# Need help simplifying a summation with binomials

## Homework Statement

"Prove that $\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=\sum_{m+n=0}^\infty s^{n+m} e^{-(\lambda+\mu)} \frac{(\lambda + \mu)^{m+n}}{(m+n)}!$

## Homework Equations

Binomial theorem: $(x+y)^n=\sum_{k=0}^n x^ky^{n-k}$
Vandermonde's identity: $\binom {n+m} m =\sum_{k=0}^m \binom n k \binom m {m-k}$

## The Attempt at a Solution

$\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \frac{\lambda^n \mu^m }{m!n!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \binom {m+n}{n} \frac{\lambda^n \mu^m }{(m+n)!}$
$=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \sum_{k=0}^n \binom m k \binom n k \frac{\lambda^n \mu^m }{(m+n)!}$

I'm afraid to go any further, because it won't get me my $(\lambda + \mu)^{n+m}$ term. If anyone has any pointers on what I should do next with this expression (or giving me another expression), or an alternate way to prove using probability-generating functions, that the sum of two independent Poisson r.v.'s have mean equal to the sum of each individual mean, then that would be much appreciated.

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DrDu
Binomial formula?

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

"Prove that $\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=\sum_{m+n=0}^\infty s^{n+m} e^{-(\lambda+\mu)} \frac{(\lambda + \mu)^{m+n}}{(m+n)}!$

## Homework Equations

Binomial theorem: $(x+y)^n=\sum_{k=0}^n x^ky^{n-k}$
Vandermonde's identity: $\binom {n+m} m =\sum_{k=0}^m \binom n k \binom m {m-k}$

## The Attempt at a Solution

$\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!}\sum_{m=0}^\infty s^m e^{-\mu}\frac{\mu^m}{m!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \frac{\lambda^n \mu^m }{m!n!}=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \binom {m+n}{n} \frac{\lambda^n \mu^m }{(m+n)!}$
$=e^{-(\lambda+\mu)}\sum_{n=0}^\infty \sum_{m=0}^\infty s^{n+m} \sum_{k=0}^n \binom m k \binom n k \frac{\lambda^n \mu^m }{(m+n)!}$

I'm afraid to go any further, because it won't get me my $(\lambda + \mu)^{n+m}$ term. If anyone has any pointers on what I should do next with this expression (or giving me another expression), or an alternate way to prove using probability-generating functions, that the sum of two independent Poisson r.v.'s have mean equal to the sum of each individual mean, then that would be much appreciated.
Does your formula $(x+y)^n=\sum_{k=0}^n x^ky^{n-k}$ work for $n = 2$ or $n = 3?$

Stephen Tashi
or an alternate way to prove using probability-generating functions, that the sum of two independent Poisson r.v.'s have mean equal to the sum of each individual mean, then that would be much appreciated.
Are you allowed to use the result that the moment generating function of a poission distribution with parameter $\lambda$ is $M(t) = e^{\lambda (e^t -1)}$ ?

Ray Vickson
Homework Helper
Dearly Missed
Are you allowed to use the result that the moment generating function of a poission distribution with parameter $\lambda$ is $M(t) = e^{\lambda (e^t -1)}$ ?
That should be $e^{\lambda (t-1)}.$

Ray Vickson
Homework Helper
Dearly Missed
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StoneTemplePython
Gold Member
2019 Award
It wasnt clear to me what $s$ is in the original post, though I now believe we're classically talking about $s \in (-1,1)$ -- though $s \in (0,1)$ really is what is of interest -- as the original post appears to already using an Ordinary Generating Function, and hence the identity to be proven comes from the fact that by stochastic independence:

$\text{left hand side} = E\big[s^{X_1}\big]E\big[s^{X_2}\big] = E\big[s^{X_1}s^ {X_2}\big] = E\big[s^{X_1 + X_2}\big] = \text{right hand side}$

OP just needs to confirm that $g(X) = s^{X}$ is a random variable and that the transform doesn't change dependencies (the fact that generating functions are in principle invertible implies this)

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Equivalently, OP's question seems to be (while using OGFs) that the convolution of two Poissons with parameters $\lambda$ and $\mu$ is a Poisson with parameter $\lambda$ and $\mu$. There's a very elegant and probabilistic argument for this that uses memorylessness and the fact that there must be some constant $\alpha \gt 0$ where $\mu \cdot t = (\alpha \lambda) \cdot t = \lambda \cdot (\alpha t)$ ...

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a less probabilistic take would be to consider properties of the exponential function and simplify. E.g. for starters

$\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!} = e^{-\lambda} \big(\sum_{n=0}^\infty \frac{(s\lambda)^n}{n!}\big)=e^{-\lambda}\big(e^{s\lambda}\big) = e^{-\lambda + s\lambda}$

and apply this process to other parts of the original equation, then simplify.

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Does your formula $(x+y)^n=\sum_{k=0}^n x^ky^{n-k}$ work for n=2n = 2 or n=3?
Oops, it should be $(x+y)^n=\sum_{k=0}^n \binom n k x^ky^{n-k}$.

a less probabilistic take would be to consider properties of the exponential function and simplify. E.g. for starters

$\sum_{n=0}^\infty s^n e^{-\lambda} \frac{\lambda^n}{n!} = e^{-\lambda} \big(\sum_{n=0}^\infty \frac{(s\lambda)^n}{n!}\big)=e^{-\lambda}\big(e^{s\lambda}\big) = e^{-\lambda + s\lambda}$

and apply this process to other parts of the original equation, then simplify.
Oh, so $G_{X+Y}(s)=G_X(s)G_Y(s)=(e^{-\lambda + s\lambda})(e^{-\mu + s\mu})=e^{(\lambda+\mu)(s-1)}=\sum_{n=0}^\infty s^n e^{-(\lambda+\mu)} \frac{(\lambda+\mu)^n}{n!}$ implies that $X+Y$ has a distribution $\text{Poiss}(\lambda+\mu)$?

StoneTemplePython
Oh, so $G_{X+Y}(s)=G_X(s)G_Y(s)=(e^{-\lambda + s\lambda})(e^{-\mu + s\mu})=e^{(\lambda+\mu)(s-1)}=\sum_{n=0}^\infty s^n e^{-(\lambda+\mu)} \frac{(\lambda+\mu)^n}{n!}$ implies that $X+Y$ has a distribution $\text{Poiss}(\lambda+\mu)$?
Yes. That's really all there is to it from the OGF standpoint. Since an OGF uses a power series in $s$ and the Poisson uses the power series for the exponential function, it should be an easy result.