0
$\begingroup$

I am considering a sample of n iid random variables, distributed according to a law F. I am interested in deriving the CDF of the i-th order statistics, $F_i(t)=\sum_{k=i}^n \binom{n}{k}F(t)^k(1-F(t))^{n-k}$, with respect to the sample size, n. Still, I do not know how to deal with deriving a sum with respect to the number of its addends and how to derive the binomial coefficient with respect to n. Suggestions?

$\endgroup$
  • $\begingroup$ Derivation of binomial coefficient is not an easy task. If you are interested in finding the derivative by yourself consider taking the derivative of the very last formula at the end of this page: en.wikipedia.org/wiki/Binomial_coefficient $\endgroup$ – Max Apr 9 '18 at 18:09
0
$\begingroup$

We can first use that the derivative operator is linear and derive the term of the sum: $ \begin{align} \frac{\partial F_i(t,n)}{\partial n} & = \sum \limits_{k=i}^n \frac{\partial \binom{n}{k}F(t)^k(1-F(t))^{n-k} }{\partial n}\\ & = \sum \limits_{k=i}^n \binom{n}{k}F(t)^kln(1-F(t))(1-F(t))^{n-k} + \frac{\partial \binom{n}{k}}{\partial n} F(t)^k(1-F(t))^{n-k} \tag1 \end{align} $ On the other hand, in order to derive the binomial coefficient we can re write it:

$ \begin{align} n \choose k & =\frac {n!}{k!\,(n-k)!} \\ & = \frac {\Gamma (n+1)}{\Gamma (k+1)\,\Gamma (n-k+1)} \\ & = \exp(\ln \Gamma (n+1)-\ln \Gamma (k+1)-\ln \Gamma (n-k+1)) \tag2 \end{align} $

Let $\psi$ be the Digamma function defined like this:

$$\psi (x)={\frac {d}{dx}}\ln {\big (}\Gamma (x){\big )}={\frac {\Gamma '(x)}{\Gamma (x)}} \tag3$$

Then,

$$\frac{\partial \binom{n}{k}}{\partial n} = \binom{n}{k}(\psi(n+1)-\psi(-k+n+1)) \tag4$$

$\endgroup$

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.