Let (\(\displaystyle A_n\)) and (\(\displaystyle B_n\)) be in A with \(\displaystyle A_n\)-->A and \(\displaystyle B_n\)-->B, with P(B)>0 and P(\(\displaystyle B_n\))>0. Show that:
a) \(\displaystyle lim_{n->\infty}P(A_n|B)\)=P(A|B)
b) \(\displaystyle lim_{n->\infty}P(A|B_n)\)=P(A|B)
c) \(\displaystyle lim_{n->\infty}P(A_n|B_n)\)=P(A|B)
So, I know that:
\(\displaystyle P(A_n|B)\)=\(\displaystyle \frac{P(B|A_n)P(A_n)}{P(B)}\)=\(\displaystyle \frac{P(B \cap A_n)}{P(B)}\)
\(\displaystyle P(A|B_n)\)=\(\displaystyle \frac{P(B_n|A)P(A)}{P(B_n)}\)=\(\displaystyle \frac{P(B_n \cap A)}{P(B_n)}\)
\(\displaystyle P(A_n|B_n)\)=\(\displaystyle \frac{P(B_n|A_n)P(A_n)}{P(B_n)}\)=\(\displaystyle \frac{P(B_n \cap A_n)}{P(B_n)}\)
\(\displaystyle P(A|B)\)=\(\displaystyle \frac{P(B|A)P(A)}{P(B)}\)=\(\displaystyle \frac{P(B \cap A)}{P(B)}\)
My problem really is how to show the limit of \(\displaystyle P(B|A_n)\), \(\displaystyle P(B_n|A)\), and \(\displaystyle P(B_n|A_n)\) is\(\displaystyle P(B|A)\). They seem obvious, but I can't figure out how to show it. I've been working on this problem for 2 weeks with no success. I know that the limits of \(\displaystyle P(A_n)\) amd \(\displaystyle P(B_n)\) are A and B, respectively. But the part in Bayes formula that has another conditional probability is throwing me off. Any ideas? Thank you!
a) \(\displaystyle lim_{n->\infty}P(A_n|B)\)=P(A|B)
b) \(\displaystyle lim_{n->\infty}P(A|B_n)\)=P(A|B)
c) \(\displaystyle lim_{n->\infty}P(A_n|B_n)\)=P(A|B)
So, I know that:
\(\displaystyle P(A_n|B)\)=\(\displaystyle \frac{P(B|A_n)P(A_n)}{P(B)}\)=\(\displaystyle \frac{P(B \cap A_n)}{P(B)}\)
\(\displaystyle P(A|B_n)\)=\(\displaystyle \frac{P(B_n|A)P(A)}{P(B_n)}\)=\(\displaystyle \frac{P(B_n \cap A)}{P(B_n)}\)
\(\displaystyle P(A_n|B_n)\)=\(\displaystyle \frac{P(B_n|A_n)P(A_n)}{P(B_n)}\)=\(\displaystyle \frac{P(B_n \cap A_n)}{P(B_n)}\)
\(\displaystyle P(A|B)\)=\(\displaystyle \frac{P(B|A)P(A)}{P(B)}\)=\(\displaystyle \frac{P(B \cap A)}{P(B)}\)
My problem really is how to show the limit of \(\displaystyle P(B|A_n)\), \(\displaystyle P(B_n|A)\), and \(\displaystyle P(B_n|A_n)\) is\(\displaystyle P(B|A)\). They seem obvious, but I can't figure out how to show it. I've been working on this problem for 2 weeks with no success. I know that the limits of \(\displaystyle P(A_n)\) amd \(\displaystyle P(B_n)\) are A and B, respectively. But the part in Bayes formula that has another conditional probability is throwing me off. Any ideas? Thank you!