Question

The variance of Bernoulli random variable is π(1-π) where π is the probability of a successful Bernoulli trial (i.e. probability that x = 1). Using the formula for the probability function (p.f.) of a Bernoulli random variable, and the formula for variance, prove that V(X) = π(1-π).

Answer 1

The variance of Bernoulli random variable is $\pi(1 - \pi)$ where $\pi$ is the probability of a successful Bernoulli trial (i.e. probability that $x = 1$). Using the formula for the probability function (p.f.) of a Bernoulli random variable, and the formula for variance, prove that $V(X) = \pi(1 - \pi)$.

Solution

From the definition of variance:

$$V(X) = E \left( \left( X - E(X) \right)^2 \right)$$

From the Expectation of Bernoulli Distribution, we have $E(X) = \pi$.

Then by definition of Bernoulli distribution:

$$\begin{array}{l} V(X) = E \left( \left( X - E(X) \right)^2 \right) = (1 - \pi)^2 * \pi + (0 - \pi)^2 * (1 - \pi) \\ = \pi - 2\pi^2 + \pi^3 + \pi^2 - \pi^3 = \pi - \pi^2 = \pi(1 - \pi) \end{array}$$

Answer: $V(X) = \pi(1 - \pi)$.