Particle ensemble performing shm, calculate amplitude pdf Consider the shm for a single particle. Then the particle's position is given by (assume zero initial phase):
$$x = a \times \sin(\omega t)$$
The infinitesimal probability of finding a particle between $x$ and $x + dx$ is:
$$dp(x) = \frac{dt}{T} = \frac{dt \times \omega}{2\pi} = \frac{dx}{2\pi \sqrt{a^{2}-x^{2}}}$$
hence the probability density function is given by (I hope that so far derivations are correct):
$$pdf(x) = \frac{1}{2\pi \sqrt{a^{2}-x^{2}}}$$
Let's now imagine that, instead of a single particle performing shm, we have an ensemble of particles performing shm with different amplitudes. Let us also assume that $pdf(x)$ is known for the whole ensemble and we call it $pdf_{E}(x)$. So, $pdf_{E}(x)$ is known. How can we calculate $pdf_{E}(a)$?
Here's my attempt:
$$pdf_{E}(x) = pdf_{E}(a \geqslant x) \times pdf(x) = \frac{pdf_{E}(a \geqslant x)}{2\pi \sqrt{a^{2}-x^{2}}}$$
then, rearranging we get:
$$pdf_{E}(a \geqslant x)=2\pi \sqrt{a^{2}-x^{2}} \times pdf_{E}(x)$$
This is where I get stuck. I'm strugling to figure out how to proceed in order to be able to extract and plot $pdf_{E}(a)$ from the above.
 A: Consider one period $T=2\pi/\omega$ and suppose that the random time $t$ at which you measure the displacement $x$ of oscillator is uniformly distributed over the period of oscillation: $$f(t)=\frac{1}{T},\quad t\in[0,T)$$ 
By symmetry the p.d.f. $g(x)$ for displacement $x$ is an odd function i.e. $g(x)=g(-x)$ for $x\in[-a,a]$, so it suffices to work out $g(x)$ for $x\in[0,a]$. A given value of $x$, except the endpoints $x=\pm a$, is obtained at two distinct time instants in $[0,T)$. For $x\in[0,a)$ let the two time instants $t_1,t_2,$ be such that $\omega t_1=\sin^{-1}(x/a)\in[0,\pi/2]$ and $\omega t_2=\sin^{-1}(x/a)\in[\pi/2,\pi]$. Then we have:
\begin{align}
g(x)dx&=f(t_1)dt+f(t_2)dt\\
g(x)&=\frac{2}{T}\frac{dt}{dx}=\frac{1}{\pi\sqrt{a^2-x^2}},\quad x\in[-a,a]
\end{align}
in which I have also included the endpoints $x=\pm a$ because the probability measure over a finite set of points is zero anyway. You may verify that $g(x)$ is correctly normalised.
Now let us consider an ensemble of oscillators. The oscillators have random amplitude $a$ whose p.d.f. is $q(a)$. Let $h(x)$ be the corresponding p.d.f. for displacement $x$; $h(x)$ is supposed given and we want to find $q(a)$. The p.d.f. $g(x)$ found previously I shall write as $g(x|a)$ which is to be read as the p.d.f. for displacement given that the oscillator has amplitude $a$. From Bayes theorem we have: $$h(x)=\int_0^\infty da~g(x|a)q(a)$$ The integral needs to be inverted somehow to obtain $q(a)$ and here is a trick which I think will work. Consider the transformation $\eta=a^2$. Then since $a\geq 0$ there is one-to-one correspondence between $\eta$ and $a$, which allows us to relate $q(a)$ to p.d.f. of $\eta$: $$q(a)=q^*(\eta)\frac{d\eta}{da}=2a~q^*(\eta)$$ Therefore $$h(x)=\int_0^\infty d\eta~\frac{1}{\pi\sqrt{\eta-x^2}}\times q^*(\eta)$$
We now similarly transform $x$. By symmetry we have $h(x)=h(-x)$, so let us limit ourselves to $x\in[0,\infty)$. Transforming to $X=x^2$ then gives: $$h(x)=h^*(X)\frac{dX}{dx}=2x~h^*(X)=2\sqrt{X}~h^*(X)$$ Therefore $$2\sqrt{X}~h^*(X)=\int_0^\infty d\eta~\frac{1}{i\pi\sqrt{X-\eta}}\times q^*(\eta)$$
We have a convolution integral above. Applying Laplace transform $\mathbf{L}\equiv\int_0^\infty ds~\exp(-sX)$ gives: $$ q^*(\eta)=\mathbf{L}^{-1}\left[ \frac{\mathbf{L}\left[ 2\sqrt{X}~h^*(X) \right]}{\mathbf{L}\left[ \frac{1}{i\pi\sqrt{X-\eta}} \right]} \right]$$ from which $q(a)$ may be obtained.
