# What can we say about $N_{i}$ where $N=N_{1}+\cdots+N_{m}$, $N\thicksim Geom(\frac{1-p}{p})$ and conditional distribution of $N_{j}$ is binomial

Suppose that the number of events $$N$$ is a Geometric random variable with mean $$\frac{1-p}{p}$$. Further suppose that each event can be classified into one of $$m$$ types with probabilities $$p_{1},p_{2},\ldots,p_{m}$$ independent of all other events. Then we consider the number of events $$N_{1},N_{2},\ldots,N_{m}$$ corresponding to event types $$1,2,\ldots,m$$, respectively,

What can we say about $$N_{i}$$?

What is the distribution of $$N_{i}$$?

Are $$N_{1},N_{2},\ldots,N_{m}$$ mutually independent random variables?

What is the sign of $$cov(N_{1},N_{2})$$?

My attempt: For fixed $$N=n$$, the conditional joint distribution of $$(N_{1},\ldots,N_{m})$$ is multinomial with parameters $$(n,p_{1},\ldots,p_{m})$$. Also, for fixed $$N=n$$, the conditional marginal distribution of $$N_{j}$$ is binomial with parameters $$(n,p_{j})$$.

I want to show that the joint pf is the product of the marginal pfs, establishing mutual independence

The joint pf of $$(N_{1},\ldots,N_{m})$$ is given by

\begin{align} \mathbb{P}(N_{1}=n_{1},\ldots,N_{m}=n_{m}) &= \mathbb{P}(N_{1}=n,\ldots,N_{m}=n_{m}|N=n)\mathbb{P}(N=n)\\ &= \left( \frac{n!}{n_{1}!\cdots n_{m}!}p_{1}^{n_{1}}\cdots p_{m}^{n_{m}}\right)(1-p)^{n}p \\ &= \mbox{I do not know how to continue} \end{align} where $$n=n_{1}+n_{2}+\cdots +n_{m}$$. Similarly, as I fail with the above, try to calculate the marginal pf of $$N_{j}$$

\begin{align} \mathbb{P}(N_{j}=n_{j}) &= \sum_{n=n_{j}}^{\infty} \mathbb{P}(N_{j}=n_{j}|N=n)\mathbb{P}(N=n)\\ &= \sum_{n=n_{j}}^{\infty} \binom{n}{n_{j}}p_{j}^{n_{j}}(1-p_{j})^{n-n_{j}}p(1-p)^{n} \\ &= \mbox{I do not know how to continue} \end{align}

• The $N_j$'s cannot be independent because, else, if you take the case $p_1=\cdots=p_m$, the sum of Geometric iid variates is a Negative Binomial $(k,p_1)$ variate. Feb 22, 2019 at 10:10

You have made a good start. To finish this working you just pull out all terms that don't depend on the summation index $$n$$ and then apply a well-known summation identity for the binomial coefficients. For all $$n_k = 0,1,2,3,...$$ you have:

\begin{align} \mathbb{P}(N_k=n_k) &= \sum_{n=0}^{\infty} \mathbb{P}(N_{k}=n_k|N=n)\mathbb{P}(N=n) \\[6pt] &= \sum_{n=0}^{\infty} \text{Bin}(n_k | n, p_j) \cdot \text{Geom}(n|p) \\[6pt] &= \sum_{n=n_k}^{\infty} \binom{n}{n_k} p_k^{n_k}(1-p_k)^{n-n_k} \cdot p(1-p)^{n} \\[6pt] &= p \Big( \frac{p_k}{1-p_k} \Big)^{n_k} \sum_{n=n_k}^{\infty} \binom{n}{n_k} \Big[ (1-p_k) (1-p) \Big]^{n} \\[6pt] &= p \Big( \frac{p_k}{1-p_k} \Big)^{n_k} \frac{[ (1-p_k) (1-p) ]^{n_k}}{[ 1 - (1-p_k) (1-p) ]^{n_k+1}} \\[6pt] &= p \cdot \frac{(p_k - p \cdot p_k)^{n_k}}{(p+p_k - p \cdot p_k)^{n_k+1}} \\[6pt] &= \frac{p}{p+p_k - p \cdot p_k} \cdot \Big( \frac{p_k - p \cdot p_k}{p+p_k - p \cdot p_k} \Big)^{n_k} \\[6pt] &= \frac{p}{p+p_k - p \cdot p_k} \cdot \Big( 1 - \frac{p}{p+p_k - p \cdot p_k} \Big)^{n_k} \\[6pt] &= \text{Geom} \Big( n_k \Big| \frac{p}{p+p_k - p \cdot p_k} \Big). \\[6pt] \end{align}

So we see that:

$$N_k \sim \text{Geom} \Big( \frac{p}{p+p_k - p \cdot p_k} \Big).$$

You can see here that the parameter in the resulting geometric distribution for $$N_k$$ is no larger than the parameter in the original geometric distribution for $$N$$. They are equal in the special case where $$p_k=1$$. This is unsurprising, given that the latter probability represents the probability that a given item in the original set of $$N$$ objects is included in the set of $$N_k$$ objects.

• You very much appreciate your answer, but I have a question, according to what you calculated What kind of distribution has $N_{k}$? Feb 22, 2019 at 6:34
• I have added further information to show the distributional family.
– Ben
Feb 22, 2019 at 7:10

This is only a long remark that is not directly related with the original question, fully addressed by the earlier answer of Ben:

The Geometric distribution is infinitely divisible. This means that, for $$X\sim\mathcal{G}(P)$$, $$X$$ can always be written as $$X=X_1+\dots+X_n\qquad X_i\stackrel{\text{iid}}{\sim}P_n$$for any $$n$$. The simplest way to prove this is to look at the moment generating function $$\varphi(z)=\mathbb{E}[e^{zX}]=\dfrac{p}{1-(1-p)e^z}$$ Indeed, $$\varphi^t$$ $$(t>0)$$ appears as a Negative Binomial $$\mathcal{NB}(t,p)$$ moment generating function (in the general definition of a Negative Binomial distribution with real parameter $$t$$).