# Finding the Fisher information given the density

For $$X_1,...,X_n$$ i.i.d. continuously distributed random variables with Lebesgue density $$f_{\theta}(x)=\theta(\theta+1)x^{\theta-1}(1-x)\mathbf{1}_{(0,1)}(x)$$ (Note: $$\mathbf{1}_{(0,1)}(x)$$ is an indicator function), How can I find the fisher information for a single observation?

For the Fisher information, I know the formula is $$I(\theta)=-E_{\theta}\left[\frac{\partial^2}{\partial\theta^2}\ln f_{\theta}(X_1)\right]$$ or alternatively $$I(\theta)=E_{\theta}\left[(\frac{\partial}{\partial\theta}\ln f_{\theta}(X_1))^2\right]$$. However, I can't solve for the partial derivative of $$ln f_{\theta}(X_1)$$ and how can I then find the expectation of this? I am very confused on this and I can't seem to find much information or examples of finding fisher information.

• The $f$ that appears in the Fisher info expression should be the joint density. Since your sample is iid, you can leave the expression as is if you include an extra factor $n$ out front. Commented Mar 9, 2023 at 14:09
• Start with the second derivative computation; what do you get? Commented Mar 9, 2023 at 14:11
• @Golden_Ratio I got $\frac{1}{\theta^2}+\frac{1}{(\theta+1)^2}$ for my second derivative. Since there is no X in this expression, is the expectation of this expression the same? Commented Mar 9, 2023 at 14:15

This is an exponential family. Therefore the second derivative of the score function is a function of $$\theta$$ only as you correctly observed. Therefore no painful integration! Actually this phenomena is characteristic of the models which are exponential families.
The Fisher information is $$\frac{1}{\theta^2}+\frac{1}{(\theta+1)^2}$$ for one observation.
The Fisher information for a random variable $$X$$ parameterized in terms of a single parameter $$\theta$$ is given by (assuming appropriate regularity conditions) $$I_X(\theta)=-\mathsf E(\partial_\theta^2 \log f_X(X|\theta)).$$ Here, the expectation is taken with respect to the random variable $$X$$. For the specific case $$f_X(x|\theta)=\theta(\theta+1)x^{\theta-1}(1-x)\mathbf{1}_{(0,1)}(x)$$ we compute the Fisher information for a single observation directly from the formula above. Since $$\log f_X(X|\theta)=\log\theta+\log(\theta+1)+(\theta-1)\log X+\log(1-X)$$ we find $$-\partial_\theta^2\log f_X(X|\theta)=\frac{1}{\theta^2}+\frac{1}{(\theta+1)^2}.$$ Notice that this expression does not contain $$X$$; thus, it is a constant and by the linearity of expectation $$I_X(\theta)=\mathsf E(-\partial_\theta^2 \log f_X(X|\theta))=\mathsf E(\theta^{-2}+(\theta+1)^{-2})=\theta^{-2}+(\theta+1)^{-2}.$$ Furhtermore, Fisher information is additive so that if we have $$n$$ i.i.d. observations of $$X$$, i.e. $$\mathbf X=\{X_1,\dots,X_n\}$$, then the information contained in all $$n$$ observations is simply $$I_{\mathbf X}(\theta)=nI_X(\theta)=\frac{n}{\theta^2}+\frac{n}{(\theta+1)^2}.$$