ $P(A)=1$ does not mean that $A$ is certain. Also, $P(A)=0$ does not mean that $A$ is impossible. Think of a shape in $R^2$ over which we have a uniform distribution. Choosing randomly any of these point is possible even though their probability is $0$. $\endgroup$
    – zoli
    Oct 14 '17 at 18:17
  • $\begingroup$ @zoli Good catch! I changed the wording, but the result shouldn't change. $\endgroup$
    – AlvinL
    Oct 14 '17 at 18:19

If $A_k\cap B_l = A_m$, then $$A_k\cap B_l\cap A_m = A_m\Longrightarrow P(B_l)P(A_k\cap A_m) = P(A_m) \Longrightarrow P(B_l)P(A_k\cap A_m) = P(B_l)P(A_k).$$ Additionally, $$A_k\cap B_l = A_m\cap A_k\Longrightarrow P(B_l)P(A_k) = P(A_m\cap A_k). $$ Necessarely $P(A_k\cap A_m)>0$, implying $P(B_l) =1$, a contradiction.

Similar contradiction for $A_k\cap B_l = B_m$.


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.