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Context:

The present question is similar to [1]. But whereas in [1] there is a single constant, $p$, raised to a power of $x$, here we have the product of $(p)^{k}\,(1-p)^{n - k}$. I have looked in books of special functions to generalize the solution found in [1] for said purposes. I have sought a solution in terms of (i) hypergeometric functions or (ii) Legendre polynomials. I have not found any generalization. Thus, my path towards a solution is halted.

Is this a tractable problem?

What is a fruitful approach?

Introductory Formalism:

Allow a binomial distribution as in [2].

Provide that its support is

${\displaystyle k\in \{0,\dots ,n\}}$, and

$p ∈ [0,1]$; and

the probability mass function, g, is

$g\left(k\right) = {\displaystyle {\frac {n!}{k! (1-k)!}} \, p^{k}\,(1-p)^{1-k}} $.

Writing the expected value as $E(\cdot)$, we have

$E(a) = a$, for constant $a$;

$E(k) = n\,p$; and

$E(a^{k}) = (1 - p + a\,p)^n$; for constant $a$ (as per [3]).

Question:

What is the expected value of the probability mass function:

$E(g) = \sum_{k=0}^{n} \textstyle \left[{n \choose k}\,p^{k}(1-p)^{n-k}\right]^2 =$ ?

Bibliography:

[1] Squared binomial coefficient

[2] https://en.wikipedia.org/wiki/Binomial_distribution

[3] Covariance of product of two functions of two binomial distributions

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Michael Levy
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    If you can apply the result from source [1], then you just need to note that your expectation can be experssed as $\displaystyle E[g(K)] = (1 - p)^{2n}\sum_{k=0}^n \binom {n} {k}^2 \left[\left(\frac {p} {1 - p}\right)^2\right]^k$ which has the same form from that source. – BGM Oct 23 '16 at 06:40

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