Using characteristic functions, find the pdf of $Z = X + Y$

Question

A R.V. $X$ has a pdf of the form $f_{X}(x) = e^{-x} u(x)$ and an independent R.V. $Y$ has a pdf of $f_{Y}(y) = 3e^{-3Y} u(y)$ using characteristic functions, find the pdf of $Z = X + Y$.

This is different from previous problems I have encountered where a joint pdf $f_{XY}$ is given and I can marginalized out $X$ and $Y$ (prove independence) and then perform convolution to find the pdf of $Z$. I am unsure how to approach this problem to use characteristic functions.

I know it should follow the form $$\int_0^{\infty} e^{itx} \lambda e^{-\lambda x } dx = \frac{\lambda}{it - \lambda} e^{(it - \lambda)x}\bigg|_0^{\infty} = \frac{\lambda}{ \lambda- it}$$

But then what?

Answer 1

You have the sum of two exponential random variables.

$$Z=X+Y$$

Therefore,

$$Z(s)=X(s)Y(s)= \frac{1}{1+s}\cdot \frac{3}{3+s}$$

The Laplace transform of an exponential random variable is $\lambda/(\lambda+s)$ and the sum of variables translates into the product of transforms.

To find the pdf of $Z$ you need to apply the inverse transform.

Answer 2

It is known that $$ f_Z (z) = \int_{-\infty}^{\infty} f_{XY}(t,z-t) dt. $$ We can show this in at least two ways. The first one is using the change-of-variable formula. The second one is using the characteristic function.

In this case, $X$ and $Y$ are independent, so $$ f_Z (z) = \int_{-\infty}^{\infty} f_{X}(t)f_{Y}(z-t) dt = 3e^{-3z}\int_{-\infty}^{\infty} e^{2t} u(t)u(z-t) dt. $$