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I have a question concerning random variables. Let us assume that we have two random variables $X$ and $Y$. Let's say $X$ is Poisson distributed with parameter $\lambda_1$, and $Y$ is Poisson distributed with parameter $\lambda_2$.

When you build the fracture from $X/Y$ and call this a random variable $Z$, how is this distributed and what is the mean? Is it $\lambda_1/\lambda_2$?

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  • $\begingroup$ I just happened to come across this when looking for references. Inference for Poisson ratio is quite straightforward, for frequentists (Nelson, 1970, "Confidence Intervals for the Ratio of Two Poisson Means and Poisson Predictor Intervals") and Bayesians alike (Lindley, 1965). No problem with zero denominators either! $\endgroup$ – Frank Tuyl Oct 23 '14 at 2:19
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    $\begingroup$ The original questioner, and others, may be interested to note that $X/(Y+1)$ has expectation value $(\lambda_1/\lambda_2) (1-e^{-\lambda_2})$. Depending on your application this may be of greater use than $X/Y$. For more details see my paper in the Journal of Analytical Atomic Spectrometry, 28, 52, called " Statistical bias in isotope ratios" w/ DOI:10.1039/C2JA10205F. $\endgroup$ – user78444 May 29 '15 at 21:36
  • $\begingroup$ This is a frequently encountered problem in Astronomy. The Bayesian solution was worked out by Park et al. (2006, Astrophysical Journal, v652, 610-628, Bayesian Estimation of Hardness Ratios: Modeling and Computations). They include background contamination in their treatment. $\endgroup$ – user78543 May 31 '15 at 13:13
  • $\begingroup$ From the abstract it is not obvious that they are dealing with the OP's question. How does this paper relate to the distribution of the ratio of two Poisson random variables? $\endgroup$ – Andy May 31 '15 at 13:31
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    $\begingroup$ ttu-ir.tdl.org/ttu-ir/bitstream/handle/2346/59954/… $\endgroup$ – kjetil b halvorsen Oct 2 '16 at 19:05
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I think you're going to have a problem with that. Because variable Y will have zero's, X/Y will have some undefined values such that you won't get a distribution.

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    $\begingroup$ +1 That's right. But (to head off possible confusion) the problem isn't just that $Y$ can equal $0$: it's that it can equal $0$ with positive probability. (For instance, a quotient of normals does have a distribution even though the denominator can equal $0$.) Thus, $X/Y$ is undefined with positive probability, making its mean (and any other moment) undefined as well. $\endgroup$ – whuber May 18 '11 at 20:06
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    $\begingroup$ +1, but in the literature on false discovery rates people have no problem with $X/Y$ where $X$ is the true positives and $Y$ is the total number of positives :-). It is always understood, by convention, that 0 out of 0 equals 0. $\endgroup$ – NRH May 18 '11 at 21:10
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    $\begingroup$ @Mark: It's probably better to ask that as a new question, and to get real specific about what you're trying to achieve. $\endgroup$ – bill_080 May 18 '11 at 23:05
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    $\begingroup$ @NRH In your case there is a strong dependence of $X$ on $Y$. That completely changes things, because now the probability of a positive:zero ratio is nil. $\endgroup$ – whuber May 19 '11 at 2:43
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    $\begingroup$ @whuber, that's of course right. Thanks for pointing that out. I was just thinking that maybe there was some unstated convention to make the problem meaningful. But from @MarkDollar's comment above it seems that that was not the case to begin with. $\endgroup$ – NRH May 19 '11 at 6:59
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By realizing that the ratio is in fact not a well defined measurable set, we redefine the ratio as a properly measurable set $$ \mathbb{P}\left[\frac{X}{Y} \leq r \right] := \mathbb{P}\left[X \leq r Y\right]\\ = \sum_{y = 0}^\infty \sum_{x=0}^{\left\lfloor ry \right\rfloor} \frac{\lambda_{2}^y }{y!}e^{-\lambda_2} \frac{\lambda_{1}^x }{x!}e^{-\lambda_1} $$ where the summation follows as long as $r > 0$, and $X$ and $Y$ are independent Poisson variables. The density follows from the Radon-Nykodym theorem.

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  • $\begingroup$ Suppose that $Y$ comes from a zero-truncated Poisson distribution. Would the answer then be: $\endgroup$ – Brash Equilibrium Sep 19 '13 at 10:50

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