I have some problems with finding this limit: $\lim_{n\to\infty}p^{n}\sum_{k \geqslant{n(p^{-1}-1)}}^{\infty}\binom{n+k-1}{n-1}(1-p)^{k}.$ I know that it is possible to find it with the help of the Central Limit Theorem. Can anyone help me? Thanks in advance!
Find the limit by using central limit theorem
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0Hint: If you can identify the probability mass function as that of a sum of $n$ i.i.d. random variables with mean $p^{-1}$, the question comes down to asking for the probability that the sum random variable exceeds its mean. – 2012-04-10
1 Answers
Imagine an infinite sequence of independent Bernoulli trials with probability $p$ of success on each trial. Let $X$ be the number of failures before the $n$ success. Then $ \Pr(X=k) = \binom{n+k-1}{n-1} p^n(1-p)^k. $ The expected number of failures before the $n$th success is $\operatorname{E}(X)= n(1-p)/p=n(p^{-1}-1)$. The variance of the number of failures before the $n$th success is $\operatorname{var}(X)=n(1-p)/p^2$, so the standard deviation is the square root of that.
$X$ is the sum of $n$ independent random variables each of which is the number of failures before the first success.
The central limit theorem therefore says $ W=\frac{X- \frac{n(1-p)}{p}}{\sqrt{\frac{n(1-p)}{p^2}}} \text{ converges in distribution to } N(0,1)\text{ as }n\to\infty. $
What you're looking for then is the probability that $X \ge \operatorname{E}(X)$, which is the same as $W\ge 0$.
In the limit, that is $\Pr(Z\ge 0) =1/2$ (where, as usual, $Z\sim N(0,1)$).
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0Thank you very much, your answer is very competent! – 2012-04-11