Question

Use the conditional variance formula to determine the variance of a geometric random variable $X$ having parameter $p$.

Short answer

The variance of a geometric random variable $X$having parameter $p$is$Var\left(X\right)=\frac{1-p}{p^2}$.

Step-by-step solution

Given Information

The variance of a geometric random variable $X$ having parameter $p$.

Explanation

Suppose a sequence of independent Bernoulli random variables ${\left(X_n\right)}_{n\ge 1}$ with the parameter of success $p$. Let $X$mark the index of first random variable which has yielded a success. We know that $Z$has Geometric distribution with parameter$p$. Using the formula for conditional variance, we can condition on random variable $X_1$. We have that

$Var\left(X\right)=E\left(Var\left(X\mid X_1\right)\right)+Var\left(E\left(X\mid X_1\right)\right)$

$E\left(Var\left(X\mid X_1\right)\right)=E\left(E\left(X^2\mid X_1\right)-E{\left(X\mid X_1\right)}^2\right)$

$=P\left(X_1=0\right)·\left(E\left(X^2\mid X_1=0\right)-E{\left(X\mid X_1=0\right)}^2\right)$+ $P\left(X_1=1\right)\left(E\left(X^2\mid X_1=1\right)-E{\left(X\mid X_1=1\right)}^2\right)$

$=(1-p)\left(E\left(X^2\right)-E(X{)}^2\right)+0$

$=(1-p)Var\left(X\right)$

Explanation

$E\left(X\mid X_1\right)=1·I_{\left(X_1=1\right)}+\left(1+\frac{1}{p}\right)I_{\left(X_1=0\right)}$

Applying the variance,

$Var\left(E\left(X\mid X_1\right)\right)=p(1-p)+{\left(1+\frac{1}{p}\right)}^{2}p(1-p)-2p(1-p)\left(1+\frac{1}{p}\right)$

$Var\left(X\right)=(1-p)Var\left(X\right)+{\left(\frac{1}{p}\right)}^{2}p(1-p)$

$Var\left(X\right)=\frac{1-p}{p^2}$

Final Answer

The variance of a geometric random variable $X$ having parameter$p$ is$Var\left(X\right)=\frac{1-p}{p^2}$.