Bayes Theorem Limits

**azdang** · September 15th 2009, 10:17 AM

Let ( $A_n$ ) and ( $B_n$ ) be in A with $A_n$ -->A and $B_n$ -->B, with P(B)>0 and P( $B_n$ )>0. Show that:

a) $lim_{n->\infty}P(A_n|B)$ =P(A|B)

b) $lim_{n->\infty}P(A|B_n)$ =P(A|B)

c) $lim_{n->\infty}P(A_n|B_n)$ =P(A|B)

So, I know that:

$P(A_n|B)$ = $\frac{P(B|A_n)P(A_n)}{P(B)}$ = $\frac{P(B \cap A_n)}{P(B)}$

$P(A|B_n)$ = $\frac{P(B_n|A)P(A)}{P(B_n)}$ = $\frac{P(B_n \cap A)}{P(B_n)}$

$P(A_n|B_n)$ = $\frac{P(B_n|A_n)P(A_n)}{P(B_n)}$ = $\frac{P(B_n \cap A_n)}{P(B_n)}$

$P(A|B)$ = $\frac{P(B|A)P(A)}{P(B)}$ = $\frac{P(B \cap A)}{P(B)}$

My problem really is how to show the limit of $P(B|A_n)$ , $P(B_n|A)$ , and $P(B_n|A_n)$ is $P(B|A)$ . They seem obvious, but I can't figure out how to show it.

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