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I have to show that

$p(k+1) = \frac{n-k}{k+1}p(k)$

So far, I have done some manipulation.

$p(k+1) = {n \choose k+1}p^{k+1}(1-p)^{n-(k+1)}$

$p(k+1) = \frac{n!}{(k+1)!(n-(k+1))!}p^{k+1}(1-p)^{n-(k+1)}$

$p(k+1) = \frac{n!}{(k+1)!(n-k-1))!}pp^k(1-p)^{n-k-1}$

$p(k+1) = \frac{n!}{(k+1)(k)!\frac{(n-k)!}{n-k}}pp^k\frac{(1-p)^{n-k}}{1-p}$

$p(k+1) = \frac{n-k}{k+1} \cdot \frac{n!}{(k)!(n-k)!}pp^k\frac{(1-p)^{n-k}}{1-p}$

It's close, but I seem to have an extra $p$ at the top, and an extra $1-p$ at the bottom. What am I missing?

fossdeep
  • 239

1 Answers1

4

If $X \sim \operatorname{Binomial}(n,p)$, then $$\Pr[X = k] = \binom{n}{k} p^k (1-p)^{n-k}, \quad k = 0, 1, \ldots, n,$$ and $$\Pr[X = k+1] = \binom{n}{k+1} p^{k+1} (1-p)^{n-k-1}, \quad k = -1, 0, 1, \ldots, n-1.$$ Consequently, $$\frac{\Pr[X = k+1]}{\Pr[X = k]} = \frac{\binom{n}{k+1}}{\binom{n}{k}} \frac{p}{1-p} = \frac{n!}{(k+1)!(n-k-1)!} \cdot \frac{k!(n-k)!}{n!} \frac{p}{1-p} = \frac{n-k}{k+1} \frac{p}{1-p},$$ for $k = 0, 2, \ldots, n-1$. The only case where the term $p/(1-p)$ drops out is if $p = 1/2$.

heropup
  • 135,869
  • I don't understand, if I rearrange my last line, I will get $p(k+1) = \frac{n-k}{k+1} \cdot \frac{n!}{(k)!(n-k)!}p^k(1-p)^{n-k} \cdot \frac{p}{1-p} = \frac{n-k}{k+1}p(k) \cdot \frac{p}{1-p}$ This is the same as $p(k+1) = \frac{n-k}{k+1}p(k)$? – fossdeep Oct 23 '15 at 21:14
  • What my answer shows is that the relationship you derived, $$\Pr[X = k+1] = \frac{n-k}{k+1} \frac{p}{1-p} \Pr[X = k],$$ is correct. You need to have the factor of $p/(p-1)$. – heropup Oct 23 '15 at 21:49
  • You're right. Thank you. – fossdeep Oct 23 '15 at 23:17