Deriving order statistics with respect to sample size

Question

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?

Answer 1

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,

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