Suppose two Poisson processes. For example, during the time interval, $\Delta t_{1} = t_{1} - t_{o} = 50\mu s$ , $x$ photons are incident on a detector with rate $\lambda_{1} = 10$x$10^4 s^{-1}$. At time point, $t_{1}$, a second process begins in which, during the time interval, $\Delta t_{2} = t_{2} - t_{1} = 50\mu s$ , $y$ photons are incident on the same detector with rate $\lambda_{2} = 6$x$10^4 s^{-1}$.

Let $X$ and $Y$ be two independent Poisson random variables described by $X$ ~ Pois($\lambda_{1}\Delta t_1$) and $Y$ ~ Pois($\lambda_{2}\Delta t_2$). And let $Z$ be a ratio distribution defined as $Z = X/(X+Y)$.

[1] What is the general distribution of $Z$ for $X+Y>0$? its standard deviation? and how are both derived?

Next, suppose we know the total number of photons, $n=x+y$ , over the time interval $\Delta t = \Delta t_{1} + \Delta t_{2} = 100\mu s$ ; e.g., $n=10$.

We would like to predict the probability distribution for observing an $(x,y)$ pair given $n$ and the knowledge that both $x$ and $y$ were drawn from Poisson distributions with rates $\lambda_1 \Delta t_1$ and $\lambda_2 \Delta t_2$, respectively.

[2] What is the new distribution for $Z|n$? its standard deviation? and how are both derived?

  • $\begingroup$ It's not clear if the fact that the two times intervals are consecutive is of any significance. Actually, the whole first paragraph seems pretty redundant, is it? And, rather, in the second paragraph, $X \approx $ Pois ($\lambda_1)$ ... here $\lambda_1$ corresponds to $\lambda_1 \Delta t_1$ in the first paragraph... $\endgroup$ – leonbloy Jun 4 '12 at 21:33
  • $\begingroup$ Conditionally on $X+Y=n$, the distribution of $X$ is binomial $(n,p)$, with $p=\lambda_1\Delta t_1/(\lambda_1\Delta t_1+\lambda_2\Delta t_2)$. $\endgroup$ – Did Jun 11 '12 at 5:12
  • $\begingroup$ @did Having difficulty visualizing the solution. How is the problem equivalent to a coin tossing experiment? If knowing $n$ simply reduces the problem to a binomial, $Z \sim X|n / (X|n + n - X|n) = X|n / n$ . Hence, $\sigma_Z = \sqrt{p(1-p)/n}$ . Does order not matter (e.g., $x$ photons before $y$)? $\endgroup$ – kgryte Jun 11 '12 at 5:38
  • $\begingroup$ See my answer below. $\endgroup$ – Did Jun 11 '12 at 5:56
  • $\begingroup$ Cross-posted on stats. $\endgroup$ – whuber Jun 11 '12 at 12:11

$Z$ is not well-defined; consider $X=Y=0$ which occurs with probability $> 0$, when $Z = 0/0$. Perhaps you should consider the distribution only for $n>0$? In that case, there is no standard distribution with a name, you just have to work through the math yourself.

Note that since X and Y are integers, Z is, except for the $0/0$ case, defined on the set of nonnegative rationals. If $\lambda_1$ and $\lambda_2$ are large, you could use a Beta distribution as a continuous approximation, based on the following:

1) The Poisson distribution approaches the Gamma distribution (with scale parameter $= 1$) as $\lambda \to \infty$, with the shape parameter $r$ of the Gamma equal to $\lambda$ (you can get to this by a) moment matching of the Poisson and Gamma, and b) observing that both distributions approach the same Gaussian as $r= \lambda \to \infty$, therefore they approach each other as well.)

2) If $x_1, x_2$ are distributed $\Gamma$ with identical scale parameters and shape parameters equal to $r_1, r_2$ respectively, then $x_1/(x_1+x_2)$ is distributed $\beta(r_1,r_2)$.

Edit (edited again) in response to followup question by OP:

The Beta approximation seems good except perhaps in the tails at $\lambda=20$, and quite good at $\lambda=50$. Quantile-quantile plots of a sample of size 10,000 from the Poisson ratio vs. the approximating Beta distribution for $\lambda=20,50$ are below. The actual distribution appears to have a slightly fatter lower tail than the Beta approximation at $\lambda=20$, and seems to be a very good fit at $\lambda=50$ (your definitions may vary.) Depending on the application, I'd say somewhere in the 20 - 50 range the Beta approximation would start to work quite well.

N <- 10000
Lambda <- 50

x1 <- rpois(N, Lambda)
x2 <- rpois(N, Lambda)
y <- sort(x1/(x1+x2))

z <- ((1:N)-0.5)/N
qqplot(qbeta(z,Lambda,Lambda), y,
       xlab = paste("Beta(",Lambda,",",Lambda,")",sep=""),
       ylab = paste("X/(X+Y) (N = ",N,")"))

enter image description here

enter image description here

  • $\begingroup$ Thanks for your answer. Yes, the distribution should only be considered for $n > 0$. I have edited this in the question. $\endgroup$ – kgryte Jun 4 '12 at 21:49
  • $\begingroup$ how large must $\lambda$ be for the Gamma approximation to be valid? I have supplied example values in the question for more context. Thanks again! $\endgroup$ – kgryte Jun 4 '12 at 22:12
  • $\begingroup$ I've added some more on that in the answer above: short version, probably around 20-50 is good, but it's application-dependent of course. $\endgroup$ – jbowman Jun 4 '12 at 22:55
  • $\begingroup$ Thanks for the plots. That helps. I am running the Q-Q plots locally, and I am not getting the same results with persistently fat tails even for large $\lambda$. Would you be able to provide the code snippets you used for your simulation? $\endgroup$ – kgryte Jun 5 '12 at 2:32
  • $\begingroup$ Huh, nor am I, now. I wonder what happened? Now I'm getting slightly thin tails at $\lambda=20$, and a very good fit at $\lambda=50$. I admit I had to recreate the code from memory, so there are several plausible explanations... I'll work on it a little more and correct the post, tomorrow my time. $\endgroup$ – jbowman Jun 5 '12 at 2:56

Let $X$ and $Y$ be independent Poisson random variables with respective rates $a$ and $b$. Then $X+Y$ is Poisson with rate $a+b$. For every $0\leqslant k\leqslant n$, $$ \mathrm P(X=k\mid X+Y=n)=\frac{\mathrm P(X=k,X+Y=n)}{\mathrm P(X+Y=n)}=\frac{\mathrm P(X=k)\mathrm P(Y=n-k)}{\mathrm P(X+Y=n)}, $$ that is, $$ \mathrm P(X=k\mid X+Y=n)=\frac{p_a(k)p_b(n-k)}{p_{a+b}(n)}, \quad \text{where}\quad p_c(i)=\mathrm e^{-c}\frac{c^i}{i!}. $$ After some easy simplifications, one gets $$ \mathrm P(X=k\mid X+Y=n)={n\choose k}p^k(1-p)^{n-k}, \quad \text{where}\quad p=\frac{a}{a+b}. $$ This means that, conditionally on $X+Y=n$, the distribution of $X$ is binomial $(n,p)$. In particular, the conditional expectation of $X$ is $np$ and its conditional variance is $np(1-p)$.

  • $\begingroup$ Thanks for this! Can Bayes' Theorem be used to `update', so to speak, or show how the distribution describing $Z$ changes in light of knowing $n$ ? Basically, what is the relation between $Z \sim \beta(\lambda_1, \lambda_2)$ and $Z|n \sim binom(n, \lambda_1 /(\lambda_1 + \lambda_2)$ ? $\endgroup$ – kgryte Jun 11 '12 at 6:45
  • $\begingroup$ But Z is not beta... $\endgroup$ – Did Jun 11 '12 at 7:22
  • $\begingroup$ Right...$Z$ can be approximated by a Beta distribution. But then is it just chance that, in approximating $Z\sim\beta(\lambda_1, \lambda_2)$ , $\sigma_Z = \sqrt{p(1-p)/(n+1)}$, and then, once conditional on $n$, $\sigma_{Z|n} = \sqrt{p(1-p)/n}\ \ $ ? $\endgroup$ – kgryte Jun 11 '12 at 8:52

Mathematica claims that there is an analytical solution for this: $$ \sum_{x=0}^\infty \sum_{y=0}^\infty P(\lambda_x,x) P(\lambda_y,y)\frac{x}{x+y} =\frac{\lambda_x+e^{\lambda_x+\lambda_y}\lambda_y}{\lambda_x+\lambda_y} $$ although I have not yet managed to prove it.

  • $\begingroup$ I think there is a mistake here, shouldn't it tend towards $\lambda_x/(\lambda_x+\lambda_y)$ for large $\lambda$s? $\endgroup$ – Lucas Jul 29 '19 at 22:26

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