Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(\Omega,\mathcal{F},P)$ be a probability space. If $A\in\cal F$ is an event with $P(A)=1$, then $$ P_{\mid A}(B)=P(B\mid A)=\frac{P(B\cap A)}{P(A)}=P(B),\quad B\in\cal F. $$ I wonder if something can be said about how "close" $P_{\mid A}$ and $P$ are, when $A\in\cal F$ is an event with probability close to $1$ and also what "close" should mean.
For example, if $P(A)=p$ and let's say that $p=0.99$, can we give a non-trivial upper bound on the maximal distance $$ \sup_{B\in\cal F}|P_{\mid A}(B)-P(B)| $$ in terms of $p$? And could other types of distances be interesting?

This is just me thinking, so anything you can add will be appreciated. Thanks.

share|cite|improve this question
up vote 4 down vote accepted

For every $B$, $\mathbb P(B\mid A)-\mathbb P(B)=b(1-a)/a-c$ with $a=\mathbb P(A)$, $b=\mathbb P(B\cap A)$ and $c=\mathbb P(B\setminus A)$. Since $0\leqslant b\leqslant a$ and $0\leqslant c\leqslant 1-a$, $$ -(1-a)\leqslant -c\leqslant \mathbb P(B\mid A)-\mathbb P(B)\leqslant b(1-a)/a\leqslant 1-a. $$ The bound $1-a$ is achieved for $B=A$, hence $$ \sup\limits_{B\in\mathcal F}\,|\mathbb P(B\mid A)-\mathbb P(B)|=1-\mathbb P(A). $$

share|cite|improve this answer
That's perfect. Exactly the kind of bound I was hoping for. Thanks. – Stefan Hansen Dec 14 '12 at 15:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.