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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}

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  • $\begingroup$ 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. $\endgroup$
    – Xi'an
    Feb 22, 2019 at 10:10

2 Answers 2

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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.

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

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