# Verify procedure to determine density of $S = S_0 \exp(X)$ where $p_X(x) = \frac{\lambda}{2}\exp(-\lambda |x |)$

Determine the density of the random variable $$S = S_0 \exp(X)$$ where $$p_X(x) = \frac{\lambda}{2}\exp(-\lambda |x |)$$ i.e. $$X$$ is a Laplace distribution with parameter $$\lambda$$, and $$S_0$$ is a constant.

In situations like this I know I should use the change of variables formula for densities. However, the solution to the above problem uses the following formula: $$p_X \, dX = p_S \, dS \Leftrightarrow p_S = p_X \frac{dX}{dS} = \frac{p_X}{S}$$ Is this formula "valid" ? What assumptions are being made here, if any, to enable this ? Does this have something to do with the inverse function theorem? Are we making special use of the definitions of $$S$$ and $$X$$ ? (I can see why $$\frac{dX}{dS} = \frac{p_X}{S}$$ that's not the problem here).

First let us note that $$\Pr(X>0)=1,$$ so $$p_S(x) = 0$$ when $$x<0.$$
For $$x>0,$$ \begin{align} p_S(x) = {} & \frac d {dx} \Pr(S\le x) \\[8pt] = {} & \frac d {dx} \Pr(S_0 e^X \le x) \\[8pt] = {} & \frac d {dx} \Pr\left( X \le \log \frac x {S_0} \right) \\[8pt] = {} & \frac d {dx} F_X\left( \log \frac x {S_0} \right) \\ & \text{((capital) F_X is the c.d.f.} \\ & \phantom{(}\text{of the random variable X)} \\[8pt] = {} & p_X\left( \log\frac x {S_0} \right) \cdot \frac d {dx}\log \frac x {S_0} \\[8pt] = {} & \frac\lambda 2 \exp\left( -\lambda \left| \log \frac x {S_0} \right| \right) \cdot \frac 1 x \\[8pt] \ldots & \text{ and now it gets} \\ & \text{ somewhat complicated:} \\ = {} & \frac \lambda 2 \left( \frac x {S_0} \right)^{-\lambda} \cdot \frac 1 x \quad \text{if } \log \frac x {S_0} \ge 0. \tag 1 \end{align} Now notice that $$\log(x/S_0)\ge0$$ precisely if $$x\ge S_0.$$
If $$0 then instead of $$(1)$$ we get $$\frac \lambda 2 \left( \frac x {S_0} \right)^\lambda \cdot \frac 1 x.$$ Thus the conditional distribution of $$X$$ given that $$X>S_0$$ is a Pareto distribution and the conditional distribution given $$X is a scaled beta distribution.