A recent statistical question asked and answered on CrossValidated derived a finite sum expression for the variance of the maximum-likelihood estimator for IID data from the Laplace distribution. The sum of interest that arose in that question is:

$$S(k) \equiv \frac{(2k+1)!}{2^k k! k!} \sum_{i=0}^k {k \choose i} \frac{(-1 /2)^{k-i} }{(2k-i+1)^3} \quad \quad \quad \text{for } k \in \mathbb{N}.$$

This is a complicated expression and I lack the mathematical skills to simplify it. I have generated this sequence of values for $k = 1, 2, 3, ..., 30$ and I have observed that the pattern appears roughly consistent with statistical intuition, but I cannot establish key properties that should hold for this sequence.

For the purposes of checking the correctness of that statistical result, and establishing better intuition about that answer, I would like to establish the following properties of this function (these are properties the function should have based on statistical reasoning relating to the problem where it was derived):

  1. Strictly positive: We should have $S(k) > 0$ for all $k \in \mathbb{N}$.

  2. Quasi-monotonicity: The sequence $S(k)$ should generally get smaller as $k$ gets larger. Early values suggest that it oscillates with bivariate frequency, but $S(k+2) < S(k)$ for all $k \in \mathbb{N}$.

  3. Asymptotic form: It should be possible to establish a simpler approximating expression for this function when $k$ is large. Based on some statistical reasoning and generated values, it is reasonable to conjucture that $S(k) \rightarrow \text{const} /k$ as $k \rightarrow \infty$, but this could be wrong. If this asymptotic form is correct, we would also like to find the limiting constant in this expression, and determine if it is some well-known mathematical constant.

  4. Simplified form? I am not all that familiar with sums of this form, so perhaps I am missing something that allows the whole expression to be simplified.

So my question is whether or not these properties hold, and if so, how you prove them. I have tried applying Stirling's approximation to simplify the expression but have not succeeded in getting any simpler results. Establishing any of the above results would be appreciated.

Update: Investigation of the asymptotic variance of the median in the related statistical question showed that there was an error in the variance expression, which omitted a factor of two; this error has not been corrected. As a result, the sum above still relates to that question, but the variance (with unit scale) is now equal to twice this sum, rather than being equal to this sum.


This is not an answer since based on numerical simulations.

As one could expect, $S(k)$ can be represented by an hypergeometric function. Using a CAS (and simplifying the nasty results) $$S(k)=(-1)^k \frac{(2k)!}{4^k(2k+1)^2(k!)^2}\, _4F_3(-2 k-1,-2 k-1,-2 k-1,-k;-2 k,-2 k,-2 k;2)$$

Using it numerically up to $k=10^7$, it seems that all your observations are correct.

Now, computing $k \,S(k)$ for large values of $k$ $$\left( \begin{array}{cc} k & k\, S(k) \\ 10^1 & 0.323227636532198 \\ 10^2 & 0.277479222683463 \\ 10^3 & 0.258876273294312\\ 10^4 & 0.252817410574130 \\ 10^5 & 0.250891736539050\\ 10^6 & 0.250282063130287\\ 10^7 & 0.250089203067797 \end{array} \right)$$ seems to show a limit around $\frac 14$ (based on the last values, Aitken acceleration would give for $k=10^8$ a value of $0.249999966472962$).

  • $\begingroup$ Despite being based on numerical simulations, this is very helpful. Thanks! $\endgroup$ – Ben - Reinstate Monica Apr 4 '18 at 7:22
  • $\begingroup$ @Ben. You are very welcome ! Glad to be of some help. $\endgroup$ – Claude Leibovici Apr 4 '18 at 7:24
  • $\begingroup$ Thanks again for the update. Looks like it is converging as expected. Interesting. $\endgroup$ – Ben - Reinstate Monica Apr 6 '18 at 5:48
  • $\begingroup$ @Ben. Be sure that the computation for $k=10^7$ took a looong time ! Cheers. $\endgroup$ – Claude Leibovici Apr 6 '18 at 6:02
  • $\begingroup$ I just found some papers that talk about the asymptotic variance of the sample median, and it looks like this convergence result is a general result that holds in a wide class of distributions. The $\tfrac{1}{4}$ constant appears to be a standard result. Excellent to have numerical confirmation of this for a complex example! $\endgroup$ – Ben - Reinstate Monica Apr 6 '18 at 6:10

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