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.

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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.