How to compute the expectation of $\frac{Y^L}{Y^L + (N-Y)^L}$ where $Y$ is Binomial(n,p)? If it is not exactly computable, then are their ways to approximate this qty?
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1$\begingroup$ For $L=2, N=n$, Mathematica gives an exact answer with the incomplete $\beta$ function: $-\frac{1}{2} i (1-p)^n \left(\frac{p}{p-1}\right)^{\left(\frac{1}{2}-\frac{i}{2}\right) n} \left((n-1) \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n-1\right)-n \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n\right)-(n-1) B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n-1\right)+n B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n\right)\right)$ $\endgroup$– user44143Commented Apr 21, 2014 at 15:39
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$\begingroup$ Mathematica also produces results for $N\neq n$ and larger values of $L$, albeit more complicated ones. $\endgroup$– EckhardCommented Apr 21, 2014 at 16:05
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$\begingroup$ When Mathematica produces answers like this, it often seems to be a restatement of the problem, and not really progress. Does this expression help if you want to approximate the value? $\endgroup$– Douglas ZareCommented Apr 21, 2014 at 17:57
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$\begingroup$ @DouglasZare, it does help for some purposes. The answer from Maple is more readable and understandable, but for large $n$ (and $L=2$), this will give you the answer more quickly and with better precision. $\endgroup$– user44143Commented Apr 21, 2014 at 20:47
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1 Answer
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It is exactly calculated in Maple by $$with(Statistics): Y := RandomVariable(Binomial(n, p)): $$ $$ M := Mean(Y^L/(Y^L+(N-Y)^L))\,assuming\, N::posint, L::posint$$ which produces $$\sum _{{\it \_t0}=0}^{n}{\frac {{{\it \_t0}}^{L}{n\choose {\it \_t0}}{ p}^{{\it \_t0}} \left( 1-p \right) ^{n-{\it \_t0}}}{{{\it \_t0}}^{L}+ \left( N-{\it \_t0} \right) ^{L}}} $$
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6$\begingroup$ This just looks like the definition of expectation. How does this help? $\endgroup$ Commented Apr 21, 2014 at 21:18
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