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Joram Soch |
BCCN Berlin |
joram.soch@bccn-berlin.de |
2022-01-20 07:19:00 -0800 |
Variance of the binomial distribution |
Probability Distributions |
Univariate discrete distributions |
Binomial distribution |
Variance |
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Wikipedia |
2022 |
Binomial distribution |
Wikipedia, the free encyclopedia |
retrieved on 2022-01-20 |
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P302 |
bin-var |
JoramSoch |
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
$$ \label{eq:bin-var}
\mathrm{Var}(X) = n p , (1-p) ; .
$$
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
$$ \label{eq:bin-var-s2}
\mathrm{Var}(X) = \mathrm{Var}(X_1) + \ldots + \mathrm{Var}(X_n) = \sum_{i=1}^{n} \mathrm{Var}(X_i) ; .
$$
With the variance of the Bernoulli distribution, we have:
$$ \label{eq:bin-var-s3}
\mathrm{Var}(X) = \sum_{i=1}^{n} p , (1-p) = n p , (1-p) ; .
$$