If $X$ is a geometric random variable, show analytically that$ P[X = n + k\mid X > n] = P [X = k]$ (Issue with reindexing a sum) So I actually found a solution online for this problem but I don't understand how to proceed past a certain step that involves reindexing a series. I have to admit I am a little bit rusty. Here's my attempt:
$$P(X=n+K)= \frac{P((X=n+k)\cap (X>n))}{P(X>n)}=\frac{P(X=n+k)}{P(X>n)}$$ If $x=n+k$ then it is certain that $x>n$. Thus the probability of the intersection is the same as the probability $P(X=n+k)$
$$\frac{P(X=n+k)}{P(X>n)}=\frac{(1-p)^{n+k-1}p}{\sum_{i=n+1}^\infty (1-p)^{i-1} p}$$
This is based off the PDF of a geometric random variable. The bottom simply sums each possibility that $X$ is greater than $n$.
I was unsure how to continue so I found a solution online. It says to reindex the sum from 1 to infinity using $j=i-n$ like so:
$${\sum_{j=1}^\infty (1-p)^{j-1+n}p}$$
The sum can then be simplified to $(1-p)^n$. I do not understand how $j$ was chosen to equal $1-n$ or why the exponent was changed to $j-1+n$. I would appreciate help understanding the reasoning.
 A: Your sum starts with $i=n+1$ and goes to $\infty$, so let's consider indexing the sum by $k=i-n$, which will start at $1$ and go to $\infty$. Here's an informal way of thinking about it.
$$\sum_{i=n+1}^\infty (1-p)^{i-1} p \stackrel{k=i-n}{=}
\sum_{n+k=n+1}^\infty (1-p)^{(n+k)-1} p = \sum_{k=1}^\infty (1-p)^{n+k-1} p$$
The reindexing is true because the sequence $(i,i+1,i+2,\dots)$ with $i=n+1$ is the same as $(n+k,n+k+1,n+k+2,\dots)$ with $k=1$.
A: How we do it?  For any infinite sequence of terms ${(a_k)}_{k\in\Bbb N}$ we have :  $${\sum_{i=n+1}^\infty a_i~{= \sum_{i-n=1}^{\infty} a_{(i-n+n)} \\= \sum_{j=1}^\infty a_{(j+n)}}}$$
Or if you prefer: $\sum\limits_{i=n+1}^\infty a_i = a_{n+1}+a_{n+2}+\cdots = \sum\limits_{j=1}^\infty a_{j+n}$
Why do we do it?  The geometric series is well known; for any $r$ such that $\lvert r\rvert< 1$ then:  $$\sum_{j=1}^\infty r^{j-1} = (1-r)^{-1}$$
So therefore:
$$\sum_{i=n+1}^\infty (1-p)^{i-1}p ~{= \sum_{i-n=1}^\infty (1-p)^{(i-n)+n-1}p \\ = \sum_{j=1}^\infty (1-p)^{j-1+n}p \\ = p(1-p)^n\sum_{j=1}^\infty (1-p)^{j-1} \\ = (1-p)^n}$$
A: Reindexing the sum is not the only way to do the problem, but knowing how to do it is worthwhile.
Here is the sum:
$$
\sum_{i\,=\,n+1}^\infty (1-p)^{i-1} p
$$
Every time $i$ increases by $1,$ the term is multiplied by the same thing: $1-p.$ Since it's the same thing every time, this is a geometric series. The common ratio is $1-p.$ The first term is what you get when $i=n+1,$ so that is $(1-p)^n p.$
If the first term is $a$ and the common ratio is $r$, and $-1<r<1,$ then we have
$$
a + ar + ar^2 + ar^3 + ar^4 + \cdots = \frac a {1-r}.
$$
So in this case we have
$$
\frac a {1-r} = \frac{(1-p)^n p}{1-(1-p)} = (1-p)^n.
$$
But this can also be done by reindexing:
\begin{align}
& \sum_{i\,=\,n+1}^\infty (1-p)^{i-1} p \\[10pt]
= {} & \underbrace{(1-p)^n p}_{i\,=\,n+1} + \underbrace{(1-p)^{n+1} p}_{i\,=\,n+2} + \underbrace{(1-p)^{n+2} p}_{i\,=\,n+3} + \cdots \\[12pt]
= {} & \overbrace{(1-p)^n p}^{j\,=\,1} + \overbrace{(1-p)^{n+1} p}^{j\,=\,2} + \overbrace{(1-p)^{n+2} p}^{j\,=\,3} + \cdots \\[12pt]
= {} & (1-p)^{1\,+\,\text{what?}} + (1-p)^{2\,+\,\text{what?}} + (1-p)^{3\,+\,\text{what?}} + \cdots 
\end{align}
The "what" has to be $n-1.$ So it's $\displaystyle \sum_{j\,=\,1}^\infty (1-p)^{j+n-1} p. $
