I saw the following claim in some book without a proof and couldn't prove it myself.

$\dfrac{d}{dp}\mathbb{P}\left(\text{Bin}\left(n,\,p\right)\leq d\right)=-n\cdot\mathbb{P}\left(\text{Bin}\left(n-1,\,p\right)=d\right)$

So far I got:

$\begin{array}{l} \dfrac{d}{dp}\mathbb{P}\left(\text{Bin}\left(n,\,p\right)\leq d\right)=\\ \dfrac{d}{dp}\sum\limits _{i=0}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)p^{i}\left(1-p\right)^{n-i}=\\ -n\cdot\left(1-p\right)^{n-1}+\sum\limits _{i=1}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)\left[ip^{i-1}\left(1-p\right)^{n-i}-p^{i}\left(n-i\right)\left(1-p\right)^{n-i-1}\right] \end{array}$

But I am not very good playing with binomial coefficients and don't know how to proceed.

  • $\begingroup$ Did you write down $$\mathbb{P}\left(\text{Bin}\left(n,\,p\right)\leq d\right)$$ as a function of $p$? What did you get? $\endgroup$ – Did Apr 3 '17 at 17:41
  • $\begingroup$ Of course I did. $P\left(\text{Bin}\left(n,\,p\right)\leq d\right)=\sum\limits _{i=0}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)p^{i}\left(1-p\right)^{n-i}$ $\endgroup$ – user25640 Apr 3 '17 at 17:56
  • $\begingroup$ And the derivative? $\endgroup$ – Did Apr 3 '17 at 18:04
  • $\begingroup$ $-n\cdot\left(1-p\right)^{n-1}+\sum\limits _{i=1}^{d}\left(\begin{array}{c} n\\ i \end{array}\right)\left[ip^{i-1}\left(1-p\right)^{n-i}-p^{i}\left(n-i\right)\left(1-p\right)^{n-i-1}\right]$ $\endgroup$ – user25640 Apr 3 '17 at 18:13

Consider the derivative of the logarithm: $$ \frac{d}{dp} \left[\log \Pr[X = x \mid p]\right] = \frac{d}{dp}\left[x \log p + (n-x) \log (1-p)\right] = \frac{x}{p} - \frac{n-x}{1-p}, $$ hence $$\frac{d}{dp}\left[\Pr[X = x \mid p]\right] = \binom{n}{x} p^x (1-p)^{n-x} \left(\frac{x}{p} - \frac{n-x}{1-p}\right) $$ and $$\begin{align*} \frac{d}{dp}\left[\Pr[X \le x \mid p] \right] &= \sum_{k=0}^x \binom{n}{k} p^k (1-p)^{n-k} \left(\frac{k}{p} - \frac{n-k}{1-p}\right) \\ &= \sum_{k=0}^x \binom{n}{k} k p^{k-1} (1-p)^{n-k} - \binom{n}{k} (n-k) p^k (1-p)^{n-1-k}. \end{align*}$$ But observe that $$\binom{n}{k}(n-k) = \frac{n!}{k!(n-k-1)!} = \frac{(k+1) n!}{(k+1)!(n-(k+1))!} = (k+1)\binom{n}{k+1},$$ hence the second term can be written $$(k+1) \binom{n}{k+1} p^{(k+1)-1} (1-p^{n-(k+1)}),$$ which is the same as the first term except the index of summation has been shifted by $1$. Therefore, the sum is telescoping, leaving $$\frac{d}{dp}\left[\Pr[X \le x \mid p]\right] = 0 - \binom{n}{x} (n-x) p^x (1-p)^{n-1-x}.$$ All that remains is to observe $$\binom{n}{x}(n-x) = \frac{n!}{x!(n-x-1)!} = \frac{n(n-1)!}{x!(n-1-x)!} = n \binom{n-1}{x},$$ therefore $$\frac{d}{dp}\left[\Pr[X \le x \mid p] \right] = -n \Pr[X^* = x \mid p],$$ where $X^* \sim \operatorname{Binomial}(n-1,p)$, as claimed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.