11.2  Key Properties of a Geometric Random Variable
11.2  Key Properties of a Geometric Random VariableOn this page, we state and then prove four properties of a geometric random variable. In order to prove the properties, we need to recall the sum of the geometric series. So, we may as well get that out of the way first.
Recall

The sum of a geometric series is:
\(g(r)=\sum\limits_{k=0}^\infty ar^k=a+ar+ar^2+ar^3+\cdots=\dfrac{a}{1r}=a(1r)^{1}\)

Then, taking the derivatives of both sides, the first derivative with respect to \(r\) must be:
\(g'(r)=\sum\limits_{k=1}^\infty akr^{k1}=0+a+2ar+3ar^2+\cdots=\dfrac{a}{(1r)^2}=a(1r)^{2}\)

And, taking the derivatives of both sides again, the second derivative with respect to \(r\) must be:
\(g''(r)=\sum\limits_{k=2}^\infty ak(k1)r^{k2}=0+0+2a+6ar+\cdots=\dfrac{2a}{(1r)^3}=2a(1r)^{3}\)
We'll use the sum of the geometric series, first point, in proving the first two of the following four properties. And, we'll use the first derivative, second point, in proving the third property, and the second derivative, third point, in proving the fourth property. Let's jump right in now!
The probability mass function:
\(f(x)=P(X=x)=(1p)^{x1} p\)
\(0<p<1\), \(x=1, 2, \ldots\) for a geometric random variable \(X\) is a valid p.m.f.
Proof
Theorem
The cumulative distribution function of a geometric random variable \(X\) is:
\(F(x)=P(X\leq x)=1(1p)^x\)
Proof
Theorem
The mean of a geometric random variable \(X\) is:
\(\mu=E(X)=\dfrac{1}{p}\)
Proof
Theorem
The variance of a geometric random variable \(X\) is:
\(\sigma^2=Var(X)=\dfrac{1p}{p^2}\)
Proof
To find the variance, we are going to use that trick of "adding zero" to the shortcut formula for the variance. Recall that the shortcut formula is:
\(\sigma^2=Var(X)=E(X^2)[E(X)]^2\)
We "add zero" by adding and subtracting \(E(X)\) to get:
\(\sigma^2=E(X^2)E(X)+E(X)[E(X)]^2=E[X(X1)]+E(X)[E(X)]^2\)
Then, here's how the rest of the proof goes: