Timeline for How to compute the expectation of $\frac{Y^L}{Y^L + (N-Y)^L}$ where Y is Binomial(n,p)
Current License: CC BY-SA 3.0
7 events
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Apr 21, 2014 at 20:47 | comment | added | user44143 | @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. | |
Apr 21, 2014 at 19:40 | answer | added | user64494 | timeline score: -5 | |
Apr 21, 2014 at 17:57 | comment | added | Douglas Zare | 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? | |
Apr 21, 2014 at 16:05 | comment | added | Eckhard | Mathematica also produces results for $N\neq n$ and larger values of $L$, albeit more complicated ones. | |
Apr 21, 2014 at 15:39 | comment | added | user44143 | For $L=2, N=n$, Mathematica gives an exact answer with the incomplete $\beta$ function: $-\frac{1}{2} i (1-p)^n \left(\frac{p}{p-1}\right)^{\left(\frac{1}{2}-\frac{i}{2}\right) n} \left((n-1) \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n-1\right)-n \left(\frac{p}{p-1}\right)^{i n} B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}+\frac{i}{2}\right) n,n\right)-(n-1) B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n-1\right)+n B_{\frac{p}{p-1}}\left(1-\left(\frac{1}{2}-\frac{i}{2}\right) n,n\right)\right)$ | |
Apr 21, 2014 at 15:31 | review | First posts | |||
Apr 21, 2014 at 15:47 | |||||
Apr 21, 2014 at 15:15 | history | asked | Rony | CC BY-SA 3.0 |