(Cross-post from Stats Stack Exchange)
I'm interested in using the Poissonization trick to solve the following problem, which I made up:
Suppose I have a categorical random variable $X$ taking values $1$, $2$, and $3$, with probability distribution $(\pi_1, \pi_2, \pi_3)$. How many samples from $X$ should I take before I have at least 5 samples from one of the categories, with probability $.9$? Or at least 5 samples from every category?
If $Y_1, Y_2, Y_3$ are independent Poisson random variables with rates $\pi_1, \pi_2, \pi_3$, then the Poissonization trick tells us that $(Y_1, Y_2, Y_3 \mid \sum Y_i = k)$ has distribution $\sim \operatorname{Mult}(k; \pi_1, \pi_2, \pi_3)$ on $\mathbb{N}^3$.
If $A\subset \mathbb{N}^3$ is the set $\{x_1, x_2, x_3 \leq 4\}$ (I'm interested in the complement of $A$), then $$\mathbb{P}\{(Y_1, Y_2, Y_3)\in A\}$$ $$ = \sum \mathbb{P}\{(Y_1, Y_2, Y_3)\in A \mid \sum Y_i =k \} \cdot \mathbb{P}(\sum Y_i = k)$$
And (this is where I get a bit shaky) I should expand in a sort of fake variable $\lambda$, representing the rate of $Y_1 + Y_2 + Y_3$ (though I know that rate is actually one), and then $\mathbb{P}\{(Y_1, Y_2, Y_3)\in A\}$ is
$$e^{\pi_1 \lambda}(1 + \frac{(\pi_1\lambda)^1}{1!} + \dotsb + \frac{(\pi_1\lambda)^4}{4!})\, \cdot \, $$ $$e^{\pi_2 \lambda}(1 + \frac{(\pi_2\lambda)^1}{1!} + \dotsb + \frac{(\pi_2\lambda)^4}{4!})\, \cdot \, $$ $$e^{\pi_3 \lambda}(1 + \frac{(\pi_3\lambda)^1}{1!} + \dotsb + \frac{(\pi_3\lambda)^4}{4!})\, \cdot \, $$
And now I should multiply this by $e^{-\lambda}$, expand in $\lambda$ somehow, and look for the coefficient on $\frac{\lambda^k}{k!}$? Is that right?
