LAPLACE TRANSFORM AND CONVOLUTION OF RANDOM VARIABLES

In a previous post we showed how useful Fourier Transform may be in computing convolution of symmetric R.V.s. The Laplace tranform plays an analogous role for the convolution of one sided non-negative R.V.s. In this post We will show it´s usefulness in handling with the convolution of Levy-Smirnov R.V.s., an alpha stable distribution.

Laplace transform:

Recall the Laplace transform pair

$\mathcal{L}[f(x)](s)=\phi(s)=\int_0^\infty e^{-sx}f(x)\,dx$ (1)

$f(x)=\mathcal{L}^{-1}[f(s)](x)=\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}\phi(s)\,ds$ (2)

Now recall from the previous post the convolution of two non-negative R.V.s

$f_{X_1+X_2}(t)=f_{S_2}(t)=\int_0^t f_{X_1}(t-u)f_{X_2}(u)\,du\,dt$

If We take the it´s Laplace transform We obtain

$\begin{aligned} \int_0^\infty e^{-st} f_{S_2}(t)\,dt&=\int_0^\infty e^{-st}\int_0^tf_{X_1}(t-u)f_{X_2}(u)\,du\,dt\\ &=\int_0^\infty f_{X_2}(u)\int_u^\infty e^{-st} f_{X_1}(t-u)\,dt\,du\\ &=\int_0^\infty f_{X_2}(u)\int_0^\infty e^{-s(t+u)} f_{X_1}(t)\,dt\,du\\ &=\int_0^\infty e^{-su}f_{X_2}(u)\,du\int_0^\infty e^{-st} f_{X_1}(t)\,dt\\ &=\phi_{X_1}(s)\phi_{X_2}(s) \end{aligned}$

If X_1 and X_2 are i.i.d we obtain

$\int_0^\infty e^{-st} f_{S_2}(t)\,dt=\phi^2(s)$

Similarly for the sum of three R.V.s

$f_{X_1+X_2+X_3}(t)=f_{S_3}(t)=f_{S_2+X_3}(t)=\int_0^t f_{S_2}(t-u)f_{X_3}(u)\,du\,dt$

Taking it´s Laplace transform

$\begin{aligned} \int_0^\infty e^{-st} f_{S_3}(t)\,dt&=\int_0^\infty e^{-st}\int_0^tf_{S_2}(t-u)f_{X_3}(u)\,du\,dt\\ &=\int_0^\infty f_{X_3}(u)\int_u^\infty e^{-st} f_{S_2}(t-u)\,dt\,du\\ &=\int_0^\infty f_{X_3}(u)\int_0^\infty e^{-s(t+u)} f_{S_2}(t)\,dt\,du\\ &=\int_0^\infty e^{-su}f_{X_3}(u)\,du\int_0^\infty e^{-st} f_{S_2}(t)\,dt\\ &=\phi_{X_3}(s)\phi_{X_2}(s)\phi_{X_1}(s)\\ &=\phi^3(s) & (\text{for i.i.d R.V.s}) \end{aligned}$

Keeping on this process We may obtain

$\int_0^\infty e^{-st} f_{S_n}(t)\,dt=\phi^n(s)$ (3)

By equation (2) We Obtain

$f_{S_n}(t)=\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}\phi^n(s)\,ds$ (4)

Levy-Smirnov distribution:

The Levy-Smirnov distribution is a continuous probability distribution for a non-negative random variable. It belongs to the family of alpha-stable distributions. Like all stable distributions except the Gaussian it is a Heavy tailed distribution, and it´s tails behave asymptotically as a power law. It´s density function is given by:

$f(x)=\frac{1}{\sqrt{2 \pi}}e^{-\frac{1}{2x}}x^{-3/2}$ (5)

Lets now compute the Laplace transform of (5).

Claim:

$\frac{1}{\sqrt{2 \pi}}\int_0^\infty e^{-sx}}e^{-\frac{1}{2x}}x^{-3/2}\,dx=e^{-\sqrt{2s}}$ (6)

Proof:

$\begin{aligned} f(s)&=\frac{1}{\sqrt{2 \pi}}\int_0^\infty e^{-sx}}e^{-\frac{1}{2x}}x^{-3/2}\,dx\\ &=\frac{1}{\sqrt{2 \pi}}\int_0^\infty e^{-sx-\frac{1}{2x}}\,x^{-3/2}\,dx\\ &=\frac{1}{\sqrt{2 \pi}}\int_0^\infty e^{-\frac{a}{2}\left(ax+\frac{1}{ax}\right)}\,x^{-3/2}\,dx &\left(a=\sqrt{2s}\right)\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}\int_0^\infty e^{-\frac{a}{2}\left(x+\frac{1}{x}\right)}\,x^{-3/2}\,dx &\left(ax \to x\right)\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}\int_0^\infty e^{-\frac{a}{2}\left(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+2\right)}\,x^{-3/2}\,dx\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}e^{-a}\int_0^\infty e^{-\frac{a}{2}\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2}\,x^{-3/2}\,dx &(1)\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}e^{-a}\int_0^\infty e^{-\frac{a}{2}\left(\frac{1}{\sqrt{x}}-\sqrt{x} \right)^2}\,x^{-1/2}\,dx &(x \to \frac{1}{x})\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}e^{-a}\int_0^\infty e^{-\frac{a}{2}\left(\sqrt{x}-\frac{1}{\sqrt{x}} \right)^2}\,x^{-1/2}\,dx &(2)\\ &=\frac{\sqrt a\,e^{-a}}{2\sqrt{2 \pi}}\int_0^\infty e^{-\frac a2 \left(\sqrt x-\frac1{\sqrt x}\right)^2}\,\left( x^{-1/2}+x^{-3/2}\right)\,dx & \left((1) + (2) \right)\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}e^{-a}\int_{- \infty}^\infty e^{-\frac{a}{2}x^2}\,dx & \left( \sqrt x-\frac1{\sqrt x} \to x\right)\\ &=\frac{\sqrt{a}}{\sqrt{2 \pi}}e^{-a}\sqrt{\frac{2 \pi}{a}}\\ &=e^{-a}\\ &=e^{-\sqrt{2s}} \qquad \blacksquare \end{aligned}$

Inverse Laplace Transform:

We now compute the Inverse Laplace Transform of the Lévy-Smirnov R.V.

$f(x)=\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds$

Lets consider the following contour integral over the contour C below

$I=\frac{1}{2 \pi i }\oint_C e^{sx}e^{-a \sqrt{s}}\,ds$

The integrand has a branch point at zero. So we consider a branch-cut along the negative real axis. Since inside the contour C, the integrand has no singularities, by Cauchy´s Theorem the contour integral equals zero! Then:

$\frac{1}{2 \pi i }\oint_C e^{sx}e^{-a \sqrt{s}}\,ds=\frac{1}{2 \pi i }\left[\int_{C_1}+\int_{C_2}+\int_{C_3}+\int_{C_4}+\int_{C_5}+\int_{C_6} \right]\,ds=0$ (7)

Lets begin by showing that the integral over C_2 vanishes in the limit as $R \to \infty$

$\begin{aligned} I_{C_2}&=\int_{C_2} e^{sx}e^{-a \sqrt{s}}\,dx\\ &=i R\int_{\pi/2}^{\pi} e^{xRe^{i \theta}}e^{-aR e^{i\frac{\theta}{2}}}\,e^{i \theta}d\theta\\ &=i R\int_{\pi/2}^{\pi} e^{xR\cos(\theta)+ixR \sin(\theta)}e^{-aR \cos\left(\frac{\theta}{2}\right)-iaR \sin\left(\frac{\theta}{2}\right)}\,e^{i \theta}d\theta\\ & \leq \Bigg|i R\int_{\pi/2}^{\pi} e^{xR\cos(\theta)+ixR \sin(\theta)}e^{-aR \cos\left(\frac{\theta}{2}\right)-iaR \sin\left(\frac{\theta}{2}\right)}\,e^{i \theta}d\theta\Bigg|\\ &\leq \Big|iR\Big| \int_{\pi/2}^{\pi}\Big|e^{xR\cos(\theta)}\Big|\Big|e^{ixR\sin(\theta)}\Big|\Big|e^{-aR \cos\left(\frac{\theta}{2}\right)}\Big|\Big|e^{-iaR \sin\left(\frac{\theta}{2}\right)}\Big|\Big|e^{i \theta}\Big|\,\Big|d \theta \Big|\\ &\leq R \int_{\pi/2}^{\pi}e^{xR\cos(\theta)-aR \cos\left(\frac{\theta}{2}\right)}\,d \theta \\ \end{aligned}$

Since in $\theta \in [\pi/2,\pi), \,\, R\cos(\theta)<0 \,\,\text{and}\,\,\cos\left(\frac{\theta}{2}\right)>0$ , as $R \to \infty$ the integral decays exponantially to zero and consequently the integral vanishes.

Similarly we have the integral in C_6

$\begin{aligned} I_{C_6} &\leq R \int_{-\pi}^{-\pi/2}e^{xR\cos(\theta)-aR \cos\left(\frac{\theta}{2}\right)}\,d \theta \\ \end{aligned}$

Since $\cos(-\theta)=\cos(\theta)$ , in ther interval $\theta \in \left[-\pi,-\frac{\pi}{2}\right)$ the integrand decays exponentially and the integral vanishes.

For the integral over C_4 we have

$\begin{aligned} I_{C_4}&=\int_{C_3} e^{sx}e^{-a \sqrt{s}}\,dx\\ &=i \epsilon\int_{\pi}^{-\pi} e^{x \epsilon e^{i \varphi}}e^{-a\epsilon e^{i\frac{\varphi}{2}}}\,e^{i \varphi}d\varphi\\ &=i \epsilon\int_{\pi}^{-\pi} e^{x \epsilon \cos(\theta)+ix \epsilon \sin(\theta)}e^{-a \epsilon\cos\left(\frac{\varphi}{2}\right)-ia \epsilon \sin\left(\frac{\varphi}{2}\right)}\,e^{i \varphi}d\varphi\\ & \leq \Bigg|i \epsilon\int_{\pi}^{-\pi} e^{x \epsilon \cos(\theta)+ix \epsilon \sin(\theta)}e^{-a \epsilon\cos\left(\frac{\varphi}{2}\right)-ia \epsilon \sin\left(\frac{\varphi}{2}\right)}\,e^{i \varphi}d\varphi\Bigg|\\ &\leq \epsilon \int_{-\pi}^{\pi}e^{x \epsilon\cos(\varphi)-a \epsilon \cos\left(\frac{\varphi}{2}\right)}\,d \varphi \\ \end{aligned}$

Taking the limitit as $\epsilon \to 0$

$\begin{aligned} \lim_{\epsilon \to 0} \Big|I_{C_4} \Big|&=\lim_{\epsilon \to 0} \epsilon \int_{-\pi}^{\pi}e^{x \epsilon\cos(\varphi)-a \epsilon \cos\left(\frac{\varphi}{2}\right)}\,d \varphi \\ &=\lim_{\epsilon \to 0} \epsilon \int_{-\pi}^{\pi} \lim_{\epsilon \to 0} \,e^{x \epsilon\cos(\varphi)-a \epsilon \cos\left(\frac{\varphi}{2}\right)}\,d \varphi \\ &= 2 \pi \lim_{\epsilon \to 0} \epsilon \longrightarrow 0 \end{aligned}$

For the integral over C_3 we let $s=s e^{i \pi}$ , then we obtain

$\begin{aligned} I_{C_3}&=\int_{C_3} e^{sx}e^{-a \sqrt{s}}\,ds\\ &=\int_{R}^{\epsilon} e^{x s e^{i \pi} }e^{-a \sqrt{s} e^{i \frac{\pi}{2}}}\,e^{i \pi}\,ds\\ &=\int_{\epsilon}^R e^{-x s -ia \sqrt{s} }\,ds\\ \end{aligned}$

Similarly for the integral over C_5 we let $s=s e^{-i \pi}$ , then we obtain

$\begin{aligned} I_{C_5}&=\int_{C_5} e^{sx}e^{-a \sqrt{s}}\,ds\\ &=\int_{\epsilon}^R e^{x s e^{-i \pi} }e^{-a \sqrt{s} e^{-i \frac{\pi}{2}}}\,e^{-i \pi}\,ds\\ &=-\int_{\epsilon}^R e^{-x s +ia \sqrt{s} }\,ds\\ \end{aligned}$

Taking the limits $\epsilon \to 0$ and $R \to \infty$ , according to (1) we have

$\begin{aligned} &\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds+\frac{1}{2 \pi i }\int_{0}^\infty e^{-x s -ia \sqrt{s} }\,ds-\frac{1}{2 \pi i }\int_{0}^\infty e^{-x s +ia \sqrt{s} }\,ds=0\\ &\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds-\frac{1}{2 \pi i }\int_{0}^\infty e^{-x s }\left(e^{i a \sqrt{s}-e^{-i a \sqrt{s}} \right)\,ds=0\\ &\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds=\frac{1}{ \pi }\int_{0}^\infty e^{-x s }\sin\left( a \sqrt{s} \right)\,ds\\ \end{aligned}$

Now

$\begin{aligned} \frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds&=\frac{1}{ \pi }\int_{0}^\infty e^{-x s }\sin\left( a \sqrt{s} \right)\,ds\\ &=\frac{2}{\pi}\int_{0}^\infty t e^{-x t^2 }\sin\left( a t \right)\,dt \qquad (\sqrt{s}=t) \end{aligned}$ (8)

Recall the result (see here)

$I(a)=\int_{0}^{\infty} \cos (a t) e^{-x t^{2}} d t=\frac{1}{2} \sqrt{\frac{\pi}{x}} e^{-\frac{a^{2}}{4 x}}$ (9)

Differentiating both sides w.r. to a

$\int_{0}^{\infty} t\sin (a t) e^{-x t^{2}} d t=\frac{a \sqrt{\pi}}{4} x^{-3/2} e^{-\frac{a^{2}}{4 x}}$ (10)

Plugging (10) in (8) we obtain

$\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-a \sqrt{s}}\,ds=\frac{a}{\sqrt{4\pi}} x^{-3/2} e^{-\frac{a^{2}}{4 x}}$ (11)

Letting $a=n\sqrt{2}$ we obtain

$\frac{1}{2 \pi i }\int_{c-i \infty}^{c+i \infty} e^{sx}e^{-n \sqrt{2s}}\,ds=\frac{n}{\sqrt{2 \pi}} x^{-3/2} e^{-\frac{n^{2}}{2 x}}$ (12)

Which is again a Levy-Smirnov density, proving it´s a stable distribution!

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