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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).

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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 $0<x<S_0$ 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 $X<S_0$ is a scaled beta distribution.

I wonder if your difficulties began with your thinking "In situations like this I know I should use the [whatever] formula" before you sufficiently understood the question.

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  • $\begingroup$ "I wonder if ..." yes, you are spot on: following what you did makes complete sense. $\endgroup$ – baibo Jan 1 at 17:16

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