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

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

$\label{eq:bin} X \sim \mathrm{Bin}(n,p) \; .$

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

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

Proof: By definition, a binomial random variable is the sum of $n$ independent and identical Bernoulli trials with success probability $p$. Therefore, the variance is

$\label{eq:bin-var-s1} \mathrm{Var}(X) = \mathrm{Var}(X_1 + \ldots + X_n)$

and because variances add up under independence, this is equal to

$\label{eq:bin-var-s2} \mathrm{Var}(X) = \mathrm{Var}(X_1) + \ldots + \mathrm{Var}(X_n) = \sum_{i=1}^{n} \mathrm{Var}(X_i) \; .$ $\label{eq:bern-var-range} 0 \leq \mathrm{Var}(X_i) \leq \frac{1}{4} \quad \text{for all} \quad i = 1,\ldots,n \; ,$

the minimum variance of $X$ is

$\label{eq:bin-var-min} \mathrm{min}\left[\mathrm{Var}(X)\right] = n \cdot 0 = 0$

and the maximum variance of $X$ is

$\label{eq:bin-var-max} \mathrm{max}\left[\mathrm{Var}(X)\right] = n \cdot \frac{1}{4} = \frac{n}{4} \; .$

Thus, we have:

$\label{eq:bin-var-int} \mathrm{Var}(X) \in \left[ 0, \; \frac{n}{4} \right] \; .$
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Metadata: ID: P304 | shortcut: bin-varrange | author: JoramSoch | date: 2022-01-27, 09:20.