假设从正态分布中抽取 5 个样本给你， {0, 3, 1, -1, 2} ，我们想使用最大似然估计来估计方差。下列哪一项最接近解？

Suppose you are given 5 samples that are drawn from a normal distribution, {0, 3, 1, -1, 2}, and we want to use maximum likelihood estimation for estimating the variance. Which of the following is closest to the solution?

Question

假设从正态分布中抽取  5  个样本给你， {0, 3, 1, -1, 2} ，我们想使用最大似然估计来估计方差。下列哪一项最接近解？

Suppose you are given 5 samples that are drawn from a normal distribution, {0, 3, 1, -1, 2}, and we want to use maximum likelihood estimation for estimating the variance. Which of the following is closest to the solution?

BlackTom AI · Accepted Answer

For maximum likelihood estimation of the variance σ^2 of a normal distribution with unknown mean, the MLE uses the sample mean and divides the sum of squared deviations by n (not n−1).

1) Relevant knowledge points and concepts
- For data x1,...,xn from N(μ,σ^2) with both μ and σ^2 unknown, the MLE for μ is the sample mean x̄ = (1/n)Σ xi.
- The MLE for the variance is σ^2_MLE = (1/n) Σ (xi − x̄)^2. Note this differs from the unbiased sample variance which uses denominator (n−1).

2) Complete reasoning process step by step
- Compute the sample mean x̄.
- Compute each deviation xi − x̄, square them, sum them.
- Divide that sum by n to get σ^2_MLE.

3) Calculations with intermediate steps
- Given samples: {0, 3, 1, −1, 2}. n = 5.
- Sum = 0 + 3 + 1 + (−1) + 2 = 5. So x̄ = 5/5 = 1.
- Deviations: 0−1 = −1; 3−1 = 2; 1−1 = 0; −1−1 = −2; 2−1 = 1.
- Squared deviations: (−1)^2 = 1; 2^2 = 4; 0^2 = 0; (−2)^2 = 4; 1^2 = 1.
- Sum of squared deviations = 1 + 4 + 0 + 4 + 1 = 10.
- σ^2_MLE = (1/n) Σ (xi − x̄)^2 = 10/5 = 2.

4) Why the correct answer is correct
Because the maximum likelihood estimator for variance divides the sum of squared deviations by n, applying that formula to these data yields 2 exactly. This is the value that maximizes the likelihood under the normal model.

5) Why the other options are incorrect
- 10: This is the total sum of squared deviations, not the variance estimate (it must be divided by n).
- 5: This equals the sample size n, not related to the variance here.
- 1: This equals the sample mean, not the variance; also not the MLE. The unbiased variance would be 10/(n−1)=10/4=2.5, so 1 is not correct in either sense.

6) Conclusion
The MLE for the variance is 2, which matches the provided correct answer.

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