# X and Y are geometric RV's with parameter p. A) P{X+Y=n}(n=1,2,...) ?

Jaxon Hamilton 2022-07-18 Answered
X and Y are geometric RV's with parameter p.
A) $P\left\{X+Y=n\right\}\left(n=1,2,...\right)$?
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Cael Cox
Step 1
I am assuming X and Y are independent. Without this assumption, it is not possible to answer your question without further information.
Step 2
So, you have to be a little more clever: Suppose we let $Z=X+Y$. We wish to find $Pr\left[Z=n\right]$. Note that we can write $Pr\left[Z=n\right]=\sum _{y=0}^{n}Pr\left[\left(X=n-y\right)\cap \left(Y=y\right)\right]=\sum _{y=0}^{n}Pr\left[X=n-y\right]Pr\left[Y=y\right].$
Now you know what each of these probabilities is, so write them out, and calculate the sum.

We have step-by-step solutions for your answer!

Step 1
For the shifted geometric distribution, of X failures before success on trial $X+1$, $X\sim \mathcal{S}\mathcal{G}\mathcal{e}\mathcal{o}\left(p\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}Pr\left(X=x\right)=p\left(1-p{\right)}^{x}$

Step 2
For the geometric distribution of $X-1$ failures before success on trial X,
$X\sim \mathcal{G}\mathcal{e}\mathcal{o}\left(p\right)\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}Pr\left(X=x\right)=p\left(1-p{\right)}^{x-1}$

We have step-by-step solutions for your answer!