The univariate Jeffreys prior is defined as any [[prior]] proportional to the square root of the [[Fisher information]] number $\pi(\theta) \propto \sqrt{I(\theta)}$ Jeffreys prior is invariant to reparameterization and a response to critiques of the [[principle of indifference]]. Jeffreys prior is a type of [[objective prior]] and can be an [[improper prior]]. To find Jeffreys prior - Find a [[likelihood]] for $\theta$ - Find the log-likelihood - Then find the [[Fisher information]] by either - find the first derivative with respect to $\theta$ and square its [[expectation]] with respect to the data - find the second derivative with respect to $\theta$ and negate its expectation (the second derivative is often easier) - Finally find a prior that is proportional to the square root of the Fisher information For multiple [[independent and identically distributed|iid]] data points $Y = Y_1, \dots, Y_n$, the Fisher information is $n$ times the marginal information. $I(\lambda) = -nE \Big[ \frac{\partial^2}{\partial^2 \theta} \ln f(y_i|\lambda) \Big]$ For example, let $Y | \lambda \sim Poisson(\lambda)$. A likelihood for the [[Poisson distribution]] can be expressed as $ f(y|\lambda) = \frac{e^{-\lambda}\lambda^y}{y!} \propto e^{-\lambda}\lambda^y $ The log-likelihood is $\ln f(y|\lambda) = -\lambda + y \ln \lambda$ The first derivative is $\frac{\partial}{\partial \lambda} \ln f(y|\lambda) = -1 + \frac{y}{\lambda}$ The second derivative is $\frac{\partial^2}{\partial^2 \lambda} \ln f(y|\lambda) = -\frac{y}{\lambda^2}$ The expectation of the Poisson distribution is $E[y] = \lambda$. The Fisher information is the negative of the expectation $I(\lambda) = -E \Big[ \frac{\partial^2}{\partial^2 \theta} \ln f(y|\lambda) \Big] = -E \Big[- \frac{y}{\lambda^2} \Big] = \frac{E[y]}{\lambda^2} = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}$ Note that in the expectation the parameter $\lambda$ is considered a constant, the expectation is with respect to the data $y$. Thus, the Jeffreys prior for a Poisson distribution is $\pi(\lambda) \propto \sqrt{I(\lambda)} = \frac{1}{\lambda^2}$