In a probability space $\big( \Omega, \mathcal{F}, P \big)$, suppose $\{E_n\}_{n\in \mathbb{N}} \subseteq \mathcal{F}$ is a sequence of mutually independent events. By Borel-Cantelli Lemma, the sequence $\big\{ P(E_n) \big\}_{n\in \mathbb{N}}$ is summable if and only if $P\big( \limsup_n E_n \big)=0$. Now for each $n\in\mathbb{N}$ define $F_n = \bigcup_{k\geq n} E_k$. Then observe that $P\big( E_n\,\vert\, F_n \big) = \dfrac{P(E_n)}{P(F_n)}$. In the case where $\big\{ P(E_n) \big\}_{n\in \mathbb{N}}$ is summable, is there a condition under which we can tell $\big\{ P(E_n\,\vert\, F_n) \big\}_{n\in \mathbb{N}}$ is summable? How about the other direction?

Intuitively, infinitely many events from $\{E_n\}_{n\in \mathbb{N}}$ occur if and only if infinitely many events from $\{F_n\}_{n\in \mathbb{N}}$ occur. However, in the case where $P\big( \limsup_nE_n \big)=0$, $F_n$, the condition in the event $\big( E_n\,\vert\, F_n \big)$ is almost impossible to happen but I cannot tell if $\big\{ P(E_n\,\vert\,F_n) \big\}_{n\in \mathbb{N}}$ is summable by using the Borel-Cantelli Lemma on the calculations. Any hints will be appreciated.

  • $\begingroup$ $P(E_n|F_n)$ does not equal $P(E_n)/P(F_n)$, instead, it equals $P(E_n, F_n)/P(F_n)$. $\endgroup$
    – jbowman
    Apr 8 at 21:27
  • $\begingroup$ As $E_n\subseteq F_n$, I have $E_n\cap F_n = E_n$. By $P\big( E_n\,,\, F_n \big)$ I suppose you mean $P\big( E_n\cap F_n \big)$? $\endgroup$ Apr 8 at 22:27
  • $\begingroup$ My mistake, I didn't read carefully enough :( $\endgroup$
    – jbowman
    Apr 8 at 22:44
  • $\begingroup$ No problems :). $\endgroup$ Apr 8 at 23:12
  • $\begingroup$ @SextusEmpiricus Since $F_n$ is measurable, $P(F_n)$ is well-defined regardless. I simply wondered if there is a "Borel-Cantelli Lemma"-like result that is for conditional probability. On the other hand, in a more general context, if every tail event either happens almost surely or almost never happens, I wonder how fast the probability of an event that is eventually impossible decreases. $\endgroup$ Apr 12 at 23:50

1 Answer 1


Here is a case where $P(E_n)$ is summable but $P(E_n|F_n)$ is not.

Say that $P(F_n) = 2^{-n}$.

For independent events $E_n$ we have

$$P(E_n) = \frac{P(F_n) - P(F_{n+1})}{ 1-P(F_{n+1})}$$

then $P(E_n) = \frac{2^{-(n+1)}}{1-2^{-(n+1)}}$ is summable

$$\sum_{n=1}^\infty P(E_n) = \sum_{n=1}^\infty \frac{1}{2^{n+1}-1} < \sum_{n=1}^\infty \frac{1}{2^{n}} = 1$$

But $P(E_n|F_n)$ has a non-zero lower limit and is not summable

$$P(E_n|F_n) = P(E_n)/P(F_n) = \frac{1/2}{1-2^{-(n+1)}} > \frac{1}{2} $$

To investigate the above example further, possibly we can use:

If we call $P(F_n) = f(n)$, then

$$P(E_n|F_n) = \frac{f(n)-f(n+1)}{f(n)(1-f(n+1))} \approx \frac{-f'(n)}{f(n)(1-f(n))}\to \frac{-f'(n)}{f(n)} = - \frac{\partial}{\partial n} \log f(n)$$

the above is very handwavy, but we could say that $f(n)$ can not be bounded from below by an exponential law.

  • $\begingroup$ Thanks for your nice example! I am surprised to see that in your example $P\big( E_n\,\vert\,F_n \big)$ actually decreases to $\dfrac{1}{2}$. Based on your answer, if I put $P\big( F_n \big) = \lambda^{-n}$ for any $\lambda\in (0, 1)$ (and then start with a large enough $n$), then $P\big( E_n\,\vert\, F_n \big)$ can converges to any number in $(0, 1)$. $\endgroup$ Apr 14 at 0:23
  • $\begingroup$ In the second part of your answer, may I ask how to show that $f(n)-f(n+1)$ is approximately equal to $f'(n)$? Although $f(n)\longrightarrow 0$ as $n$ approaches to zero, by my observation above, $P\big( E_n \,\vert\, F_n \big)$ can only possibly and eventually stay in $(0, 1)$. $\endgroup$ Apr 14 at 0:29
  • $\begingroup$ @SanaeKochiya It is a Taylor series approximation. I often use this approximation with a derivative when there is an difference between points to gain some intuition about the problem. I am actually not very sure whether it works well with these exponential laws and you do get errors. If we apply the example then we get $\log(2) \approx 0.301$ whereas the exact computation is $0.5$. $\endgroup$ Apr 14 at 6:48

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