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WolframAlpha: PDF(X+Y) where X and Y are iid Poisson(lambda)

Posted 9 years ago
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I'm trying to play around with two random variables and I can't get wolfram alpha to parse my input.

I've tried something really simple that like `PDF(X+Y) where X and Y are iid Poisson(lambda)` trying to get wolfram alpha to tell me that this is Poisson(2*lambda) or even just give me the joint cdf or pdf.

all I can get out is `Using closest Wolfram|Alpha interpretation: Poisson(lambda)` I really want the distribution of `X/(X+Y) where X and Y are iid Poisson(lambda)` but I figured I would start small.

any ideas what I could try to help it interpret my input?
2 Replies
thanks seth.

I've tried a few other ways too.  Unfortunately this has also been timing out on me.  (maybe if i restrict the domain to be x > 0 and y > 0 it would work)
P[ X/(X+Y) < f ]  --> exp(-2*lambda) * sum_{y=0}^{+inf} sum_{x=0}^{floor[ f/(1-f) * y ]} lambda^(x+y)/(x!*y!)
So, I have bad news.  Even if you could develop a syntax that would persuade WolframAlpha to try to compute the PDF of distribution of `X/(X+Y) where X and Y are iid Poisson(lambda)` , WA is going to time out on the computation.  I just tried running it on actual Mathematica for a few minutes on a reasonably powerful machine and got nowhere.  Also, the domain of the Poisson distribution includes 0.  So, don't we have a division by zero problem in your transformation? (which might be why Mathematica is unhappy when it tries to compute the PDF).
POSTED BY: Seth Chandler
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