Looking at the Fisher information matrix for a simple linear model, such as here, I do not understand how to use the matrix to compute confidence intervals. There are multiple examples on the internet showing how to obtain the matrix, but I suppose that since it is thought to be trivial, it is nowhere shown how to use it.

My understanding after some reading, is that confidence intervals for a parameter under a normal distribution should be computed from $CI_{\theta} = \theta \pm 1.96 * \sqrt{\frac{1}{I(\theta)}}$.

For example, if I wanted to compute the 95% confidence interval for $\beta_1$ in the model $yi=\beta_0+\beta_1x_i+\epsilon_i$, how should I proceed given the matrix $\left[ \begin{matrix} \tfrac{n}{\sigma^2} & \tfrac{\sum_i^n x_i}{\sigma^2} & 0\\ \tfrac{\sum_i^n x_i}{\sigma^2} & \tfrac{\sum_i^n x_i^2}{\sigma^2} & 0\\ 0 & 0 & \tfrac{n}{2\sigma^4} \end{matrix} \right]$?


1 Answer 1


You have three parameters in your problem, so $\theta = (\beta_0, \beta_1, \sigma^2)$. $I(\theta)$ is a matrix and you cannot "divide by" $I(\theta)$, as in the formula in your second paragraph.

What you need instead is to take the inverse of $I(\theta)$. If you are looking for the confidence interval e.g. of $\beta_1$, you would take the element $[2,2]$ of $I(\theta)^{-1}$. and plug it in place of $1/I(\theta)$ in your formula. You would have then

$$CI_{\beta_1} = \hat\beta_1 \pm 1.96\sqrt{{I(\theta)^{22}}}$$

where $I(\theta)^{22}$ denotes the element $[2,2]$ of $I(\theta)^{-1}$.

  • $\begingroup$ So If I was trying to determine the intervals for beta0, I would use element [1,1] ? $\endgroup$
    – Dom Jo
    Commented Nov 28, 2019 at 17:27
  • $\begingroup$ Is it taking the square root of element [2,2] or taking square root of the inverse matrix and then taking element [2,2] $\endgroup$
    – Dom Jo
    Commented Nov 28, 2019 at 18:59
  • $\begingroup$ @Dom Jo: The [2,2] element of the inverse. I have edited my answer, there was a mistake. $\endgroup$
    – F. Tusell
    Commented Feb 29, 2020 at 18:59

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