# Suppose that X is a geometric random variable with parameter (probability of success) p. Show that Pr(X>a+b∣X>a)=Pr(X>b).

Geometric Distribution Probability Problem
Suppose that X is a geometric random variable with parameter (probability of success) p.
Show that $Pr\left(X>a+b\mid X>a\right)=Pr\left(X>b\right)$.
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ruinsraidy4
Step 1
There are two ways to get $Pr\left(X>n\right)$, the easy way and the hard way. Since there is something to be learned from both, we do it both ways.
The hard way: The probability that $X=n+1$ is the probability of n failures in a row, then success. This is $\left(1-p{\right)}^{n}p$
Similarly, the probability that $X=n+2$ is the probability of $n+1$ failures in a row, then success. This has probability $\left(1-p{\right)}^{n+1}p$.
Continue, and add up. There is a $\left(1-p{\right)}^{n}p$ "in" each term. Take it out as a common factor. So we get $\left(1-p{\right)}^{n}p\left(1+\left(1-p\right)+\left(1-p{\right)}^{2}+\cdots \right).$
The geometric series inside the brackets has sum $\frac{1}{1-\left(1-p\right)}=\frac{1}{p}$
Step 2
The easy way: We have $X>n$ precisely if we get n failures in a row, probability $\left(1-p{\right)}^{n}$.
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koraby2bc
Step 1
Here's how to handle an infinite geometric series when the index starts at n instead of 0:
$\begin{array}{rl}\sum _{k=n}^{\mathrm{\infty }}a{x}^{k}& =a{x}^{n}+a{x}^{n+1}+a{x}^{n+2}+a{x}^{n+3}+\cdots \\ & =\left(a{x}^{n}\right)+\left(a{x}^{n}\right)x+\left(a{x}^{n}\right){x}^{2}+\left(a{x}^{n}\right){x}^{3}+\cdots \\ & =b+bx+b{x}^{2}+b{x}^{3}+\cdots \end{array}$
Now it starts at index 0. And of course b is $a{x}^{n}$.
It looks as if you can do the rest.
Second method: You mentioned that the probability of "success" is p. That means the probability of success on each trial is p.
If X is defined as the number of trials needed to get one success, then the event $X>n$ is the same as the event of failure on all of the first n trials, so that probability of that is $\left(1-p{\right)}^{n}$.
If X is defined as the number of trials needed to get one failure, then the event $X>n$ is the same as the event of success on all of the first n trials, so the probability of that is ${p}^{n}$.