Statistics - Laplacian Characteristic Function

Question

I was asked to find the characteristic function of the Laplacian random variable. And, from that, to find the mean of it.

Im having trouble solving this. That's what I have done so far, but it doesn't match the answer on the book:

Laplacian R.V.: $f(x)={\alpha}/2*e^{-\alpha|x|}$

To find the characteristic function, I have done the following steps: $CF(w)=\alpha/2\int_{0}^{\infty} e^{jwx} e^{-\alpha x} dx + \alpha/2\int_{-\infty}^{0} e^{jwx} e^{\alpha x} dx$

$CF(w)=\alpha^2/\alpha^2+j^2w^2$

I don't understand why in my answer I have the $j$ imaginary number. It does not appear on the answer of the book.

Answer 1

You did two things wrong. First of all, you have $j$ in your answer because you did not notice that $j^2=1$. But before that, you did the integrals wrong, or much more likely, you did the integrals right but you got wrong the step where $$ \frac{1}{\alpha-i\omega}-\frac{1}{\alpha+i\omega}=\frac{2\alpha}{\alpha^2+\omega^2} $$ Note the denominator.

Answer 2

One may observe that, for $\alpha>0$, $w \in \mathbb{R}$, one has

$$ \int_0^\infty \cos(wx)\:e^{-\alpha x}dx=\Re \int_0^\infty e^{-(\alpha -wj) x}dx=\Re \:\frac1{\alpha-wj}=\frac{\alpha}{\alpha^2+w^2} \tag1 $$

Then one gets $$ \begin{align} CF(w)=&\alpha/2\int_{0}^{\infty} e^{jwx} e^{-\alpha x} dx + \alpha/2\int_{-\infty}^{0} e^{jwx} e^{\alpha x} dx \\\\=&\alpha/2\int_{0}^{\infty} e^{jwx} e^{-\alpha x} dx + \alpha/2\int_{0}^{\infty} e^{-jwx} e^{-\alpha x} dx \\\\=&\alpha\int_{0}^{\infty} \cos(wx) \:e^{-\alpha x} dx \\\\=&\frac{\alpha^2}{\alpha^2+w^2} \end{align} $$ where we have used $(1)$ in the last step.