What am I doing wrong in calculating Fisher Information of Triangular Distribution?

Question

I am trying to find Jeffrey's prior for the Triangular distribution which has the following probability density function:

$$f(x\mid \theta) = \begin{cases} \dfrac{2x}{\theta} & : x \leq \theta , x\in[0,1], \theta \in [0,1] \\[6pt] \dfrac{2(1-x)}{1-\theta} & : x > \theta , x\in[0,1], \theta \in [0,1] \end{cases} $$

Jeffrey's prior is given by $\pi(\theta) \propto \sqrt{I(\theta)}$ where $I(\theta)$ is the Fisher Information for parameter $\theta$. My professor said that the Jeffrey's prior should be $\beta(1/2,1/2)$. So, essentially I need $I(\theta) \propto \frac{1}{\theta(1-\theta)}$ since the kernel of $\beta(1/2,1/2)$ is $\frac{1}{\theta^{1/2}(1-\theta)^{1/2}}$

So, I begin to calculate $I(\theta) = \mathbb{E}_\theta[(\frac{\partial}{\partial \theta}\log(f(x\mid \theta))^2]$.

$\log(f(x\mid \theta)) = \left\{ \begin{array}{lr} \log{2x} - \log{\theta} & : x \leq \theta , x\in[0,1], \theta \in [0,1] \\ \log{2(1-x)} - \log(1-\theta) & : x > \theta , x\in[0,1], \theta \in [0,1] \end{array} \right. $

$\Rightarrow \frac{\partial}{\partial \theta}\log(f(x\mid \theta) = \left\{ \begin{array}{lr} -\frac{1}{\theta} : x \leq \theta , x\in[0,1], \theta \in [0,1] \\ \frac{1}{(1-\theta)} : x > \theta , x\in[0,1], \theta \in [0,1] \end{array} \right. $

$\Rightarrow \mathbb{E}_\theta[(\frac{\partial}{\partial \theta}\log(f(x\mid\theta))^2] = \left\{ \begin{array}{lr} \frac{1}{\theta^2} : x \leq \theta , x\in[0,1], \theta \in [0,1] \\ \frac{1}{(1-\theta)^2} : x > \theta , x\in[0,1], \theta \in [0,1] \end{array} \right. $

which is certainly not proportional to $\frac{1}{\theta(1-\theta)}$ for all $\theta$.

What am I doing wrong?

Answer 1

The crux of the mistake is right here:

$$\frac{\partial}{\partial \theta}\log(f(x\mid \theta) = \left\{ \begin{array}{lr} -\frac{1}{\theta} : x \leq \theta , x\in[0,1], \theta \in [0,1] \\ \frac{1}{(1-\theta)} : x > \theta , x\in[0,1], \theta \in [0,1] \end{array} \right. = -\frac{1}{\theta}I(X\leq\theta) + \frac{1}{1-\theta}I(X>\theta)$$

So, you have

$$\mathbb{E}_\theta[(\frac{\partial}{\partial \theta}\log(f(x\mid\theta))^2] = \frac{1}{\theta^2} P(X\leq \theta) + \frac{1}{(1-\theta)^2}[1-P(X\leq \theta)] = \frac{1}{\theta} + \frac{1}{1-\theta} = \frac{1}{(1-\theta)\theta}$$

which gives the desired Jeffrey's prior.