Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter
𝑝
𝐿
(
𝑥
1
,
.
.
.
,
𝑥
𝑇
)
=
∏
𝑡
=
1
𝑇
𝑝
𝑥
𝑡
(
1
−
𝑝
)
1
−
𝑥
𝑡
.
If you estimate the parameter
𝑝
using a Maximum Likelihood estimator, you obtain the point estimate
𝑝
̂
=
1
𝑇
∑
𝑡
=
1
𝑇
𝑥
𝑡
, which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are
𝔼
(
𝑥
𝑡
)
=
𝑝
,
𝕍
(
𝑥
𝑡
)
=
𝑝
(
1
−
𝑝
)
.
Using this information, what is the variance of the estimator
𝕍
(
𝑝
̂
)
?

Question

Consider the likelihood of an i.i.d. sample from a Bernoulli population with parameter 
𝑝
𝐿
(
𝑥
1
,
.
.
.
,
𝑥
𝑇
)
=
∏
𝑡
=
1
𝑇
𝑝
𝑥
𝑡
(
1
−
𝑝
)
1
−
𝑥
𝑡
.
If you estimate the parameter 
𝑝
 using a Maximum Likelihood estimator, you obtain the point estimate 
𝑝
̂
=
1
𝑇
∑
𝑡
=
1
𝑇
𝑥
𝑡
, which corresponds to the sample mean. We know that for a Bernoulli random variable the expected value and the variance are
𝔼
(
𝑥
𝑡
)
=
𝑝
,
𝕍
(
𝑥
𝑡
)
=
𝑝
(
1
−
𝑝
)
.
Using this information, what is the variance of the estimator 
𝕍
(
𝑝
̂
)
?

BlackTom AI · Accepted Answer

We start by considering what we know about the Bernoulli(p) population and the estimator p-hat. The data x1, ..., xT are i.i.d. Bernoulli(p), and p-hat = (1/T) ∑ xt is the sample mean. For a Bernoulli random variable, E[xt] = p and Var(xt) = p(1−p). Since p-hat is the average of T independent copies, its variance is Var(p-hat) = Var((1/T) ∑ xt) = (1/T^2) ∑ Var(xt) by independence, which simplifies to (1/T^2) · T · p(1−p) = p(1−p)/T.

Now let's evaluate each option:

Option A: The variance of p-hat is V(p-hat) = p^2.
This is incorrect because p^2 would be the variance of a degenerate indicator that always equals p, not the variance of the sample mean. It ignores the variability introduced by the Bernoulli trials and the sample size T. So this option misunderstands the role of T and the variability across samples.

Option B: The variance of p-hat is V(p-hat) = p(1−p)·T.
This increases with T, which cannot be right for a sample mean of independent observations. As you collect more data, the estimator should become more precise, not less precise. The missing division by T shows why this overstates the uncertainty.

Option C: The variance of p-hat is V(p-hat) = p(1−p).
This would equal the variance of a single Bernoulli xt, not the average of T such observations. Since we average T independent draws, the dispersion shrinks by a factor of T, so this option ignores the averaging effect.

Option D: The variance of p-hat is V(p-hat) = E(p^2).
This expression is not even in terms of p(1−p) and does not reflect the true variance formula. E(p^2) is a fixed quantity unrelated to the sampling variance of p-hat and would be incorrect for this estimator.

Option E: All the answers are incorrect.
Given the calculation above, none of the alternatives A through D correctly match Var(p-hat) = p(1−p)/T. Since we identified the correct formula in the setup, this blanket statement would be incorrect if any of A–D were correct, but they are not. Therefore, this option would be chosen only if none of the prior formulas matched the known result, which is not the case here.

Putting it together, the correct variance for p-hat is p(1−p)/T, which corresponds to the standard result for the sample mean of i.i.d. Bernoulli trials.

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