Cumulative distribution function for geometric random variable

Question

For geometric random variable $f(k)=(1-p)^{k-1}p$.

\begin{align} F(k) = P(X\leq x) &= 1-P(x>k) \\ &= 1 - \sum_{i=k+1}p(1-p)^{i-1} \\ &= 1 - (1-p)^k \sum_{i=1}^ \infty p(1-p)^{i-1} \\ &= 1 - (1-p)^k \end{align}

I would appreciate a breakdown of these steps. Especially why do we take $1-P(x>k)$ and what operations are preformed on the summation sign? Stating the obvious is very welcone, my mathematical background is quite limited.

Answer 1

The main idea is using the geometric series $$\sum_{i=0}^{\infty}a \, r^i = \frac{a}{1-r}$$ $P(X \leq k)$ is a finite sum, while $P(X>k)$ is an infinite series, specifically a geometric series. So to utilize the geometric series expression, instead of looking at $P(X \leq k)$ one looks at the equivalent $1-P(X>k)$.

Then $P(X>k)=\sum_{i=k+1}^{\infty}p(1-p)^{i-1}$ simplifies as done in your expression. I personally find it easier to look at the first few terms written out:

\begin{align}\sum_{i=k+1}^{\infty}p(1-p)^{i-1} &= p(1-p)^k + p(1-p)^{k+1} + p(1-p)^{k+2} + \dots \\ &= p(1-p)^k \left\{1+(1-p)+(1-p)^2+\ldots\right\}\\ &= p(1-p)^k \frac{1}{1-(1-p)} = (1-p)^k\end{align}

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Added to answer the questions in the comments:

$P(X>k)$ is the probability of $X$ taking values greater than $k$ so:

\begin{align} P(X>k) & = P(X=k+1)+P(X=k+2)+P(X=k+3)+\dots \\ & = p(1-p)^{k+1-1} + p(1-p)^{k+2-1}+ p(1-p)^{k+3-1}+\dots \end{align}

This is the same as writing $\sum_{i=k+1}^{\infty}p(1-p)^{i-1}$.

$1+(1-p)+(1-p)^2+\ldots$ is the geometric series with $a=1$ and $r=1-p$. Plugging them in the formula $\frac{a}{1-r}$ to get $\frac{1}{1-(1-p)}$.

Answer 2

Actually, we don't need the knowledge of geometric series to prove this.

$P(X>k)$ is the probability of the event where the first k tosses/trials... result in tails/failures. Therefore, the probability is $(1-p)^k$. As the cumulative distribution function is the complement of $P(X>k)$, we have the final result = $1 - (1-p)^k$