Variance: Max of Bernoulli Variables

Let $X_1, X_2$ be independent and identically distributed (i.i.d.) Bernoulli($p$) random variables, where $0 < p < 1$. We need to compute the variance of $Y = \max(X_1, X_2)$.

Bernoulli: Properties of $X_1$ and $X_2$

The properties of $X_1$ and $X_2$ are:

• $P(X_i=1) = p$ and $P(X_i=0) = 1-p$ for $i=1, 2$.
• The variance of a single Bernoulli variable is $\text{Var}(X_i) = p(1-p)$.

Max: Distribution of $Y = \max(X_1, X_2)$

The variable $Y$ takes values in $\{0, 1\}$. We determine its probability distribution:

• The event $Y=0$ occurs only if both $X_1=0$ and $X_2=0$. Due to independence:

  $ P(Y=0) = P(X_1=0 \text{ and } X_2=0) = P(X_1=0) P(X_2=0) = (1-p)(1-p) = (1-p)^2 $

• The event $Y=1$ occurs if at least one of $X_1$ or $X_2$ is 1. This is the complement of the event $Y=0$.

  $ P(Y=1) = 1 - P(Y=0) = 1 - (1-p)^2 $

  Expanding and simplifying this expression:

  $ P(Y=1) = 1 - (1 - 2p + p^2) = 2p - p^2 = p(2-p) $

Therefore, $Y$ follows a Bernoulli distribution with parameter $p' = P(Y=1) = p(2-p)$.

Variance: Calculation for $Y$

The variance of a Bernoulli random variable with parameter $\theta$ is given by $\text{Var}(\text{Bernoulli}(\theta)) = \theta(1-\theta)$.

Applying this formula to $Y$, we have:

$ \text{Var}(Y) = p'(1-p') $

Substitute $p' = p(2-p)$:

$ \text{Var}(Y) = [p(2-p)] [1 - p(2-p)] $

We simplify the second term: $1 - p(2-p) = 1 - 2p + p^2 = (1-p)^2$.

Substituting this back into the variance formula:

$ \text{Var}(Y) = [p(2-p)] [(1-p)^2] $

$ \text{Var}(Y) = p(2-p)(1-p)^2 $

Result: Final Variance

The variance of $\max(X_1, X_2)$ is $p(2-p)(1-p)^2$.