CDF
\[
\small{
\begin{aligned}
F_X(k)=\text{P}(X\leq k)&=\sum_{i=1}^k (1-p)^{i-1}p \\
&=p\cdot\frac{1-(1-p)^k}{1-(1-p)} \\
&=1-(1-p)^k \\
\end{aligned}
}
\]
\[
\small{
\begin{aligned}
\text{An easier proof:} \; F_X(k)=\text{P}(X\leq k)&=1-\text{P}(X>k) \\
&=1-\text{P}(\text{all $k$ tosses are tails})\\
&=1-(1-p)^k \\
\end{aligned}
}
\]
Expectation & variance
\[
p_X(k)=\text{P}(X=k)=(1-p)^{k-1}p, \;\;\;\;\;\text{for } k=1, 2, \cdots
\]
\[
\begin{aligned}
\text{E}[X]&=\sum_{k=1}^{\infty} k (1-p)^{k-1}p = \frac{1}{p} \\
\\
\text{var}(X)&=\text{E}[X^2]-\big(\text{E}[X]\big)^2=\frac{1-p}{p^2} \\
\end{aligned}
\]
Given that a coin has been tossed \(t\) times with all tails, the probability of needing more than another \(s\) tosses to see the first heads, is the same as the probability of simply needing more than \(s\) tosses to see the first heads originally.
\[\small{
\text{P}(X>t+s|X>t)=\text{P}(X>s) \\
\text{for all nonnegative integers $t$ and $s$}
}\]
It’s called memoryless because the past (the last \(t\) coin toss were all tails) has no bearing on its future (the probability of needing at least \(s\) more tosses before seeing the first heads).
\[
\small{
\begin{aligned}
\text{P}(X>t+s|X>t)&=\frac{\text{P}\big((X>t+s) \cap (X>t)\big)}{\text{P}(X>t)} \\
\\
&=\frac{\text{P}(X>t+s)}{\text{P}(X>t)} \\
\\
&=\frac{(1-p)^{t+s}}{(1-p)^t} \\
\\
&=(1-p)^s =\text{P}(X>s) \\
\end{aligned}
}
\]
