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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.

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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 $$ \sup\limits_{B\in\mathcal F}\,|\mathbb P(B\mid A)-\mathbb P(B)|=1-\mathbb P(A). $$

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