Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the
distribution function of $Z=X+Y$ – how to break into cases?
I first found the density functions:
$$
f_x(t) =\begin{cases}
1 && t\in[0,1] \\
0 && else
\end{cases}
$$
$$
f_y(t) =\begin{cases}
\frac{1}{3} && t\in[-1,2] \\
0 && else
\end{cases}
$$
Now:
$ F_z(t)=P(Z\leq t)=P(X+Y\leq t)=\int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$
But now I am stuck with breaking the result into cases.
How could it be done? I simply can't understand how to break that into cases when we have two differently distributed variables?
With one variable I would usually draw the function and then break to cases according to it's behavior, but how could it be done here?
Thanks
Best Answer
By convolution theorem we get
$$f_Z(z) = \int_{0}^{1} f_X(x) f_Y(z-x)\ \text{dx}.$$
In order to make non vanishing integrand we should have $-1 \leq z-x \leq 2$ i.e. $z-2 \leq x \leq z+1$. Observe that $-1 \leq z \leq 3 .$ Now there are three cases to consider
$(1)$ $z-2 < 0 \leq z+1 < 1.$
$(2)$ $z-2 < 0 < 1 \leq z+1.$
$(3)$ $0 \leq z-2 \leq 1 < z+1.$
If you analyze each of these cases you will find that the probability density function $f_Z$ of $Z = X + Y$ is defined as follows $:$
$$f_Z(z) = \begin{cases} \frac 1 3(z+1) & \text{for $-1 \leq z < 0$} \\ \frac 1 3 & \text{for $0 \leq z < 2$} \\ \frac 1 3 (3-z) & \text{for}\ 2 \leq z \leq 3 \\ 0 & \text{elsewhere} \end{cases}$$