Index: The Book of Statistical ProofsProbability Distributions ▷ Univariate discrete distributions ▷ Bernoulli distribution ▷ Range of variance

Theorem: Let $X$ be a random variable following a Bernoulli distribution:

$\label{eq:bern} X \sim \mathrm{Bern}(p) \; .$

Then, the variance of $X$ is necessarily between 0 and 1/4:

$\label{eq:bern-var-range} 0 \leq \mathrm{Var}(X) \leq \frac{1}{4} \; .$

Proof: The variance of a Bernoulli random variable is

$\label{eq:bern-var} X \sim \mathrm{Bern}(p) \quad \Rightarrow \quad \mathrm{Var}(X) = p \, (1-p)$

which can also be understood as a function of the success probability $p$:

$\label{eq:bern-var-p} \mathrm{Var}(X) = \mathrm{Var}(p) = -p^2 + p \; .$

The first derivative of this function is

$\label{eq:dVar-dp} \frac{\mathrm{d}\mathrm{Var}(p)}{\mathrm{d}p} = -2 \, p + 1$

and setting this deriative to zero

$\label{eq:dVar-dp-0} \begin{split} \frac{\mathrm{d}\mathrm{Var}(p_M)}{\mathrm{d}p} &= 0 \\ 0 &= -2 \, p_M + 1 \\ p_M &= \frac{1}{2} \; , \end{split}$

we obtain the maximum possible variance

$\label{eq:bern-var-max} \mathrm{max}\left[\mathrm{Var}(X)\right] = \mathrm{Var}(p_M) = -\left( \frac{1}{2} \right)^2 + \frac{1}{2} = \frac{1}{4} \; .$

The function $\mathrm{Var}(p)$ is monotonically increasing for $0 < p < p_M$ as $\mathrm{d}\mathrm{Var}(p)/\mathrm{d}p > 0$ in this interval and it is monotonically decreasing for $p_M < p < 1$ as $\mathrm{d}\mathrm{Var}(p)/\mathrm{d}p < 0$ in this interval. Moreover, as variance is always non-negative, the minimum variance is

$\label{eq:bern-var-min} \mathrm{min}\left[\mathrm{Var}(X)\right] = \mathrm{Var}(0) = \mathrm{Var}(1) = 0 \; .$

Thus, we have:

$\label{eq:bern-var-int} \mathrm{Var}(p) \in \left[ 0, \; \frac{1}{4} \right] \; .$
Sources:

Metadata: ID: P303 | shortcut: bern-varrange | author: JoramSoch | date: 2022-01-27, 09:03.