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Given that x_1, ..., x_n are i.i.d. random variables with N(0, 6^2), calculate σ^2 MLE for 3 ≤ σ^2 ≤ 4.

User Arry
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Final answer:

The maximum likelihood estimator (MLE) for the variance, σ^2, can be obtained by taking the derivative of the log-likelihood function and setting it equal to zero. For a normal distribution with known mean, the MLE for the variance is given by: σ^2 MLE = (1/n) * Σ(x_i - μ)^2. To calculate σ^2 MLE for 3 ≤ σ^2 ≤ 4, we can substitute the values of n and μ into the formula to calculate the MLE for each value of σ^2 within the given range.

Step-by-step explanation:

The maximum likelihood estimator (MLE) for the variance, σ^2, can be obtained by taking the derivative of the log-likelihood function and setting it equal to zero. For a normal distribution with known mean, the MLE for the variance is given by:



σ^2 MLE = (1/n) * Σ(x_i - μ)^2



Given that x_1, ..., x_n are i.i.d. random variables with a normal distribution N(0, 6^2), where n is the sample size, μ is the known mean (0 in this case), and σ is the standard deviation (6 in this case), we can substitute these values into the MLE formula to calculate the MLE for σ^2.



To calculate σ^2 MLE for 3 ≤ σ^2 ≤ 4, we need to evaluate the MLE formula for these values of σ^2:



σ^2 MLE (when σ^2 = 3) = (1/n) * Σ(x_i - μ)^2



σ^2 MLE (when σ^2 = 4) = (1/n) * Σ(x_i - μ)^2



Substituting the values of n and μ into the formula, we can calculate the MLE for each value of σ^2 within the given range.

User Christian Palmer
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