# Poisson Distribution of sum of two random independent variables $X$, $Y$

$X \sim \mathcal{P}( \lambda)$ 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 7:03
• All I've learned in the definition of a Poisson Random Variable, is there a simpler way? – user82004 Nov 25 '13 at 7:07
• If they are independent. – Did Nov 25 '13 at 8:14
• Doesn’t it suffice that their covariance vanishes? – Michael Hoppe Feb 1 '18 at 7:09

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 = k-i , X =i)\\ &= \sum_{i=0}^k P(Y = k-i)P(X=i)\\ &= \sum_{i=0}^k e^{-\mu}\frac{\mu^{k-i}}{(k-i)!}e^{-\lambda}\frac{\lambda^i}{i!}\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \frac{k!}{i!(k-i)!}\mu^{k-i}\lambda^i\\ &= e^{-(\mu + \lambda)}\frac 1{k!}\sum_{i=0}^k \binom ki\mu^{k-i}\lambda^i\\ &= \frac{(\mu + \lambda)^k}{k!} \cdot e^{-(\mu + \lambda)} \end{align*} Hence, $X+ Y \sim \mathcal P(\mu + \lambda)$.

• Thank you! but what happens if they are not independent? – user31280 Oct 25 '12 at 20:20
• In general we can't say anything then. It depends on how they depend on another. – martini Oct 25 '12 at 20:22
• Thank you! it's very simple and I feel like a complete idiot. – user31280 Oct 25 '12 at 20:40
• Nice 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. – javadba Aug 30 '14 at 20:59
• @LiorA Yes. k! included to combine with the rest and simplify as intended, so 1/k! is included to compensate. – Rolazaro Azeveires Jan 7 '18 at 14:23

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 $X+Y\sim\mathrm{po}(\lambda+\mu)$.

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 (1-t)}\\ \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 (1-t)}\\ \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 (1-t)}e^{-\mu (1-t)}\\ &= e^{-(\lambda+\mu) (1-t)}\\ \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($\lambda+\mu$)

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 $$Pr(X+Y=k)=\sum_{i=0}^kPr(Y=k-i)Pr(X=i)$$ which is \begin{align} Pr(X+Y=k)&=\sum_{i=0}^k\frac{e^{-\lambda_y}\lambda_y^{k-i}}{(k-i)!}\frac{e^{-\lambda_x}\lambda_x^i}{i!}\\ &=e^{-\lambda_y}e^{-\lambda_x}\sum_{i=0}^k\frac{\lambda_y^{k-i}}{(k-i)!}\frac{\lambda_x^i}{i!}\\ &=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k\frac{k!}{i!(k-i)!}\lambda_y^{k-i}\lambda_x^i\\ &=\frac{e^{-(\lambda_y+\lambda_x)}}{k!}\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i \end{align} The sum part is just $$\sum_{i=0}^k{k\choose i}\lambda_y^{k-i}\lambda_x^i=(\lambda_y+\lambda_x)^k$$ 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 $Po(\lambda_y+\lambda_x)$.

• 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 6:55

### 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 $\mathcal{P}(\lambda)=e^{\lambda(e^t-1)}$(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^t-1)}e^{\mu(e^t-1)}\\&=e^{(\lambda+\mu)(e^t-1)} \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^t-1)} \end{align}

hint: $\sum_{k=0}^{n} P(X = k)P(Y = n-k)$

• why this hint, why the sum? This is what I don't understand – user31280 Oct 25 '12 at 20:22
• adding two random variables is simply convolution of those random variables. That's why. – jay-sun Oct 25 '12 at 20:24
• gotcha! Thanks! – user31280 Oct 25 '12 at 20:31
• adding two random variables is simply convolution of those random variables... Sorry but no. – Did Feb 13 '13 at 6:28
• There is no usual sense for convolution of random variables. Either convolution of distributions or addition of random variables. – Did Feb 13 '13 at 6:51

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 successdul has a negligible probability. Then we are done.