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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.

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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 $0.249999966472962$).

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  • $\begingroup$ 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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