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The Laplace distribution has PDF $f(x) = \frac{1}{2}e^{-|x|}$ for all real x.

Let $X$~$Expo(1)$ and $S$ be a random sign (1 or -1, with equal probabilities), with $S$ and $X$ independent. Find the PDF of $SX$ (by first finding the CDF), and compare the PDF of $SX$ and the Laplace PDF.

I know that the CDF of $X$~$Expo(1)$ is $1 - e^{-x}$ and that the PDF is $e^{-x}$. I also know that $S$~$Unif[-1,1]$. How do I find the CDF of $SX$? So that I can find the PDF of $SX$.

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2 Answers 2

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When $x \le 0$, you have $$\Pr(SX \le x) = \Pr(S=-1)\Pr(X \ge -x) = \tfrac12 e^x$$

while when $x \gt 0$, you have $$\Pr(SX \le x) = \Pr(S=-1)\Pr(X \ge 0) + \Pr(S=1)\Pr(X \le x) = \tfrac12 +\tfrac12(1- e^x) = 1 - \tfrac12 e^{-x}$$

so those combined are the cumulative distribution function which you can differentiate.

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By the Law of Total Probability, and because of the independence of $S,X$, we have $$\begin{align} f_{SX} (\lvert x\rvert) & = \mathsf P(S{=}1)\,f_X(\lvert x\rvert) \,+\, \mathsf P(S{=}{-1})f_X(-\lvert x\rvert)\\[1ex] & = \tfrac 1 2 \mathsf e^{-\lvert x\rvert} \quad[0\leq \lvert x\rvert] \end{align}$$

Likewise $$\begin{align} f_{SX} (-\lvert x\rvert) & = \mathsf P(S{=}1)\,f_X(-\lvert x\rvert) \,+\, \mathsf P(S{=}{-1})f_X(\lvert x\rvert)\\[1ex] & = \tfrac 1 2 \mathsf e^{-\lvert x\rvert} \quad[0\leq \lvert x\rvert] \end{align}$$

So then clearly: $\;f_{SX}(x) = \tfrac 1 2 \mathsf e^{-\lvert x\rvert} \quad [x\in\Bbb R]$

That is all.

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