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The question I'm posing arises in the so-called "family-size process" studied by D.Kendall. It is an example of an open migration process, but describing carefully this model will take too much time, so let me go directly to my problem.

For $j \in \mathbb{N}$ define $n_j \sim \mbox{Poisson}(\alpha_j)$, where $\displaystyle \alpha_j:=\frac{\nu}{\lambda j} \left (\frac{\lambda}{\mu} \right )^j$. All these processes are independent of each other and each random variable $n_j$ can be thought as the number of families of size $j$.

Define also the following random variables:

$\displaystyle N:=\sum_j n_j$ is the total number of families;

$\displaystyle M:=\sum_j j n_j$ is the total number of individuals.

Could you help me in proving that $\displaystyle E(N|M)=\sum_{i=1}^M \frac{\rho}{\rho+i-1}$ where $\displaystyle \rho:=\frac{\nu}{\lambda}$? Thanks a lot!

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Is the formula really independent of $\mu$? –  Yuval Filmus May 18 '11 at 22:57
    
As far as I know yes, the formula should be independent of $\mu$ –  Ale Zok May 18 '11 at 23:18
    
You say "All these processes are independent of each other". Did you mean all of the random variables $n_j$ are independent of each other? –  Michael Hardy Aug 17 '11 at 3:31

1 Answer 1

Let $\beta = \lambda/\mu$. Let $P(x,y)$ be the generating function where the coefficient of $x^N y^M$ is the probability to get $N,M$. For an individual random variable, it is well-known that the generating function is $\exp(\alpha_j(xy^j-1))$. So $$ P = \exp\left( \sum_{j=1}^\infty \frac{\rho}{j} \beta^j (xy^j - 1) \right). $$ Let's evaluate the constant term: $$ \sum_{j=1}^\infty \frac{\rho}{j} \beta^j = -\rho\log(1-\beta). $$ Substituting this, and using the Taylor expansion of log once again, we get $$ P = (1-\beta)^\rho \exp\left( x \sum_{j=1}^\infty \frac{\rho}{j} (\beta y)^j \right) = (1-\beta)^\rho \exp(-x\rho\log(1-\beta y)). $$ In order to get expectation with respect to $x$, we differentiate with respect to $x$ and substitute $x = 1$. First, $$ \frac{\partial}{\partial x} P = (1-\beta)^\rho \exp(-x\rho\log(1-\beta y)) (-\rho \log(1-\beta y)).$$ Now substitute $x=1$ to get a univariate generating function $Q(y)$: $$ Q = \rho (1-\beta)^\rho (1-\beta y)^{-\rho} \sum_{j=1}^\infty \frac{\beta^j y^j}{j}. $$ We can expand $(1-\beta y)^{-\rho}$ into a Taylor series as well: $$ Q = \rho (1-\beta)^\rho \sum_{i=0}^\infty \binom{i+\rho-1}{\rho-1} \beta^i y^i \sum_{j=1}^\infty \frac{\beta^j y^j}{j}. $$ We can read off the coefficient of $y^M$: $$ \rho (1-\beta)^\rho \sum_{j=1}^M \binom{M-j+\rho-1}{\rho-1} \beta^{M-j} \frac{\beta^j}{j} = \rho (1-\beta)^\rho \beta^M \sum_{j=1}^M \frac{\binom{M-j+\rho-1}{\rho-1}}{j}. $$ That's as far as I got...

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At least we see why the $\beta$ drops out. Indeed, you also need to divide by $(1-\beta)^\rho \beta^M \binom{M+\rho-1}{\rho-1} \; ,$ to get the conditional expectation. –  Raskolnikov Oct 16 '11 at 9:46

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