Past exam of the thermal physics course of the University of Oxford / 2016 / 6

Normalize the probability density function on $x\ge 0$:For $y=\sqrt{x}\ge 0$, the change of variables is $x=y^2$, $dx=2ydy$. Conservation of probability gives the transformed probability density functionIt is a Rayleigh distribution with scale $\sigma =1/\sqrt{2\alpha }$. Its derivative is $g^{\prime }(y)=2\alpha e^{-\alpha y^2}(1-2\alpha y^2)$, so it rises from zero, has one maximum, and then decays:The expected value follows by putting $z=\sqrt{\alpha }y$:Here ${\int }_0^{\infty }{z}^2{e}^{-z^2}dz=\sqrt{\pi }/4$, obtained from the Gaussian integral by integration by parts. In particular $⟨\sqrt{x}⟩\ne \sqrt{⟨x⟩}$, and the mode changes under a nonlinear variable transformation because a density includes its Jacobian determinant.