Consider a Normal population with mean
𝜇
and variance
𝜎
2

Question

Consider a Normal population with mean 
𝜇
 and variance 
𝜎
2
<
∞
. We collect a sample of 
𝑇
=
3
 independent and identically distributed observations
𝑥
=
(
7
,
10
,
4
)
In class we have shown that the standardized log-likelihood of the sample for this case is
1
𝑇
log
(
𝐿
(
𝑥
,
𝜇
,
𝜎
2
)
)
=
−
1
2
log
(
2
𝜋
)
−
1
2
log
(
𝜎
2
)
−
1
𝑇
∑
𝑡
=
1
𝑇
(
𝑥
𝑡
−
𝜇
)
2
2
𝜎
2
.
Then the Maximum Likelihood estimate of the variance 
𝜎
2
 is:

BlackTom AI · Accepted Answer

We start by restating the problem in our own words to ensure clarity: we have a Normal population with unknown mean μ and variance σ^2, a sample x = (7, 10, 4) of size T = 3, and the standardized log-likelihood expression is given as

(1/T) log L(x, μ, σ^2) = -1/2 log(2π) - 1/2 log(σ^2) - [1/(2σ^2)] * (1/T) ∑ (x_t - μ)^2.

We are asked to find the maximum likelihood estimate of σ^2 based on this sample. The key steps revolve around two standard moves in MLE for a normal model: (1) compute the MLE of μ, and (2) plug that into the log-likelihood to maximize with respect to σ^2.

Option-by-option analysis follows:

Option A: σ̂^2 = 6.
Prestige of this choice lies in computing the sample variance around the MLE of μ. First, the MLE for μ with a normal model is the sample mean: μ̂ = (7 + 10 + 4)/3 = 21/3 = 7. Then compute the average squared deviation around μ̂: (1/3)[(7−7)^2 + (10−7)^2 + (4−7)^2] = (1/3)[0 + 9 + 9] = 18/3 = 6. This yields σ̂^2 = 6. If we accept μ̂ = 7 (as MLE for μ), this is exactly the result for the MLE of σ^2. This option is consistent with the standard MLE procedure for σ^2 in a normal model when μ is unknown.

Option B: There is not enough information to compute the value of σ̂^2.
This claim is incorrect. We do have the entire sample and can compute μ̂ = 7, then the variance around μ̂ as shown above. Therefore, the data provided are sufficient to determine σ̂^2, and no extra information is required for the MLE.

Option C: σ̂^2 = 8.
This would be the result if the average squared deviation around μ̂ were 8, i.e., if (1/3) ∑ (x_t − μ̂)^2 = 8. But as computed previously, ∑ (x_t − μ̂)^2 = (0)^2 + (3)^2 + (−3)^2 = 0 + 9 + 9 = 18, and (1/3) of that is 6, not 8. Hence this value arises from an incorrect calculation of the deviations or an incorrect μ̂, and is not the MLE given the data.

Option D: σ̂^2 = 9.
This would correspond to (1/3) ∑ (x_t − μ̂)^2 = 9, i.e., total sum of squares equal to 27. However, with μ̂ = 7, the total sum of squared deviations is 18, not 27, so 9 as the average squared deviation is not supported by the data. This miscounts the spread around the mean and is therefore incorrect.

Option E: σ̂^2 = 15.
Similarly, 15 would imply a much larger dispersion: (1/3) ∑ (x_t − μ̂)^2 = 15, which conflicts with the actual calculation of 6. The discrepancy indicates a misapplication of the MLE equations or using an incorrect mean value, so this choice is incorrect.

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