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Addition of two independent Poisson random variables
by Sangchun Han, PhD student in ECE
Question:
Let $ \mathbf{Z}=\mathbf{X}+\mathbf{Y} $ where $ \mathbf{X} $ and $ \mathbf{Y} $ are independent Poisson random variables with means $ \lambda $ and $ \mu $, respectively.
- a) Find the pmf of $ \mathbf{Z} $.
- b) Show that the conditional pmf of $ \mathbf{X} $ conditioned on the event $ \left\{ \mathbf{Z}=n\right\} $ is binomially distributed, and determine the parameters of binomial distribution ($ n $ and $ p $).
Solution
(a)
Find the pmf of $ \mathbf{Z} $.
According to the characteristic function of Poisson random variable
$ \Phi_{\mathbf{X}}(\omega)=e^{-\lambda\left(1-e^{i\omega}\right)},\Phi_{\mathbf{Y}}(\omega)=e^{-\mu\left(1-e^{i\omega}\right)}. $
$ \mathbf{X} $ and $ \mathbf{Y} $ are independent $ \Longrightarrow \mathbf{X} $ and $ \mathbf{Y} $ are uncorrelated $ \Longrightarrow e^{i\omega\mathbf{X}} $ and $ e^{i\omega\mathbf{Y}} $ are uncorrelated.
$ \Phi_{\mathbf{Z}}(\omega)=E\left[e^{i\omega\mathbf{Z}}\right]=E\left[e^{i\omega\left(\mathbf{X}+\mathbf{Y}\right)}\right]=E\left[e^{i\omega\mathbf{X}}e^{i\omega\mathbf{Y}}\right]=E\left[e^{i\omega\mathbf{X}}\right]\cdot E\left[e^{i\omega\mathbf{Y}}\right]=e^{-\lambda\left(1-e^{i\omega}\right)}\cdot e^{-\mu\left(1-e^{i\omega}\right)}=e^{-\left(\lambda+\mu\right)\left(1-e^{i\omega}\right)}. $
Now, we know that $ \mathbf{Z} $ is a Poisson random variable with mean $ \lambda+\mu $.
$ \therefore p_{\mathbf{Z}}(k)=\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{k}}{k!}. $
(b)
Show that the conditional pmf of $ \mathbf{X} $ conditioned on the event $ \left\{ \mathbf{Z}=n\right\} $ is binomially distributed, and determine the parameters of binomial distribution ($ n $ and $ p $).
$ P_{\mathbf{X}}\left(\mathbf{X}|\left\{ \mathbf{Z}=n\right\} \right)= P\left(\left\{ \mathbf{X}=k\right\} |\left\{ \mathbf{Z}=n\right\} \right)=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Z}=n\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)}=\frac{P\left(\left\{ \mathbf{X}=k\right\} \cap\left\{ \mathbf{Y}=n-k\right\} \right)}{P\left(\left\{ \mathbf{Z}=n\right\} \right)} $
$ =\frac{\frac{e^{-\lambda}\lambda^{k}}{k!}\cdot\frac{e^{-\mu}\mu^{n-k}}{\left(n-k\right)!}}{\frac{e^{-\left(\lambda+\mu\right)}\left(\lambda+\mu\right)^{n}}{n!}}=\left(\frac{n!}{k!\left(n-k\right)!}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k} $
$ =\left(\begin{array}{c} n\\ k \end{array}\right)\left(\frac{\lambda}{\lambda+\mu}\right)^{k}\left(\frac{\mu}{\lambda+\mu}\right)^{n-k}\;,\; k=0,\,1,\,2,\,\cdots. $
This is a binomial pmf $ b(n,p) $ with parameters $ n $ and $ p=\frac{\lambda}{\lambda+\mu} $.
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