Past exam of the biological physics course of the University of Oxford / 2015 / 5 / d / Solution

Taking $k_2\to k_1=k$ in the preceding probability density gives $p_2(t)=k^{2}t{e}^{-kt}$. Convolving once more with an independent exponential distribution givesMore generally, suppose $p_n(s)=k^n{s}^{n-1}{e}^{-ks}/(n-1)!$. ThenSince $p_1(t)=k{e}^{-kt}$, this proves the Erlang distribution formula by mathematical induction. Normalization follows either from the convolution or ${\int }_0^{\infty }{t}^{n-1}{e}^{-kt}dt=(n-1)!/k^n$.For $n>1$, differentiating $\log p_n$ gives $(n-1)/t-k=0$, and the derivative changes from positive to negative. For $n=1$ the maximum occurs at the boundary $t=0$. Source dimensional qualification: the printed expression without $dt$ is a probability density, with units of inverse time; it cannot itself be a dimensionless interval probability.

In the right panel of the original illustration in d, $p_2/k$ rises linearly near zero and peaks at $kt=1$ with height $1/e$. The curve $p_3/k$ rises quadratically and peaks later, at $kt=2$, with lower height $2e^{-2}$. Both have exponential tails, but the latter has the larger expected value $3/k$ and variance $3/k^2$, compared with $2/k$ and $2/k^2$. This distinguishes both the onset and the width of the requested sketches.