Problem source

In this question, you may use the result 
\[
\displaystyle \int_0^\infty \frac{t^m}{(t+k)^{n+2}} \; \mathrm{d}t
=\frac{m!\, (n-m)!}{(n+1)! \, k^{n-m+1}}\;,
\]
where $m$ and $n$ are positive integers with $n\ge m\,$, and where $k>0\,$. 
 
The random variable $V$ has density function 
\[ 
\f(x) 
= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \quad \quad (0 \le x < \infty) \;,
\] 
where $a$ is a positive integer. Show that  $\displaystyle C = \frac{(2a+1)!}{a! \, a!}\;$. 
 
Show, by means of a suitable substitution, that 
\[ 
\int_0^v \frac{x^a}{(x+k)^{2a+2}} \; \mathrm{d}x 
= \int_{\frac{k^2}{v}}^\infty \frac{u^a}{(u+k)^{2a+2}} \; \mathrm{d}u  
\] 
and deduce that the median value of $V$ is  $k$. 
Find the expected value of $V$. 
 
The random variable $V$ represents the speed of a randomly chosen gas molecule. The time taken for such a particle to travel a fixed distance $s$ is given by the random variable $\ds T=\frac{s}{V}$. 
 
Show that 
\begin{equation}
\mathbb{P}( T < t) = \ds \int_{\frac{s}{t}}^\infty \frac{C \, k^{a+1} \,  
x^a}{(x+k)^{2a+2}}\; \mathrm{d}x 
\tag{$ *$}
\end{equation}
and hence find the density function of $T$. 
You may find it helpful to make the substitution $\ds u = \frac{s}{x}$ 
in the integral $(*)$. 
 
Hence show that the product of the median time and the median speed is equal to the distance $s$, but that the product of the expected time and the expected speed is greater than $s$.

Solution source

\begin{align*}
&& f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\
\Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\
&&&= \int_0^{\infty}  \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\
&&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\
&&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\
&&&= C \frac{a!a!}{(2a+1)!} \\
\Rightarrow && C &= \frac{(2a+1)!}{a!a!}
\end{align*}

\begin{align*}
&& I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\
u = k^2/x, \d x = -k^2u^{-2}  \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\
&&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\
&&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\
\end{align*}

At the median we want a value $M$ such that $M = k^2/M$ ie $M = k$

\begin{align*}
&& \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\
&&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\
&&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\
&&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\
&&&= \frac{k(a+1)}{a} = \frac{a+1}a k
\end{align*}


\begin{align*}
&& \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\
&&&= \mathbb{P}(V > \frac{s}{t}) \\
&&&= \int_{s/t}^{\infty} f(x) \d x \\
&&&=  \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\
\\
\Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\
&&&=  \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\
&&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\
&&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\
&&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\
&&&=  \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}}
\end{align*}

Therefore $T$ follows the same distribution, but with parameter $s/k$ rather than $k$. In particular it has median $s/k$ (and the product of the medians is $s$). 

However, the product of the expected time and expected speed is $\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s$