Given $X$ an exponentially distributed random variable with parameter $\lambda$ and $Y$ a uniformly distributed random variable between $-C$ and $C$. $X$ and $Y$ are independent. I'm supposed to calculate the distribution of $X + Y$ using convolution.
Does anyone have any tips on how to do this?
I understand that the convolution is represented by
$$\int_{-\infty}^{+\infty} f_1(x) \cdot f_2(z-x)dx\tag {1}$$
and so given $$f_1(x) = \begin{cases}
\lambda e^{-\lambda x} &, x\ge 0\\ 0 &, x < 0 \end{cases}$$
and $$f_2(y) = \begin{cases} \frac{1}{2C} &, y \in [-C,C]\\ 0 & , \text{otherwise}\end{cases}$$
(1) becomes: $$\int_{-\infty}^{+\infty} \lambda e^{-\lambda x}\cdot \frac{1}{2C} dx$$
but I don't know how to procede from here. How do I choose which intervals of $z$ to integrate?
Best Answer
Your final integral is incorrect; where is $z$ - it needs to be in your integral limits?
It is probably easier to calculate
$$\int_{-\infty}^{+\infty}f_1(z-x)\cdot f_2(x)dx= \begin{cases} \int_{-C}^{+C}\lambda e^{-\lambda (z-x)}\cdot \frac{1}{2C} &, z \ge x\\ 0 &, z\lt x \end{cases}$$
On second thoughts - your way is better
$$\begin{align}\int_{-\infty}^{+\infty}f_1(x)\cdot f_2(z-x)dx&= \begin{cases} \int_{z+C}^{z-C}\lambda e^{-\lambda x}\cdot \frac{1}{2C} dx &,z \ge C \\ \int_{z+C}^0\lambda e^{-\lambda x}\cdot \frac{1}{2C} dx &,C \ge z \ge -C \\ 0 &,z\lt -C \end{cases} \end{align} $$