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Re: computing expectations for probability distributions in mathematica
- To: mathgroup at smc.vnet.net
- Subject: [mg105297] Re: computing expectations for probability distributions in mathematica
- From: dh <dh at metrohm.com>
- Date: Thu, 26 Nov 2009 06:15:29 -0500 (EST)
- References: <hekuoc$9an$1@smc.vnet.net>
per wrote:
> 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)).
>
Hi,
you can split your sum into 3 sums:
x=0; y=1..n
y=0; x=1..n;
x=1..n; y=1..n
However, are you sire that you can simply eliminate the case x=0 and
y=0? Otherwise the expectation value does not exist.
Daniel
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