$X \sim \mathcal{P}( \lambda) $ and $Y \sim \mathcal{P}( \mu)$ meaning that $X$ and $Y$ are Poisson distributions. What is the probability distribution law of $X + Y$. I know it is $X+Y \sim \mathcal{P}( \lambda + \mu)$ but I don't understand how to derive it.

Try using the method of moment generating functions :) – Samuel Reid Nov 25 '13 at 07:03

All I've learned in the definition of a Poisson Random Variable, is there a simpler way? – Nov 25 '13 at 07:07

9If they are `independent`. – Did Nov 25 '13 at 08:14

Doesn’t it suffice that their covariance vanishes? – Michael Hoppe Feb 01 '18 at 07:09
7 Answers
This only holds if $X$ and $Y$ are independent, so we suppose this from now on. We have for $k \ge 0$: \begin{align*} P(X+ Y =k) &= \sum_{i = 0}^k P(X+ Y = k, X = i)\\ &= \sum_{i=0}^k P(Y = ki , X =i)\\ &= \sum_{i=0}^k P(Y = ki)P(X=i)\\ &= \sum_{i=0}^k e^{\mu}\frac{\mu^{ki}}{(ki)!}e^{\lambda}\frac{\lambda^i}{i!}\\ &= e^{(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(ki)!}\mu^{ki}\lambda^i\\ &= e^{(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{ki}\lambda^i\\ &= \frac{(\mu + \lambda)^k}{k!} \cdot e^{(\mu + \lambda)} \end{align*} Hence, $X+ Y \sim \mathcal P(\mu + \lambda)$.
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13In general we can't say anything then. It depends on how they depend on another. – martini Oct 25 '12 at 20:22

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2Nice derivation: specifically the transformation of (a) the i/k factorials and (b) the mu/lambda polynomials into the binomial form of the polynomial power expression. – WestCoastProjects Aug 30 '14 at 20:59

ty for that answer but what i don't get is how come u've taken 1/k! outside the sum? is it to say that k! *1/k! would be 1? – LiorA Aug 08 '17 at 10:01

1@LiorA Yes. k! included to combine with the rest and simplify as intended, so 1/k! is included to compensate. – Rolazaro Azeveires Jan 07 '18 at 14:23

1If the sum is indeed poisson, and we conversely say that X, Y MUST be independent? You said it only holds if independent at the beginning. I am struggling to come up with a proof here – Charlie Tian Mar 16 '18 at 15:01
Another approach is to use characteristic functions. If $X\sim \mathrm{po}(\lambda)$, then the characteristic function of $X$ is (if this is unknown, just calculate it) $$ \varphi_X(t)=E[e^{itX}]=e^{\lambda(e^{it}1)},\quad t\in\mathbb{R}. $$ Now suppose that $X$ and $Y$ are independent Poisson distributed random variables with parameters $\lambda$ and $\mu$ respectively. Then due to the independence we have that $$ \varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)=e^{\lambda(e^{it}1)}e^{\mu(e^{it}1)}=e^{(\mu+\lambda)(e^{it}1)},\quad t\in\mathbb{R}. $$ As the characteristic function completely determines the distribution, we conclude that $X+Y\sim\mathrm{po}(\lambda+\mu)$.
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You can use Probability Generating Function(P.G.F). As poisson distribution is a discrete probability distribution, P.G.F. fits better in this case.For independent X and Y random variable which follows distribution Po($\lambda$) and Po($\mu$). P.G.F of X is \begin{equation*} \begin{split} P_X[t] = E[t^X]&= \sum_{x=0}^{\infty}t^xe^{\lambda}\frac{\lambda^x}{x!}\\ &=\sum_{x=0}^{\infty}e^{\lambda}\frac{(\lambda t)^x}{x!}\\ &=e^{\lambda}e^{\lambda t}\\ &=e^{\lambda (1t)}\\ \end{split} \end{equation*} P.G.F of Y is \begin{equation*} \begin{split} P_Y[t] = E[t^Y]&= \sum_{y=0}^{\infty}t^ye^{\mu}\frac{\mu^y}{y!}\\ &=\sum_{y=0}^{\infty}e^{\mu}\frac{(\mu t)^y}{y!}\\ &=e^{\mu}e^{\mu t}\\ &=e^{\mu (1t)}\\ \end{split} \end{equation*}
Now think about P.G.F of U = X+Y. As X and Y are independent, \begin{equation*} \begin{split} P_U(t)=P_{X+Y}(t)=P_X(t)P_Y(t)=E[t^{X+Y}]=E[t^X t^Y]&= E[t^X]E[t^Y]\\ &= e^{\lambda (1t)}e^{\mu (1t)}\\ &= e^{(\lambda+\mu) (1t)}\\ \end{split} \end{equation*}
Now this is the P.G.F of $Po(\lambda + \mu)$ distribution. Therefore,we can say U=X+Y follows Po($\lambda+\mu$)
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In short, you can show this by using the fact that $$Pr(X+Y=k)=\sum_{i=0}^kPr(X+Y=k, X=i).$$
If $X$ and $Y$ are independent, this is equal to $$ Pr(X+Y=k)=\sum_{i=0}^kPr(Y=ki)Pr(X=i) $$ which is $$ \begin{align} Pr(X+Y=k)&=\sum_{i=0}^k\frac{e^{\lambda_y}\lambda_y^{ki}}{(ki)!}\frac{e^{\lambda_x}\lambda_x^i}{i!}\\ &=e^{\lambda_y}e^{\lambda_x}\sum_{i=0}^k\frac{\lambda_y^{ki}}{(ki)!}\frac{\lambda_x^i}{i!}\\ &=\frac{e^{(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k\frac{k!}{i!(ki)!}\lambda_y^{ki}\lambda_x^i\\ &=\frac{e^{(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k{k\choose i}\lambda_y^{ki}\lambda_x^i \end{align} $$ The sum part is just $$ \sum_{i=0}^k{k\choose i}\lambda_y^{ki}\lambda_x^i=(\lambda_y+\lambda_x)^k $$ by the binomial theorem. So the end result is $$ \begin{align} Pr(X+Y=k)&=\frac{e^{(\lambda_y+\lambda_x)}}{k!}(\lambda_y+\lambda_x)^k \end{align} $$ which is the pmf of $Po(\lambda_y+\lambda_x)$.
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Moderator notice: This answer was moved here as a consequence of merging two questions. This explains the small differences in notation. The OP's $\lambda$ is $\lambda_x$ here, and OP's $\mu$ is $\lambda_y$. Otherwise there is no difference. – Jyrki Lahtonen Apr 23 '15 at 06:55
Using Moment Generating Function.
If $X \sim \mathcal{P}(\lambda)$, $Y \sim \mathcal{P}(\mu)$ and S=X+Y.
We know that MGF(Moment Generating Function) of $\mathcal{P}(\lambda)=e^{\lambda(e^t1)}$(See the end if you need proof)
MGF of S would be
$$\begin{align}
M_S(t)&=E[e^{tS}]\\&=E[e^{t(X+Y)}]\\&=E[e^{tX}e^{tY}]\\&=E[e^{tX}]E[e^{tY}]\quad \text{given }X,Y\text{ are independent}\\&=e^{\lambda(e^t1)}e^{\mu(e^t1)}\\&=e^{(\lambda+\mu)(e^t1)}
\end{align}$$
Thus S is a Poisson Distribution with parameter $\lambda+\mu$.
MGF of Poisson Distribution
If $X \sim \mathcal{P}(\lambda)$, then by definition Probability Mass Function is
$$\begin{align}
f_X(k)=\frac{\lambda^k}{k!}e^{\lambda},\quad k \in 0,1,2....
\end{align}$$
It's MGF is
$$\begin{align}
M_X(t)&=E[e^{tX}]\\&=\sum_{k=0}^{\infty}\frac{\lambda^k}{k!}e^{\lambda}e^{tk}\\&=e^{\lambda}\sum_{k=0}^{\infty}\frac{\lambda^ke^{tk}}{k!}\\&=e^{\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}\\&=e^{\lambda}e^{\lambda e^t}\\&=e^{\lambda e^t\lambda}\\&=e^{\lambda(e^t1)}
\end{align}$$
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hint: $\sum_{k=0}^{n} P(X = k)P(Y = nk)$
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adding two random variables is simply convolution of those random variables. That's why. – jaysun Oct 25 '12 at 20:24


*adding two random variables is simply convolution of those random variables*... Sorry but no. – Did Feb 13 '13 at 06:28

@Did I meant in the usual pdf sense and assumes independence of course. – jaysun Feb 13 '13 at 06:48

2There is no `usual sense` for convolution of random variables. Either convolution of distributions or addition of random variables. – Did Feb 13 '13 at 06:51
Here's a much cleaner solution:
Consider a two Poisson processes occuring with rates $\lambda$ and $\mu$, where a Poisson process of rate $r$ is viewed as the limit of $n$ consecutive Bernoulli trials each with probability $\frac{r}{n}$, as $n\to\infty$.
Then $X$ counts the number of successes in the trials of rate $\lambda$ and $Y$ counts the number of successes in the trials of rate $\mu$, so the total number of successes is the same as if we had each trial succeed with probability $\frac{\lambda + \mu}{n}$, where we take $n$ to be large enough so that the event where the $i$th Bernoulli trial in both processes are successful has a negligible probability. Then we are done.
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Your answer applies to MERGE case instead of SUM, since if X succeeds and Y succeeds, the result is 2. – Tengerye May 01 '20 at 08:55

If we consider the case where that Poisson process runs from time $t=0$ to $t=1$, then the number of successes is Pois($r$) by construction. As $n\to\infty$, the number of trials that succeed in both Poisson processes should be $\left(\frac{\lambda}{n}\right)\left(\frac{\mu}{n}\right)n + O_p\left(\sqrt{n\left(\frac{\mu\lambda}{n^2}\right)\left(1\frac{\mu\lambda}{n^2}\right)}\right)\to 0$ by CLT. – Anon May 03 '20 at 06:17

Formally, we'd have to use Slutsky's Theorem and Continuous Mapping Theorem fully justify why this convergence in probability means that $X+Y\sim Pois(\mu+\lambda)$ (where we let $X_n$ and $Y_n$ be the number of successes using $n$ trials with their respective probabilities, so $X_n\to X$ and $Y_n\to Y$ in distribution), but the intuition that "the Bernoulli trial in both processes are successful has negligible probability" should be enough to give a good idea why the statement is true. – Anon May 03 '20 at 06:27