# Let X and Y are independent random variables following geometric distribution with parameter p. Find the distribution of X given that X+Y=n.

Conditional probability distribution with geometric random variables
Let X and Y are independent random variables following geometric distribution with parameter p. Find the distribution of X given that $X+Y=n$.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

ahem37
Step 1
I will assume that X, Y are the numbers of trials until the first success, where the probability of success on any trial is p. Then $Pr\left(X=j\right)=Pr\left(Y=j\right)=\left(1-p{\right)}^{j-1}p$.
Let A be the event $X=i$ and let B be the event $X+Y=n$. Then by the definition of conditional probability, we have $Pr\left(A|B\right)=\frac{Pr\left(A\cap B\right)}{Pr\left(B\right)}.$
The probability of $A\cap B$ is easy to compute. It is $Pr\left(X=i\right)Pr\left(Y=n-i\right)$. This is $\left(1-p{\right)}^{i-1}p\left(1-p{\right)}^{n-i-1}p$, which simplifies to $\left(1-p{\right)}^{n-2}{p}^{2}$.
Step 2
For the probability that $X+Y=n$, we can use the fact that $Pr\left(X+Y=n\right)=Pr\left(X=1\right)Pr\left(Y=n-1\right)+Pr\left(X=2\right)Pr\left(Y=n-2\right)+\cdots +Pr\left(X=n-1\right)Pr\left(Y=1\right).$.
This simplifies to $\left(n-1\right)\left(\left(1-p{\right)}^{n-2}{p}^{2}$.
Divide. We get $\frac{1}{n-1}$. The conditional probability is discrete uniform. One can save a little time in the calculation of $Pr\left(X+Y=n\right)$ by noting that $X+Y$ has negative binomial distribution: it is the number of trials until the second success.

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