I am currently reading a paper and puzzling about a certain statement. I have a ratio



$D_{T-4}+...+D_{T}\sim\text{Poisson}(\alpha(H_{T-4}+...+H_{T}))$, $D_{T-3}+...+D_{T+1}\sim\text{Poisson}(\alpha(H_{T-3}+...+H_{T+1}))$,

and $\alpha$ denotes an unknown parameter (around 1), $D_{t}$ and $H_{t}$ denote the actual and the expected number of deaths during year t, respectively.

Later, the author states, that the "ratio follows a (scaled and translated) Poisson distribution which is well approximated by a normal distribution".

Correct me, if I'm wrong, but don't we have a problem with having two poisson distributions as a ratio (i.e. the sums of actual deaths), since $D_{T-4}+...+D_{T}$ may be zero with positive probability? I read at


that you can handle the ratio of Poisson random variables with a "method of truncation" and that the normal distribution may be a good approximation for "large means" although it says that further investigation on what "large is" is needed.

But even by avoiding the undefined values of the ratio, it doesn't seem clear to me why the ratio should follow a Poisson distribution?! What am I missing?


I missed the point, that the author, although not explicit written, seems to speak about the ratio conditioned on $\{D_{t}\}_{t\leq T}$, i.e. the distribution of the ratio conditioned on data up to and including $T$. Could it be that he isn't claiming that the ratio is "really" Poisson distributed, but rather approximately distributed as

$\frac{D_{T-3}+...+D_{T}}{D_{T-4}+...+D_{T}}+\frac{\text{Poisson}(\alpha H_{T+1})}{\alpha(H_{T-4}+...+H_{T})}$?

(Note that the $H$-ratio is replaced by 1 as an approximation, $D_{T-4}+...+D_{T}$ is replaced by it's expectation $\alpha(H_{T-4}+...+H_{T})$ and $D_{T+1}\sim\text{Poisson}(\alpha H_{T+1})$.)

That being the case, would it be reasonable to approximate the (conditioned) distribution of the ratio by a normal distribution?

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    $\begingroup$ Which paper? I'd like to see what is said. Note that many terms in the numerator and denominator are in common; I don't think that resolves the issues, though - but it may simplify things quite a bit. $\endgroup$ – Glen_b Feb 5 '15 at 11:06
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    $\begingroup$ It can be found at actuaries.org/lyon2013/papers/LIFE_Jarner_Moeller.pdf on pages 17-19. $\endgroup$ – Stats_L Feb 5 '15 at 12:38
  • $\begingroup$ Look at my edit what I missed, Glen_b. $\endgroup$ – Stats_L Feb 5 '15 at 16:10
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    $\begingroup$ A ratio of Poisson variables can't be Poisson. Suppose z=x/y. If x=2 and y=3 then z is not an integer and so cannot follow a Poisson distribution. $\endgroup$ – conjectures Feb 5 '15 at 16:37
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    $\begingroup$ Yes, the title of this thread is now kind of misleading. Did you read my edit? Since I believe the distribution of the ratio conditioned on data up to and including T is meant, there is no Poisson distributed r.v. in the denominator anymore as I tried to point out. But still there is a denominator which means the ratio can't be Poisson distributed in a strict sense. Maybe the "scaled and translated" statement tries to point this out and the author claims that the ratio though can be approximated by a normal distribution? Nevertheless, I clearly need further help. $\endgroup$ – Stats_L Feb 5 '15 at 16:58

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