Find the PDF of exponential distribution from its characteristic function

Question

I know that the characteristics function of the exponential distribution is as following: $$ \phi_x(t) =\frac{\lambda}{(\lambda -it)}$$ Also, I know that the pdf of the exponential distribution is: $$f_x(x)=\lambda e^{-\lambda x}$$

Moreover, I know that the relation ship between the pdf and the characteristics function can be describe as following: $$ f_x(x)= \int_0^\infty e^{-itx} \phi_x(t) $$ $$ f_x(x)= \int_0^\infty e^{-itx} \frac{\lambda}{(\lambda -it)} $$

However, I can't compute the last equation to find the exactly pdf that i already mentioned before. Could you guys help me to solve this integral. I used wolfram, but without any result.

Thanks .

Answer 1

${\displaystyle\int_\mathbb{R}|\phi(t)|\mathrm{d}t = \infty}$, so ${\displaystyle f(x) \neq \frac{1}{2\pi}\int_\mathbb{R}e^{-itx}\phi(t)\mathrm{d}t}$. Also note that the pdf of the exponential distribution is not uniformly continuous.

Instead, we can consider the cdf.

\begin{align} F(b) - F(a) &= \frac{1}{2\pi}\int_\mathbb{R}\frac{e^{-ita} - e^{-itb}}{it}\frac{\lambda}{\lambda - it}\mathrm{d}t\\ &= \frac{\lambda}{2\pi}\int_\mathbb{R}\frac{e^{-ita} - e^{-itb}}{t(i\lambda + t)}\mathrm{d}t. \end{align}

The residue at $0$ is $0$ and the residue at $-i\lambda$ is ${\displaystyle\frac{i(e^{-\lambda a} - e^{-\lambda b})}{\lambda}}$. Therefore, $F(b) - F(a) = {\displaystyle\frac{-\lambda}{2\pi}2\pi i\Bigg[0 + \frac{i(e^{-\lambda a} - e^{-\lambda b})}{\lambda}\Bigg]}$, where the negative sign is due to the contour's clockwise orientation. The expression simplifies to $e^{-\lambda a} - e^{-\lambda b}$.

Knowing $F(0) = 0$, $F(x) = 1 - e^{-\lambda x}$ for $x \geq 0$ and so $f(x) = \lambda e^{-\lambda x}, x > 0$.

Answer 2

If you start defining the probability density function as $$ f(x) = \lambda e^{-\lambda x} \Theta(x), $$ with $\lambda \in \mathbb{R}$, $\lambda>0$ and $\Theta(x)$ the Heaviside step function since we only want to take care of $x>0$. We can take its characteristic function performing the inverse Fourier transform of it, where $$ \phi(t) = \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} e^{itx}f(x) \mathrm{d}x = \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{itx} \lambda e^{-\lambda x} \mathrm{d}x = \frac{1}{\sqrt{2\pi}}\frac{\lambda}{\lambda - it}, $$ as written. We can invert the process computing the Fourier transform of the characteristic function $\phi(t)$, where $$ \begin{aligned} f(x) &= \frac{1}{\sqrt{2\pi}} \int_\mathbb{R} e^{-ixt}\phi(t) \mathrm{d}t = -\frac{\lambda e^{-\lambda x}}{2} \left[\mathrm{sgn}(x)\left(\mathrm{sgn}(|\mathrm{Re}(\lambda)|) - 1\right) -2\,\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right)\Theta(x\,\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right))\right] \\ &= \lambda e^{-\lambda x} \Theta(x), \end{aligned} $$ since $\mathrm{sgn}\left(\mathrm{Re}(\lambda)\right) = 1$.