# Thread: Product of two discrete distributions is one of the same family?

1. ## Product of two discrete distributions is one of the same family?

Hi, I have this problem with which I could do with some help.

Given the marginal pdfs shown below, find the pdf of X + Y (given X and Y are independent).

$p_{X}(k) = \exp{(- \lambda)} \cdot \frac{\lambda^k}{k!}$

and

$p_{Y}(k) = \exp{(- \mu)} \cdot \frac{\mu^k}{k!}$

where $k = 1, 2, 3,\ldots$

I know from the solution manual this should be a distribution of the same form, namely

$p_{X+Y}(z) = \exp{-(\lambda + \mu)} \cdot \frac{(\lambda + \mu)^z}{z!}$

where $z = 1, 2, 3,\ldots$

But I don't see how this was arrived at. I'm especially miffed as to what z stands for here. Is $z = k_Y + k_X$ or what?

Many thanks in advance to anyone who can clarify this for me.

2. ## Re: Product of two discrete distributions is one of the same family?

Hello,

This is the sum, not the product you're looking for.

And k can be equal to 0. It's a Poisson distribution.

Think about what values can X+Y take. The same : 0,1,2,...
So for k=0,1,2,..., compute $P(X+Y=k)=\sum_{j=0}^k P(X=j;Y=k-j)$ (try to think why it's this way, then use the independence)

3. ## Re: Product of two discrete distributions is one of the same family?

Hi,
I see what you are saying I think. I should have said the distribution of the sum of two independent random variables. But to find the distribution for the sum X+Y, i.e. the joint, one has to take the sum over the product of the two distributions (that's where independence comes in), right ? i.e. for Z = X +Y canonically

$p_{Z}(z) = \sum_{all x} p_{X}(x) \cdot p_{Y}(z-x)$

so does this translate for our example as

$p_{X + Y = Z}(z)= \sum_{all k} \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}$

?

But I cant massage this form into the desired form. Am I setting things up incorrectly here?

And here the notation implies that k is the same for both initial distributions. Should the notation not more correctly be $k_Y = 1,2, \ldots$ and $k_X = 1,2, \ldots$

So that $z = k_Y + k_X$

Or am I missing something (as I guess is much more likely) ?

Thanks again.

4. ## Re: Product of two discrete distributions is one of the same family?

Originally Posted by Mathsdog
Hi,
I see what you are saying I think. I should have said the distribution of the sum of two independent random variables. But to find the distribution for the sum X+Y, i.e. the joint, one has to take the sum over the product of the two distributions (that's where independence comes in), right ? i.e. for Z = X +Y canonically

$p_{Z}(z) = \sum_{all x} p_{X}(x) \cdot p_{Y}(z-x)$

so does this translate for our example as

$p_{X + Y = Z}(z)= \sum_{all k} \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}$

?

But I cant massage this form into the desired form. Am I setting things up incorrectly here?

And here the notation implies that k is the same for both initial distributions. Should the notation not more correctly be $k_Y = 1,2, \ldots$ and $k_X = 1,2, \ldots$

So that $z = k_Y + k_X$

Or am I missing something (as I guess is much more likely) ?

Thanks again.
you are on the right track.

$p_{X + Y = Z}(z)= \sum_{k=0}^z \exp{(-\mu) \cdot\frac{\mu^k}{k!} \cdot \exp{(-\lambda) \cdot\frac{\lambda^{(z-k)}}{(z-k)!}$

$=e^{-(\mu+\lambda)} \sum_{k=0}^z \dfrac{z!}{z!}\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$

now use the binomial theorem

5. ## Re: Product of two discrete distributions is one of the same family?

Hi, Thanks for your post again,

If I multiply by $e^{-(\mu+\lambda)}\sum_{k=0}^z\dfrac{z!}{z!} \cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$ by $\frac{(1-\mu)^{z-k}}{(1-\mu)^{z-k}}$

then the sum over the binomial distribution with parameters (z, k) (equalling 1) falls out and I am left with

$e^{-(\mu+\lambda)} \cdot \frac{\lambda^{z-k}}{(1-\mu)^{z-k} \cdot z!}$

So I reckon I need to show that

$\frac{\lambda^{z-k}}{(1-\mu)^{z-k}} = (\lambda + \mu)^{z}$

in order to arrive at the final expression

$p_{X+Y}(z)= e^{-(\lambda + \mu)} \cdot \frac{(\lambda+\mu)^z}{z!}$

But I can't quite see how this final step works. Wanna put me out of my misery?

Thanks, MD

6. ## Re: Product of two discrete distributions is one of the same family?

Originally Posted by Mathsdog
Hi, Thanks for your post again,

If I multiply by $e^{-(\mu+\lambda)}\sum_{k=0}^z\dfrac{z!}{z!} \cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$ by $\frac{(1-\mu)^{z-k}}{(1-\mu)^{z-k}}$

then the sum over the binomial distribution with parameters (z, k) (equalling 1) falls out and I am left with

$e^{-(\mu+\lambda)} \cdot \frac{\lambda^{z-k}}{(1-\mu)^{z-k} \cdot z!}$

So I reckon I need to show that

$\frac{\lambda^{z-k}}{(1-\mu)^{z-k}} = (\lambda + \mu)^{z}$

in order to arrive at the final expression

$p_{X+Y}(z)= e^{-(\lambda + \mu)} \cdot \frac{(\lambda+\mu)^z}{z!}$

But I can't quite see how this final step works. Wanna put me out of my misery?

Thanks, MD
$e^{-(\mu+\lambda)} \sum_{k=0}^z \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$

multiplying by z!/z! gives:

$=e^{-(\mu+\lambda)} \sum_{k=0}^z \dfrac{z!}{z!}\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$

$=\dfrac{e^{-(\mu+\lambda)}}{z!} \sum_{k=0}^z z!\cdot \dfrac{\mu^k \lambda^{z-k}}{k!(z-k)!}$

$=\dfrac{e^{-(\mu+\lambda)}}{z!} \sum_{k=0}^z \dfrac{z!}{{k!(z-k)!}}\cdot \mu^k \lambda^{z-k}$

$=\dfrac{e^{-(\mu+\lambda)}}{z!} \underbrace{\sum_{k=0}^z \dbinom{z}{k}\cdot \mu^k \lambda^{z-k}}_{\text{This by binomial theorem is equal to} (\lambda+\mu)^z}$

$=\dfrac{e^{-(\mu+\lambda)}}{z!} (\lambda+\mu)^z}$