Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter p
L(x1,...,xT)=
T
∏

t=1pxt(1−p)1−xt.
If you estimate the parameter p using a Maximum Likelihood estimator, you obtain the point estimate
ˆ
p
=
1

T
∑
T
t=1
xt, which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are
𝔼(xt)=p,
𝕍(xt)=p(1−p).
Using this information, what is the variance of the estimator 𝕍(
ˆ
p
)?

Question

Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter p
L(x1,...,xT)=
T
∏

t=1pxt(1−p)1−xt.
If you estimate the parameter p using a Maximum Likelihood estimator, you obtain the point estimate 
ˆ
p
=
1

T
∑
T
t=1
xt, which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are
𝔼(xt)=p,
𝕍(xt)=p(1−p).
Using this information, what is the variance of the estimator 𝕍(
ˆ
p
)?

BlackTom AI · Accepted Answer

We start by recalling the setup: we have an i.i.d. sample from a Bernoulli(p) population, with xt ∈ {0,1} for t = 1,...,T, and the MLE for p is the sample mean ˆp = (1/T) Σ xt.

First, note a useful property of independent Bernoulli trials: the sum Σ xt follows a Binomial(T, p) distribution. Therefore, the sample mean ˆp = (1/T) Σ xt is simply the average of T independent Bernoulli(p) indicators.

Next, we compute the variance of ˆp using the variance of a sum of independent variables. Since Var(Σ xt) = Σ Var(xt) for independent xt, and Var(xt) = p(1 − p) for a Bernoulli(p) variable, we have Var(Σ xt) = T p(1 − p).

Then, because ˆp = (1/T) Σ xt, apply the scaling rule Var(aX) = a^2 Var(X) with a = 1/T. This gives Var(ˆp) = (1/T^2) Var(Σ xt) = (1/T^2) · T p(1 − p) = p(1 − p)/T.

Thus, the variance of the maximum likelihood estimator ˆp decreases linearly with the number of observations T and is equal to p(1 − p) divided by T.

Some common misconceptions to avoid: thinking the variance is p(1 − p) independent of T ignores the averaging effect of taking the mean; or mistaking it for p(1 − p)/√T or p(1 − p) · T would misrepresent how the uncertainty shrinks as more data are collected. The correct form is clearly p(1 − p)/T, reflecting both the Bernoulli variability and the averaging over T observations.

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