$P\left( n,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ Disaggregating Tail of Poisson I have a Poisson tail $P\left( x,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ which is sum of two independent Poisson distribution with rate $\lambda_1$ and $\lambda_2$. I am trying to write the distribution tail  (probability of $x$ events or more) in terms of two Poisson tails. So I am trying to disaggregate it. 
where: $P\left( x,\lambda T \right)=\sum\limits_{i=x}^{\infty }{p\left( i,\lambda T \right)}$
Thanks!
 A: If $X$ is Poisson with mean $a+b$, a way to exhibit $X_a$ and $X_b$ independent and Poisson with respective means $a$ and $b$ such that $X=X_a+X_b$ is to consider an i.i.d. sequence $(U_n)_{n\geqslant1}$ of Bernoulli random variables with $P[U_n=1]=a/(a+b)$ and $P[U_n=0]=b/(a+b)$ for every $n\geqslant1$, independent of $X$, and to define
$$
X_a=\sum_{n=1}^XU_n,\qquad X_b=\sum_{n=1}^X(1-U_n).
$$
This is called thinning, as explained in about every textbook on Poisson processes, and can be checked by elementary computations: define $X_a$ and $X_b$ as above, compute the distribution of $(X_a,X_b)$, note that obviously, $X=X_a+X_b$ (for starters, show that the distribution of $X_a$ is Poisson with mean $a$).
A: If I understand your question, yes you can. It is a well known result that the sum of two independent Poisson random variables is another Poisson random variable whose parameter is the sum of the constituent parameters. One can show this in a straight forward way, but it is easier to do so via mgfs:
If $X_1 \sim $Poisson$(\lambda_1)$ and $X_2 \sim $Poisson$(\lambda_2)$ is independent of $X_1$ then the mgf for $X_i$ is $\mathbb{E}[e^{sX_i}] = e^{\lambda_i(e^s-1)}$ and so
$$
  \mathbb{E}[e^{s(X_1 + X_2)}]
= \mathbb{E}[e^{sX_1}]\mathbb{E}[e^{s X_2}]
= e^{\lambda_1(e^s-1)}e^{\lambda_2(e^s-1)}
= e^{(\lambda_1 + \lambda_2)(e^s-1)}.
$$
