Let $P$ be a probability measure on $\Omega$ with $|\Omega| \le\aleph_0$. Then there is no sequence $A_i$ consisting of independent events with $P(A_i)=p \; \forall i$.


Proof by contradiction:

Let $A_i$ be a sequence of independent events with $P(A_i)=p \; \forall i$, whereas $0<p<1$.

$P(\bigcup A_i)=\sum_{i=1}^{n=\aleph_0}P(A_i)=np=\infty$.

Now, because $\Omega = \bigcup A_i$, it should $P(\Omega)=P(\bigcup A_i)$ but $P(\Omega)=1 \neq P(\bigcup A_i)=\infty$

So there exists no such sequence $A_i$

  • 1
    $\begingroup$ Is the assumption that the $A_i$ are disjoint? Note that this is not the same as "independent"...indeed the two properties are more or less incompatible. $\endgroup$ – lulu Apr 25 '18 at 19:27
  • $\begingroup$ "stochastically independent" actually $\endgroup$ – verzweifeltzumquadrat Apr 25 '18 at 19:31
  • $\begingroup$ That's the same as "independent". But your calculation, stating that the probability of the union is the sum of the probabilities, requires that the events be mutually exclusive. $\endgroup$ – lulu Apr 25 '18 at 19:33
  • $\begingroup$ And independent doesn't implicate disjoint in general? Do you know of any (correct) alternative to this proof? Thanks! $\endgroup$ – verzweifeltzumquadrat Apr 25 '18 at 19:39
  • $\begingroup$ Never exasperate. To have an example how things may work in some "simple" space, which is not at most countable, let us consider the space $X=[0,1)$, where each element $x$ is written in binary representation. We consider the $\sigma$-algebra generated by the events $A_n$, where $A_n$ is the set of all $x\in X$ having the $n$.th decimal equal to zero. We give $A_n$ the probability $1/2$. Then we make all these $A$'s independent, by claiming, as it already happens, that an event having fixed digits on positions $k_1, \dots,k_r$ has probability $2^{-r}$. (Your argument ignores cardinality.) $\endgroup$ – dan_fulea Apr 25 '18 at 19:40

The key point: consider a point $\omega \in \Omega$ and let $\psi_{\omega}$ be the associated probability. for each $i$, $\omega$ is either in $A_i$ or its complement. so, if our sequence existed, we get $\psi_{\omega}$ less than or equal to an infinite product of $p's$ and $(1-p)'s$. Hence $\psi_{\omega}=0\;\forall \omega\in \Omega$. But by countable additivity that is impossible.

  • $\begingroup$ Thanks! How do we know that $\psi_{\omega}=0\;\forall \omega\in \Omega$? $\endgroup$ – verzweifeltzumquadrat Apr 25 '18 at 20:26
  • $\begingroup$ Consider the first $n$ events. Say $\omega$ is in exactly $k$ of them, so it is in the complement of exactly $n-k$. Applying independence to the probability of the intersection shows that $\psi_{\omega}≤ p^k\times (1-p)^{n-k}$. if we let $\rho=\min (p,1-p)$ that shows that $\psi_{\omega}≤\rho^n$. As this must hold $\forall n$ we see that $\psi_{\omega}=0$. $\endgroup$ – lulu Apr 25 '18 at 20:43
  • $\begingroup$ Correction: that should have read $\rho=\max (p,1-p)$ not min. $\endgroup$ – lulu Apr 25 '18 at 20:59

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