Find asymptotical variance.

Question

Consider $X_{i}$ are independent and have exponential distribution $exp(a)$.

Find asymptotical variance of $(k! / \bar{X^{k}})^{1/k}$, where $\bar{X} = n^{-1} \sum X_{i}$

As for me it looks like using Stirling approximation. But I don't know can we use it or not?

Answer 1

If $k>0$, then the random variable of interest $(k!/\bar{X}^k)^{1/k}$ simplifies to $k!^{1/k}/\bar{X}$. The general form for the exact variance is $\frac{a^2 n^2 (k!)^{2/k}}{(n-1)^3-(n-1)^2} = \frac{a^2 n^2 (k!)^{2/k}}{(n-2) (n-1)^2}$ when $n>2$. I've taken the lazy/efficient way below using Mathematica.

var[a_, n_] := Module[{mean, jointDistribution, reciprocalMean},
  mean = Sum[x[i], {i, n}]/n;
  jointDistribution = 
   Table[x[i] \[Distributed] ExponentialDistribution[a], {i, n}];
  reciprocalMean = 
   TransformedDistribution[k!^(1/k)/mean, jointDistribution];
  Variance[reciprocalMean]]

var[a, 3]
(* 9/4 a^2 (k!)^(2/k) *)
var[a, 4]
(* 8/9 a^2 (k!)^(2/k) *)
var[a, 5]
(* 25/48 a^2 (k!)^(2/k) *)
var[a, 6]
(* 9/25 a^2 (k!)^(2/k) *)
var[a, 7]
(* 49/180 a^2 (k!)^(2/k) *)