Probability density function of difference of two independent random variables in the Laplace domain

Question

I find clear that the probability density function of the sum of two independent random variables is the convolution between the probability density functions of the addends and, for this reason, in the Laplace domain, the convolution "becomes" a product. If I had a difference rather than a sum, what can we say about the relative formulation in the Laplace domain?

Answer 1

Let

$$f(t) := \begin{cases} e^{-t} & \text{if } t \geq 0\\ 0 & \text{if } t < 0\end{cases}$$

Convolving $f$ with its own reflection,

$$\begin{aligned} f (t) * f (-t) &= \int_{-\infty}^{\infty} f (\tau) f(\tau - t) \, {\rm d} \tau\\ &= e^t \int_{\max(0,t)}^{\infty} e^{-2\tau} \, {\rm d} \tau\\ &= \begin{cases} \frac 12 e^{-t} & \text{if } t \geq 0\\ \frac 12 e^{t} & \text{if } t < 0\end{cases}\\ &= \frac 12 e^{-|t|}\end{aligned}$$

which is positive and whose integral over the real line is $1$. It is a Laplace distribution.

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In the Laplace domain

Let

$$F (s) := \mathcal{L} \left\{ f (t) \right\} = \frac{1}{s+1}, \qquad \Re\{s\} > - 1$$

and note that

$$\mathcal{L} \left\{ f (-t) \right\} = - \left(\frac{1}{s-1}\right) = \frac{1}{1-s} = F(-s), \qquad \Re\{s\} < 1$$

Hence,

$$\mathcal{L} \left\{ f (t) * f (-t) \right\} = F (s) F (-s) = -\frac 12 \left(\frac{1}{s-1}\right) + \frac 12 \left(\frac{1}{s+1}\right), \qquad -1 < \Re\{s\} < 1$$