Can a function be split into sub-function to prove it is a probability mass function? And how to find variance of such function? 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? 
 A: 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
$$\mu_1 = (1-\pi)\lambda_1 + \pi\lambda_2$$
and
$$\mu_2 = (1-\pi)(\lambda_1+\lambda_1^2) + \pi(\lambda_2 +\lambda_2^2).$$
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").
A: 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.
A: 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)
A: Probably simply no, you cannot split the pieces and treat both as Poisson, as the two individual pieces:

*

*Do not sum to 1 so are not valid PMFs (may be neither relevant nor true)

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

Reference:
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)
