 Solution

ID: past-exam-of-the-thermal-physics-course-of-the-university-of-oxford/2016/6/solution

x \geq 0

1 = N \int_{0}^{\infty} e^{- αx} d x = \frac{N}{α}, N = α, ⟨ x ⟩ = α \int_{0}^{\infty} x e^{- αx} d x = \frac{1}{α} .

For

y = x \geq 0

x = y^{2}

d x = 2 y d y

. Conservation of probability gives the transformed probability density function

g (y) = 2 α y e^{- α y^{2}}, y \geq 0, g (y) = 0, y < 0.

σ = 1/ 2 α

g^{'} (y) = 2 α e^{- α y^{2}} (1 - 2 α y^{2})

, so it rises from zero, has one maximum, and then decays:

y_{mode} = \frac{1}{2 α} .

The expected value follows by putting

z = α y

⟨ y ⟩ = 2 α \int_{0}^{\infty} y^{2} e^{- α y^{2}} d y = \frac{2}{α} \int_{0}^{\infty} z^{2} e^{- z^{2}} d z = \frac{π}{2 α} .

Here

\int_{0}^{\infty} z^{2} e^{- z^{2}} d z = π /4

, obtained from the Gaussian integral by integration by parts. In particular

⟨ x ⟩ \neq = ⟨ x ⟩

, and the mode changes under a nonlinear variable transformation because a density includes its Jacobian determinant.

 New to topics? Read the docs here!