You need to use the Hessian matrix of the log-likelihood function

When you find the MLE from the log-likelihood function, typically its first derivative is going to be zero (or at least, approximately zero if you did things numerically). This occurs because the MLE typically falls at a critical point of the log-likelihood function. Consequently, the value $\Delta \mathcal{L} (\hat{\boldsymbol{\theta}})$ is not particularly useful, other than to check if your numerical solution ended at or near a critical point.

To form a confidence interval for the parameter vector $\boldsymbol{\theta}$ (or the individual parameters that are elements of this vector) you need to use the second derivative of the log-likelihood function, which is the Hessian matrix (the matrix of second-order partial derivatives of the log-likelihood function):

$$\mathbf{H}(\boldsymbol{\theta}) \equiv \bigg[ \frac{\partial^2 \mathcal{L}}{\partial \theta_i \partial_j} \bigg]_{i,j}.$$

Under certain regularity conditions (which basically require the MLE to fall at a critical point and require the log-likelihood to be sufficiently smooth) there is a classic theorem on the behaviour of the MLE which says that for large $n$ you have:$^\dagger$

$$\sqrt{n} (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \overset{\text{Approx}}{\sim} \text{N}(\mathbf{0}, \mathbf{H}^{-1}(\boldsymbol{\theta})).$$

This approximate distributional result can be used to form a confidence region for the entire parameter vector or individual (marginal) confidence intervals for the individual parameter values.

CI for the parameter vector: Taking $\chi_{n, \alpha}^2$ to be the critical point of the chi-squared distribution with $n$ degrees-of-freedom and upper tail $\alpha$, you can form the following confidence region for the parameter vector:

$$\text{CI}(1-\alpha) \equiv \bigg\{ \boldsymbol{\theta} \bigg| (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}})^\text{T} \mathbf{H}^{-1}(\hat{\boldsymbol{\theta}}) (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \leqslant \chi_{n, \alpha}^2 \bigg\}.$$

For large $n$ you have $\mathbb{P}(\boldsymbol{\theta} \in \text{CI}(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

CI for individual parameters: Taking $z_{\alpha/2}$ to be the critical point of the standard normal distribution with upper tail $\alpha/2$, you can form the following confidence region for an individual parameter:

$$\text{CI}_k(1-\alpha) \equiv \bigg[ \hat{\theta}_k - z_{\alpha/2} \hat{\text{se}}_{k}, \hat{\theta}_k + z_{\alpha/2} \hat{\text{se}}_{k} \bigg]. \quad \quad \quad \hat{\text{se}}_{k} \equiv \sqrt{\mathbf{H}^{-1}(\hat{\boldsymbol{\theta}})_{k,k}}.$$

For large $n$ you have $\mathbb{P}(\theta_k \in \text{CI}_k(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

$^\dagger$ For large $n$ you have $\hat{\boldsymbol{\theta}} \approx \boldsymbol{\theta}$ (due to consistency of the MLE under the regularity conditions) so the result is sometimes stated with the estimated parameter as the argument to the Hessian matrix. Both approximating results are valid for large $n$.

You need to use the Hessian matrix of the log-likelihood function

Note on notation: I am going to use slightly different notation to you, to conform with standard notation in the field. I will denote the log-likelihood function as $\ell$ (instead of $\mathcal{L}$) and the differentiation operator as $\nabla$ (instead of $\Delta$).

When you find the MLE from the log-likelihood function, typically its first derivative is going to be zero (or at least, approximately zero if you did things numerically). This occurs because the MLE typically falls at a critical point of the log-likelihood function. Consequently, the value $\nabla \ell (\hat{\boldsymbol{\theta}})$ is not particularly useful, other than to check if your numerical solution ended at or near a critical point.

To form a confidence interval for the parameter vector $\boldsymbol{\theta}$ (or the individual parameters that are elements of this vector) you need to use the second derivative of the log-likelihood function, which is the Hessian matrix (the matrix of second-order partial derivatives of the log-likelihood function):

$$\mathbf{H}(\boldsymbol{\theta}) \equiv \nabla^2 \ell(\boldsymbol{\theta}) = \bigg[ \frac{\partial^2 \ell}{\partial \theta_i \partial \theta_j} \bigg]_{i,j}.$$

Under certain regularity conditions (which basically require the MLE to fall at a critical point and require the log-likelihood to be sufficiently smooth) there is a classic theorem on the behaviour of the MLE which says that for large $n$ you have:$^\dagger$

$$\sqrt{n} (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \overset{\text{Approx}}{\sim} \text{N}(\mathbf{0}, \mathbf{H}^{-1}(\boldsymbol{\theta})).$$

This approximate distributional result can be used to form a confidence region for the entire parameter vector or individual (marginal) confidence intervals for the individual parameter values.

CI for the parameter vector: Taking $\chi_{n, \alpha}^2$ to be the critical point of the chi-squared distribution with $n$ degrees-of-freedom and upper tail $\alpha$, you can form the following confidence region for the parameter vector:

$$\text{CI}(1-\alpha) \equiv \bigg\{ \boldsymbol{\theta} \bigg| (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}})^\text{T} \mathbf{H}^{-1}(\hat{\boldsymbol{\theta}}) (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \leqslant \chi_{n, \alpha}^2 \bigg\}.$$

For large $n$ you have $\mathbb{P}(\boldsymbol{\theta} \in \text{CI}(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

CI for individual parameters: Taking $z_{\alpha/2}$ to be the critical point of the standard normal distribution with upper tail $\alpha/2$, you can form the following confidence region for an individual parameter:

$$\text{CI}_k(1-\alpha) \equiv \bigg[ \hat{\theta}_k - z_{\alpha/2} \hat{\text{se}}_{k}, \hat{\theta}_k + z_{\alpha/2} \hat{\text{se}}_{k} \bigg]. \quad \quad \quad \hat{\text{se}}_{k} \equiv \sqrt{\mathbf{H}^{-1}(\hat{\boldsymbol{\theta}})_{k,k}}.$$

For large $n$ you have $\mathbb{P}(\theta_k \in \text{CI}_k(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

$^\dagger$ For large $n$ you have $\hat{\boldsymbol{\theta}} \approx \boldsymbol{\theta}$ (due to consistency of the MLE under the regularity conditions) so the result is sometimes stated with the estimated parameter as the argument to the Hessian matrix. Both approximating results are valid for large $n$.

You need to use the Hessian matrix of the log-likelihood function

When you find the MLE from the log-likelihood function, typically its first derivative is going to be zero (or at least, approximately zero if you did things numerically). This occurs because the MLE typically falls at a critical point of the log-likelihood function. Consequently, the value $\Delta \mathcal{L} (\hat{\boldsymbol{\theta}})$ is not particularly useful, other than to check if your numerical solution ended at or near a critical point.

To form a confidence interval for the parameter vector $\boldsymbol{\theta}$ (or the individual parameters that are elements of this vector) you need to use the second derivative of the log-likelihood function, which is the Hessian matrix (the matrix of second-order partial derivatives of the log-likelihood function):

$$\mathbf{H}(\boldsymbol{\theta}) \equiv \bigg[ \frac{\partial^2 \mathcal{L}}{\partial \theta_i \partial_j} \bigg]_{i,j}.$$

Under certain regularity conditions (which basically require the MLE to fall at a critical point and require the log-likelihood to be sufficiently smooth) there is a classic theorem on the behaviour of the MLE which says that for large $n$ you have:$^\dagger$

$$\sqrt{n} (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \overset{\text{Approx}}{\sim} \text{N}(\mathbf{0}, \mathbf{H}^{-1}(\boldsymbol{\theta})).$$

This approximate distributional result can be used to form a confidence region for the entire parameter vector or individual (marginal) confidence intervals for the individual parameter values. Specifically, taking $\chi_{n, \alpha}^2$ to be the critical point of the chi-squared distribution with $n$ degrees-of-freedom and upper tail $\alpha$, you can form the following confidence region for the parameter vector:

$$\text{CI}_n(1-\alpha) \equiv \bigg\{ \boldsymbol{\theta} \bigg| (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}})^\text{T} \mathbf{H}^{-1}(\hat{\boldsymbol{\theta}}) (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \leqslant \chi_{n, \alpha}^2 \bigg\}.$$

For large $n$ you have $\mathbb{P}(\boldsymbol{\theta} \in \text{CI}_n(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

$^\dagger$ For large $n$ you have $\hat{\boldsymbol{\theta}} \approx \boldsymbol{\theta}$ (due to consistency of the MLE under the regularity conditions) so the result is sometimes stated with the estimated parameter as the argument to the Hessian matrix. Both approximating results are valid for large $n$.

You need to use the Hessian matrix of the log-likelihood function

When you find the MLE from the log-likelihood function, typically its first derivative is going to be zero (or at least, approximately zero if you did things numerically). This occurs because the MLE typically falls at a critical point of the log-likelihood function. Consequently, the value $\Delta \mathcal{L} (\hat{\boldsymbol{\theta}})$ is not particularly useful, other than to check if your numerical solution ended at or near a critical point.

To form a confidence interval for the parameter vector $\boldsymbol{\theta}$ (or the individual parameters that are elements of this vector) you need to use the second derivative of the log-likelihood function, which is the Hessian matrix (the matrix of second-order partial derivatives of the log-likelihood function):

$$\mathbf{H}(\boldsymbol{\theta}) \equiv \bigg[ \frac{\partial^2 \mathcal{L}}{\partial \theta_i \partial_j} \bigg]_{i,j}.$$

Under certain regularity conditions (which basically require the MLE to fall at a critical point and require the log-likelihood to be sufficiently smooth) there is a classic theorem on the behaviour of the MLE which says that for large $n$ you have:$^\dagger$

$$\sqrt{n} (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \overset{\text{Approx}}{\sim} \text{N}(\mathbf{0}, \mathbf{H}^{-1}(\boldsymbol{\theta})).$$

This approximate distributional result can be used to form a confidence region for the entire parameter vector or individual (marginal) confidence intervals for the individual parameter values.

CI for the parameter vector: Taking $\chi_{n, \alpha}^2$ to be the critical point of the chi-squared distribution with $n$ degrees-of-freedom and upper tail $\alpha$, you can form the following confidence region for the parameter vector:

$$\text{CI}(1-\alpha) \equiv \bigg\{ \boldsymbol{\theta} \bigg| (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}})^\text{T} \mathbf{H}^{-1}(\hat{\boldsymbol{\theta}}) (\boldsymbol{\theta} - \hat{\boldsymbol{\theta}}) \leqslant \chi_{n, \alpha}^2 \bigg\}.$$

For large $n$ you have $\mathbb{P}(\boldsymbol{\theta} \in \text{CI}(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

CI for individual parameters: Taking $z_{\alpha/2}$ to be the critical point of the standard normal distribution with upper tail $\alpha/2$, you can form the following confidence region for an individual parameter:

$$\text{CI}_k(1-\alpha) \equiv \bigg[ \hat{\theta}_k - z_{\alpha/2} \hat{\text{se}}_{k}, \hat{\theta}_k + z_{\alpha/2} \hat{\text{se}}_{k} \bigg]. \quad \quad \quad \hat{\text{se}}_{k} \equiv \sqrt{\mathbf{H}^{-1}(\hat{\boldsymbol{\theta}})_{k,k}}.$$

For large $n$ you have $\mathbb{P}(\theta_k \in \text{CI}_k(1-\alpha) | \boldsymbol{\theta}) \approx 1-\alpha$, which means that the confidence interval has roughly the correct coverage probability for large $n$.

$^\dagger$ For large $n$ you have $\hat{\boldsymbol{\theta}} \approx \boldsymbol{\theta}$ (due to consistency of the MLE under the regularity conditions) so the result is sometimes stated with the estimated parameter as the argument to the Hessian matrix. Both approximating results are valid for large $n$.