Question

Use the conditional variance formula to find the variance of a geometric random variable.

Short answer

The variance of a geometric random variable with probability of success \(p\) can be found using the conditional variance formula. First, find the expected value \(E(X) = \frac{1}{p}\) and the second moment \(E(X^2) = \frac{2}{p^2}\). Then, use the variance formula: \(Var(X) = E(X^2) - [E(X)]^2\), which gives \(Var(X) = \frac{1}{p^2}\).

Step-by-step solution

1. Understanding Geometric Random Variables and Variance Formula

A geometric random variable, denoted as \(X\), represents the number of trials needed to get the first success in independent Bernoulli trials. Let \(p\) be the probability of success in each trial. The probability mass function (PMF) of a geometric random variable is given by: \[ P(X = k) = (1 - p)^{k-1} p \] where \(k\) is the number of trials. The variance formula for a random variable is given by: \[ Var(X) = E(X^2) - [E(X)]^2 \] Our goal is to derive the expressions for the expected value \(E(X)\) and the second moment \(E(X^2)\), and then use the formula to find the variance.

2. Find the Expected Value (E(X)) of a Geometric Random Variable

The expected value of a geometric random variable can be found with: \[ E(X) = \sum_{k=1}^{\infty} kP(X = k) \] Substituting the PMF in the formula above, we get: \[ E(X) = \sum_{k=1}^{\infty} k (1 - p)^{k-1} p \] Let's evaluate the sum: \[E(X) = p\sum_{k=1}^\infty k(1-p)^{k-1}\] \[E(X) = p(1-p)^0 + 2p(1-p)^1 + 3p(1-p)^2 + \cdots\] Now, let's consider \((1 - p)E(X)\): \[(1 - p) E(X) = 0 + p(1-p)^0 + 2p(1-p)^1 + 3p(1-p)^2 + \cdots\] We can now subtract \((1 - p)E(X)\) from \(E(X)\) to obtain: \[pE(X) = p + (1 - p)^1 + (1-p)^2 + (1-p)^3 + \cdots\] This is the geometric series with a common ratio of \((1 - p)\): \[pE(X) = \frac{p}{1 - (1 - p)} = \frac{p}{p} = 1\] Therefore, \(E(X) = \frac{1}{p}\).

3. Find the Second Moment (E(X^2)) of a Geometric Random Variable

To find the second moment, we consider: \[E(X^2) = \sum_{k=1}^{\infty} k^2(1 - p)^{k-1} p \] This is slightly trickier to obtain directly. However, we can use the technique of differentiation. Recall that for \(|r| < 1\), \[\sum_{k=0}^{\infty} r^k = \frac{1}{1 - r} \] Now, differentiate both sides of the above equation with respect to r: \[\sum_{k=1}^{\infty} kr^{k-1} = \frac{1}{(1 - r)^2} \] Differentiate once more: \[\sum_{k=2}^{\infty} k(k-1)r^{k-2} = \frac{2}{(1 - r)^3} \] Substitute \(r = 1 - p\), and multiply both sides by \(p^2\): \[E(X^2) = \frac{2p^2}{(2p)^3} = \frac{2}{p^2}\]

4. Calculate the Variance of a Geometric Random Variable

Now that we have the expected value and the second moment, we can compute the variance of a geometric random variable using the formula: \[ Var(X) = E(X^2) - [E(X)]^2 \] Substitute the values: \[ Var(X) = \frac{2}{p^2} - \left(\frac{1}{p}\right)^2 \] Finally, \[ Var(X) = \frac{2}{p^2} - \frac{1}{p^2} = \frac{1}{p^2} \] So, the variance of a geometric random variable is \(\frac{1}{p^2}\).