By setting the derivatives of the log likelihood function (1.54) with respect to 𝜇 and 𝜎2
equal to zero, verify the results (1.55) and (1.56).

To verify the results (1.55) and (1.56) by setting the derivatives of the log-likelihood function with respect to \(\mu\) and \(\sigma^2\) equal to zero, we first need to start from the log-likelihood function itself, and then compute the derivatives.

Assume we are dealing with a normal (Gaussian) distribution for a sample \(\{x_1, x_2, \dots, x_n\}\), where the parameters \(\mu\) (mean) and \(\sigma^2\) (variance) are unknown. The log-likelihood function \(L(\mu, \sigma^2)\) for the normal distribution is given by:

\[
L(\mu, \sigma^2) = -\frac{n}{2} \log(2\pi) - \frac{n}{2} \log(\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^{n} (x_i - \mu)^2
\]

### Step 1: Derivative with respect to \(\mu\)

To find the maximum likelihood estimate (MLE) for \(\mu\), we take the derivative of the log-likelihood function with respect to \(\mu\):

\[
\frac{\partial L}{\partial \mu} = \frac{1}{\sigma^2} \sum_{i=1}^{n} (x_i - \mu)
\]

Set this equal to zero:

\[
\frac{1}{\sigma^2} \sum_{i=1}^{n} (x_i - \mu) = 0
\]

Multiplying both sides by \(\sigma^2\):

\[
\sum_{i=1}^{n} (x_i - \mu) = 0
\]

This simplifies to:

\[
\sum_{i=1}^{n} x_i = n \mu
\]

Solving for \(\mu\), we get:

\[
\mu = \frac{1}{n} \sum_{i=1}^{n} x_i
\]

This is the maximum likelihood estimate of the mean, which corresponds to result (1.55).

### Step 2: Derivative with respect to \(\sigma^2\)

Now, we take the derivative of the log-likelihood function with respect to \(\sigma^2\):

\[
\frac{\partial L}{\partial \sigma^2} = -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4} \sum_{i=1}^{n} (x_i - \mu)^2
\]

Set this equal to zero:

\[
-\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4} \sum_{i=1}^{n} (x_i - \mu)^2 = 0
\]

Multiplying through by \(2\sigma^4\):

\[
-n\sigma^2 + \sum_{i=1}^{n} (x_i - \mu)^2 = 0
\]

This simplifies to:

\[
\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (x_i - \mu)^2
\]

This is the maximum likelihood estimate of the variance, which corresponds to result (1.56).

### Conclusion

By setting the derivatives of the log-likelihood function with respect to \(\mu\) and \(\sigma^2\) equal to zero, we verified the results (1.55) and (1.56) for the maximum likelihood estimates of the mean and variance for a normal distribution.

---

Do you want further details or clarifications on any step?  
Here are some related questions to explore further:

1. How does the MLE for \(\mu\) compare with the sample mean?
2. What assumptions are made about the normal distribution in deriving these results?
3. How does the result for \(\sigma^2\) change if we were to use the unbiased sample variance estimator?
4. Can we generalize this procedure for distributions other than the normal distribution?
5. What role does the log-likelihood function play in the context of Bayesian inference?

**Tip**: When working with likelihood functions, taking the log of the likelihood simplifies calculations, especially when dealing with products of probabilities.

By setting the derivatives of the log likelihood function (1.54) with respect to 𝜇 and 𝜎^2 equal to zero, verify the results (1.55) and (1.56).

Explore how to verify the maximum likelihood estimates (MLE) for the mean (μ) and variance (σ^2) in a normal distribution by setting the derivatives of the log-likelihood function to zero. Includes detailed derivations and key statistical formulas, suitable for advanced students.

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