For a likelihood $p(y | \theta)$ and pdf $f(y)$:

Suppose that a likelihood is location invariant i.e.

$p(y | \theta) = f(y - \theta)$

Show that the Jeffreys prior is of the form $p(θ) ∝ 1$.

I understand that we have to use the Fisher information to solve this, but am confused about the idea of location invariance.

Fisher = $\sqrt{-E[d^2/(d\theta)^2 \log (f(y - \theta))]}$

which has to be proportional to $1$, but how do I use $f(y - \theta)$ in this equation for Fishers information?

  • $\begingroup$ This is an exercise and should have the self-study tag. You should also provide more details on why you cannot prove that the information is constant. $\endgroup$
    – Xi'an
    Jun 11, 2020 at 8:54
  • $\begingroup$ I don't really understand how this location invariance comes into this idea. I can back track and realize that the log of the second derivative of f(y - theta) must be a constant, but I can't see why that's the case. $\endgroup$ Jun 12, 2020 at 5:49

1 Answer 1


When considering the Fisher information $$\mathfrak{I}=-\int_\mathcal X\frac{\partial^2 \log(f(x-\theta))}{\partial\theta^2} f(x-\theta)\,\text{d}x$$ we have that $$\frac{\partial^2 \log(f(x-\theta))}{\partial\theta^2}=-\frac{\partial}{\partial\theta}\frac{f'(x-\theta)}{f(x-\theta)}=\frac{f'(x-\theta)}{f(x-\theta)}+\frac{f''(x-\theta)^2}{f(x-\theta)^2}$$ and making the change of variable $y=x-\theta$ (with Jacobian one) in the integral(s) \begin{align*} \int_\mathcal X\frac{\partial^2 \log(f(x-\theta))}{\partial\theta^2} f(x-\theta)\,\text{d}x &= \int \frac{f'(x-\theta)}{f(x-\theta)} \,\text{d}x + \int\frac{f''(x-\theta)^2}{f(x-\theta)^2} \,\text{d}x\\ &= \int \frac{f'(y)}{f(y)} \,\text{d}y + \int\frac{f''(y)^2}{f(y)^2} \,\text{d}y \end{align*} which is constant in $\theta$.


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