Re: computing expectations for probability

*To*: mathgroup at smc.vnet.net*Subject*: [mg105300] Re: [mg105288] computing expectations for probability*From*: danl at wolfram.com*Date*: Fri, 27 Nov 2009 06:27:38 -0500 (EST)*References*: <200911260404.XAA09561@smc.vnet.net>

> hi all, > > i am trying to see if there is a simplified expression for a > particular random variable using mathematica. i have two independent > binomial random variables, X and Y, and i'd like to compute the > expectation of X/(X+Y). more formally, X is distributed Binomial(p1, > n) and Y is distributed Binomial(p2, n). note that p1 and p2 can be > different, but the n's are the same. i would like to compute E(X/(X > +Y)). since the ratio is undefined when X+Y = 0, i would like to > assume that X+Y >= 1. > > is there a way to compute this expectation and see if it has a > simplified analytic form subject to this constraint? > > i defined my function to compute the probability of this random > variable as follows: > > myvar[x_, y_] := > Binomial[n, x]*p1^x*(1 - p1)^(n - x)*Binomial[n, y]* > p2^y*(1 - p2)^(n - y)*(x/(x + y)) > > now i just want to take the expectation of this. since the value of X > ranges from 0 to N and the value of Y ranges from 0 to N, it should be > possible to simply take a sum over X from 0 to N and sum of Y from 0 > to Y to see what this expression evaluates to and whether it has a > simple analytic form. > > the trick is to take the sum with the constraint of X + Y >= 1 -- how > can i express this in Mathematica? > > i am also open to computing the expectation of X/(X+Y) by assuming > they are independent Poissons if that makes the math easier, i.e. > using poisson approximation to binomial. > > thank you for your help. > > > >> > where X, Y are independent but p1 and p2 could be different. I am >> > trying to compute the mean E(X/(X+Y)) but i am stuck. is there an easy >> > form for this mean? i looked it up in several probability books but >> > could not get it. >> >> > i am willing to make either a Poisson approximation or a Normal >> > approximation to thebinomialfor my application if i could compute >> > this mean. if i assume that X and Y are distributed Poisson, i know >> > that E(X+Y) has a nice form as a function of the rates of the two >> > poissons, but i still do not know how to get the distribution of X/(X >> > +Y) so that i can compute the expectation E(X/(X+Y)). The business about starting at 0 or 1 is probably not relevant. I'm not able to get Sum to do the thing analytically. Sum[j*PDF[BinomialDistribution[n, p1], j]* PDF[BinomialDistribution[n, p2], k]/(j + k), {j, n}, {k, n}, Assumptions -> {0 < p1 < 1, 0 < p2 < 1}] Sum[((1 - p1)^(-j + n) p1^j (1 - p2)^(-k + n) p2^ k Binomial[n, j] Binomial[n, k])/(j + k), {j, n}, {k, n}, Assumptions -> {0 < p1 < 1, 0 < p2 < 1}] Nor could I get an analytic approximation using Integrate. But I think the alteration below gives the Poisson approximation you have in mind. In[111]:= approx = Sum[j*PDF[PoissonDistribution[mu1], j]* PDF[PoissonDistribution[mu2], k]/(j + k), {j, Infinity}, {k, Infinity}] Out[111]= (E^(-mu1 - mu2) (-E^mu1 mu1 + E^(mu1 + mu2) mu1 + mu2 - E^mu1 mu2))/(mu1 + mu2) In[119]:= approx2 = Simplify[approx /. {mu1 -> n*p1, mu2 -> n*p2}] Out[119]= (E^(-n (p1 + p2)) (E^(n (p1 + p2)) p1 + p2 - E^(n p1) (p1 + p2)))/(p1 + p2) A quick sanity check is to plug in explicit values for the parameters and compare to explicit summing of the Binomial case. In[127]:= eval1 = approx2 /. {n -> 14, p1 -> 1/2, p2 -> 1/3} Out[127]= (6 (1/3 - (5 E^7)/6 + E^(35/3)/2))/(5 E^(35/3)) In[128]:= eval2 = With[{n = 14, p1 = 1/2, p2 = 1/3}, Sum[j*PDF[BinomialDistribution[n, p1], j]* PDF[BinomialDistribution[n, p2], k]/(j + k), {j, n}, {k, n}]] Out[128]= 1886578214434897919/3143703825373593600 In[129]:= N[{eval1, eval2}] Out[129]= {0.5906, 0.600113} With p1=1/10 and p2=4/5 I get for the numeric values {0.1111, 0.105788} which again seem respectably close to one another. Daniel Lichtblau Wolfram Research

**References**:**computing expectations for probability distributions in mathematica***From:*per <perfreem@gmail.com>