I have a question that requires to prove if the following function whether is it a PMF with poisson random variable. The function is as follows...

$f(x) = \pi \frac {\lambda_1^x}{x!} e^{-\lambda_1} + (1-\pi) \frac {\lambda_2^x}{x!}e^{-\lambda_2} $

where $x\epsilon \mathrm X = \{0,1,...\}, \pi \epsilon (0,1), \lambda_1, \lambda_2 > 0, \lambda_1 \neq \lambda_2$

Can I split $f(x)$ into 2 function $\sum_{x=0}^{\infty}\pi \frac {\lambda_1^x}{x!} e^{-\lambda_1}$ and $\sum_{x=0}^{\infty} (1-\pi) \frac {\lambda_2^x}{x!}e^{-\lambda_2}$ to prove that it is a PMF and hence continue with calculating it's expectation and variance using this 2 sub-function?

I've found out the expectation of $f(x)$, which is $\mathbb E[X] = \mathbb E[X_1] + \mathbb E[X_2];$ where, $\mathbb E[X_1] = \sum_{x=0}^{\infty}x\pi \frac{\lambda_1^x}{x!}e^{-\lambda_1} = \pi\lambda_1$
$\mathbb E[X_2]=\sum_{x=0}^{\infty} x(1-\pi)\frac{\lambda_2^x}{x!}e^{-\lambda_2} = (1-\pi)\lambda_2$

Now my question is how am I suppose to carry to find out what is the variance as the variance i've got is different from the expectation I got above. For a Poisson random variable, the expectation should be the same as variance. Am I right? I have also tried using moment generating function. For the 1st differential, I got the same result as my expectation. But I couldn't further differentiate the 2nd time to get my variance as all the remaining terms are constant terms. Could I have some suggestion of how should I carry on from where?

  • $\begingroup$ Why not? What problem(s) do you run into when you carry out the calculations? $\endgroup$ – whuber Sep 11 '13 at 17:06
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    $\begingroup$ For this $f(x)$ to be a pmf, each of $f(0), f(1), f(2), \ldots$ must be nonnegative (true by inspection here) and the sum $\sum_{i=0}^\infty f(i)$ must equal $1$. If you know the series $$e^\lambda = \sum_{i=0}^\infty \frac{\lambda^i}{i!},$$ you can apply this to $\sum_i f(i)$ to show that $f(\cdot)$ is indeed a pmf (provided that $\pi \in [0,1]$) but that it is not a Poisson pmf except when $\pi = 0$ or $\pi = 1$. $\endgroup$ – Dilip Sarwate Sep 11 '13 at 18:12
  • $\begingroup$ You may be interested in (Eisenberger 1971). Reference:Eisenberger, I. "Estimating the Parameters of the Distribution of a Mixture of Two Poisson Populations." Deep Space Network Progress Report 3 (1971): 94-97. $\endgroup$ – Avraham Sep 11 '13 at 18:18
  • $\begingroup$ Although your edit changes the question, the edit is already answered here. Note that the variance is not a non-central moment of the distribution, but is an algebraic combination of such moments. $\endgroup$ – whuber Sep 12 '13 at 5:41
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    $\begingroup$ Your calculation of $E[X]$ results in the correct value, but the meanings you ascribe to various parts of your answer are not correct. What you call $E[X_i]$ is not the expectation of any random variable, and certainly not the expectation of a Poisson random variable with parameter $\lambda_i$. Nor is $X$ equal to the sum $X_1+X_2$ from which the relationship $E[X]=E[X_1]+E[X_2]$ that you state would naturally follow. If you will correct these misconceptions (see whuber's answer for the correct formulation of the relationship between $X$ and the $X_i$), you will make more progress. $\endgroup$ – Dilip Sarwate Sep 12 '13 at 13:20

As per whuber's wishes, I am expanding my comment on the OP's question into a full-fledged answer. Whether the answer is using statistical ideas only or not is a matter for the readership to judge.

Let $\pi \in [0,1]$. Then the function $$f(x) = \begin{cases}(1-\pi)e^{-\lambda_1}\frac{\lambda_1^n}{n!} + \pi e^{-\lambda_2}\frac{\lambda_2^n}{n!}, &x = \text{nonnegative integer}~ n,\\ 0, &\text{otherwise,} \end{cases}$$ is a probability mass function since $f(x) \geq 0$ for all $x$ and $$\begin{align}\sum_{n=0}^\infty (1-\pi)e^{-\lambda_1}\frac{\lambda_1^n}{n!} + \pi e^{-\lambda_2}\frac{\lambda_2^n}{n!} &= (1-\pi)\sum_{n=0}^\infty e^{-\lambda_1}\frac{\lambda_1^n}{n!} + \pi \sum_{n=0}^\infty e^{-\lambda_2}\frac{\lambda_2^n}{n!}\\ &= (1-\pi) + \pi\\ &= 1 \end{align}$$ since the two summands on the right are recognizable as the pmfs of Poisson random variables. The pmf $f(x)$ is called a mixture pmf (of two Poisson random variables). Let $X$ denote the random variable with this mixture pmf. Then, $X$ is not a Poisson pmf except for the extremal value $\pi=0$ and $\pi=1$ in which case we get $X$ is Poisson with parameters $\lambda_1$ and $\lambda_2$ respectively. A proof of the assertion of non-Poissonity is as follows.

Note that $E[X]= (1-\pi)\lambda_1 + \pi \lambda_2$ wnich can be obtained by straightforward summations just as in computing the mean of a Poisson random variable. Now suppose that $X$ is indeed a Poisson random variable with parameter $\lambda$. So we have $$E[X] = \lambda = (1-\pi)\lambda_1 + \pi \lambda_2$$ which, as a function of $\pi$ varies linearly from $\lambda_1$ at $\pi=0$ to $\lambda_2$ at $\pi = 1$. Now, according to the mixture pmf, $$P(X=0) = (1-\pi)e^{-\lambda_1} + \pi e^{-\lambda_2}\tag{1}$$ whereas the alleged Poissonity of $X$ gives us that $$P(X=0) = e^{-\lambda} = e^{-((1-\pi)\lambda_1 + \pi \lambda_2)}.\tag{2}$$ The right sides of these two expressions are not equal and so the assumption that $X$ is a Poisson random variable is not tenable. How do we know that the right sides of $(1)$ and $(2)$ are not equal? Let $Y$ be a discrete random variable taking on values $\lambda_1$ and $\lambda_2$ with probabilities $(1-\pi)$ and $\pi$ respectively. Then $$E[Y] = (1-\pi)\lambda_1 + \pi \lambda_2$$ and so, for the convex function $e^{-x}$ we have, by Jensen's inequality that $$e^{-E[Y]} = e^{-((1-\pi)\lambda_1 + \pi \lambda_2)} \leq E[e^{-Y}] = (1-\pi)e^{-\lambda_1} + \pi e^{-\lambda_2}$$ with equality occurring only at the end points because the straight line through the points $(0,e^{-\lambda_1})$ and $(1,e^{-\lambda_2})$ is strictly above the curve $e^{-((1-\pi)\lambda_1 + \pi \lambda_2)}$ for $\pi \in (0,1)$. That is, the right side of $(2)$ is smaller than the right side of $(1)$ except when $\pi=0$ or $\pi = 1$, and so $X$ is not a Poisson random variable except in these extreme cases.

Note added in response to OP's comment and query

Since $X$ is not a Poisson random variable, it is not necessary that its mean equal its variance as is the case for Poisson random variables. The variance of $X$ can be calculated most easily as indicated in whuber's answer. Begin with the fact that for a Poisson random variable $W$ with parameter $\mu$ (and hence mean $E[W] = \mu$), $$E[W^2] = \sum_{n=0}^\infty n^2e^{-\mu}\frac{\mu^n}{n!} = \mu^2+\mu.$$ and so $$\begin{align} E[X^2] &= \sum_{n=0}^\infty n^2\left[(1-\pi)e^{-\lambda_1}\frac{\lambda_1^n}{n!} + \pi e^{-\lambda_2}\frac{\lambda_2^n}{n!}\right]\\ &= (1-\pi)\sum_{n=0}^\infty n^2e^{-\lambda_1}\frac{\lambda_1^n}{n!} + \pi\sum_{n=0}^\infty n^2e^{-\lambda_2}\frac{\lambda_2^n}{n!}\\ &= (1-\pi)(\lambda_1^2 + \lambda_1) + \pi(\lambda_2^2 +\lambda_2) \end{align}$$ and so $$\begin{align} \text{var}(X) &= E[X^2] - (E[X])^2\\ &= (1-\pi)(\lambda_1^2 + \lambda_1) + \pi(\lambda_2^2 +\lambda_2)- ((1-\pi)\lambda_1 + \pi \lambda_2)^2\\ &= [(1-\pi)\lambda_1 + \pi\lambda_2] + (1-\pi)\lambda_1^2 + \pi\lambda_2^2 - ((1-\pi)\lambda_1 + \pi \lambda_2)^2\\ &= [(1-\pi)\lambda_1 + \pi\lambda_2] + \pi(1-\pi)(\lambda_1-\lambda_2)^2 \end{align}$$ as already pointed out to you by whuber.

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  • $\begingroup$ Dear Dilip, I did the same to prove f(x) is a PMF. But how m I suppose to find out the $\mathbb Var[X]$ as the result I got is different from with the $\mathbb E[X]$ $\endgroup$ – melyong Sep 12 '13 at 4:23
  • $\begingroup$ For the last part ($var(x)$) how did u get from step 3 to 4? Also did $\lambda$ 1 and 2 got swapped? For me, $\lambda_2$ is the one with the $(1-\pi)$ term $\endgroup$ – Jiew Meng Sep 16 '13 at 14:25

The left hand side, $f$, is a convex mixture of two probability mass functions (pmf), whence it is a pmf. However, because its variance exceeds its mean, it cannot be the pmf of a Poisson distribution.

Consider a bivariate random variable $(U, X)$ where $U$ is a Bernoulli$(\pi)$ variable and, conditional on the value of $U$, $X$ either has a distribution with probability mass function $f_0$ when $U=0$ or a distribution with pmf $f_1$ when $U=1$. The marginal distribution of $X$ is a mixture of $f_0$ and $f_1$. Let $f$ be the the probability mass function of the marginal distribution and let $x$ be a possible outcome. According to the definitions of pmf and marginal distributions,

$$\eqalign{ f(x) &= \Pr(X=x) \\ &= \Pr(X=x|U=0)\Pr(U=0) + \Pr(X=x|U=1)\Pr(U=1) \\ &= (1-\pi)f_0(x) + \pi f_1(x). }$$

In the question, $f_0$ is the pmf of a Poisson$(\lambda_1)$ distribution and $f_1$ is the pmf of a Poisson$(\lambda_2)$ distribution. Therefore $f$ is a valid pmf.

Is $f$ the pmf of some Poisson distribution, say with parameter $\lambda$? There are many ways to check. Because a Poisson distribution depends on a single parameter, whenever we obtain two numerical properties of the distribution they must have a definite relationship. The best-known relationship in a Poisson distribution is that the variance equals the mean. The first two moments of the mixture are

$$\mu_1 = (1-\pi)\lambda_1 + \pi\lambda_2$$


$$\mu_2 = (1-\pi)(\lambda_1+\lambda_1^2) + \pi(\lambda_2 +\lambda_2^2).$$

From these we discover that the variance of the mixture is

$$\mu_2 - \mu_1^2 = \mu_1 + \pi(1-\pi)(\lambda_1-\lambda_2)^2.$$

The right hand side is strictly greater than the mean $\mu_1$ for $0\lt\pi\lt 1$ and $\lambda_1\ne\lambda_2$ because under these conditions $\pi\gt 0,$ $1-\pi\gt 0$, and $(\lambda_1-\lambda_2)^2\gt 0,$ whence the excess over $\mu_1$ on the right hand side is strictly positive. Therefore the mixture cannot be Poisson (it is "over dispersed").

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  • $\begingroup$ Anyone unsure about this calculation of the variance of a mixture can see it worked it out in gory detail at stats.stackexchange.com/questions/69399/…. The basic idea is very simple, though, as explained in an answer at stats.stackexchange.com/questions/32699. $\endgroup$ – whuber Sep 11 '13 at 20:54
  • $\begingroup$ As I said in my comment on the question, the mixture cannot be Poisson except in the trivial cases $\pi=0$ or $\pi = 1$, when it is really not a mixture at all but just one of the constituent Poisson random variables. $\endgroup$ – Dilip Sarwate Sep 11 '13 at 21:19
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    $\begingroup$ @Dilip Yes, you said correctly and I would have been glad to see your comment (more fully fleshed out) posted as an answer. For the record, so that we have a valid answer in this thread, here I have proven the result and explained (using statistical ideas) how one would go about finding such a proof in the first place. $\endgroup$ – whuber Sep 11 '13 at 21:22
  • $\begingroup$ I fleshed out my comment into an answer. $\endgroup$ – Dilip Sarwate Sep 12 '13 at 12:53

The sum of Poisson Random Variables is a Poisson (http://www.proofwiki.org/wiki/Sum_of_Independent_Poisson_Random_Variables_is_Poisson). Hence you can prove they are individually Poissons to show that their sum is a Poisson.

Expectation: E[A+B] = E[A] + E[B] where A,B are R.V.s

Variance: Var[A+B] = Var[A] + Var[B] + 2Cov(A,B)

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    $\begingroup$ Unfortunately, this answer is not relevant to the question asked which is about a mixture distribution (weighted sum of two Poisson pmfs with weights summing to $1$) and not about the sum of two Poisson random variables. $\endgroup$ – Dilip Sarwate Sep 11 '13 at 18:05
  • $\begingroup$ Sorry missed that point $\endgroup$ – Budhapest Sep 11 '13 at 18:09

Probably simply no, you cannot split the pieces and treat both as Poisson, as the two individual pieces:

  1. Do not sum to 1 so are not valid PMFs (may be neither relevant nor true)
  2. Are dependent on each other through $\pi$

For further reading, perhaps see the Eisenberger paper mentioned in the comments on the question and (Wang et al 2006).


@whuber's answer is far better, as usual.


Kui Wang, Kelvin K.W. Yau, Andy H. Lee, Geoffrey J. McLachlan, Two-component Poisson mixture regression modelling of count data with bivariate random effects, Mathematical and Computer Modelling, Volume 46, Issues 11–12, December 2007, Pages 1468-1476, ISSN 0895-7177, http://dx.doi.org/10.1016/j.mcm.2007.02.003. (http://www.sciencedirect.com/science/article/pii/S0895717707001100)

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  • $\begingroup$ What do you mean by the "individual pieces"? The pieces I see in the question are Poisson pmfs; there's nothing invalid about them. $f$ describes a mixture of two Poisson variates (provided $0\lt\pi\lt 1$). $\endgroup$ – whuber Sep 11 '13 at 19:57
  • $\begingroup$ If $\sum_{x=0}^\infty \frac{\lambda^xe^{-\lambda}}{x!} = 1$ then $(1-\pi)\sum_{x=0}^\infty \frac{\lambda^xe^{-\lambda}}{x!} \ne 1$. Each piece without the weight is certainly Poisson, but with the weight the PMF <> 1, and the two are not independent since $\pi$ appears in both, no? Please correct me if I have made an error. $\endgroup$ – Avraham Sep 11 '13 at 20:00
  • $\begingroup$ That strikes me as being less than relevant. What matters here is that $\sum_{x=0}^\infty \frac{\lambda^xe^{-\lambda}}{x!}$ describes a pmf. Thus the question concerns whether a convex linear combination of two pmfs is a pmf. What is ambiguous is whether the OP is also asking whether such a mixture of Poisson distributions can be Poisson; I suspect they are not asking this but it might seem they are because of difficulties with English. $\endgroup$ – whuber Sep 11 '13 at 20:02
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    $\begingroup$ @whuber OK, now we're swimming way beyond my capabilities and understanding, but per this entry a convex linear combination of characteristic functions is also a characteristic function. Since there is a one-to-one relationship between characteristic functions and CDFs, this would mean that the convex linear combination does define a PMF. There is no guarantee that PMF is Poisson, though. $\endgroup$ – Avraham Sep 11 '13 at 20:18

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