# 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?

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

• when you wrote $z-y$, did you mean $t-y$? – Keen-ameteur Feb 11 at 15:17
• @Keen-ameteur yes, my fault – superuser123 Feb 11 at 15:19
• Use convolution theorem. – Dbchatto67 Feb 11 at 15:21
• @Dbchatto67 Isn't there another way? I don't have in with convolutions. – superuser123 Feb 11 at 15:22
• The range of $z$ is from -1 to 3. I think the distribution of $z$ can be written as a piece-wise continuous function in three parts, with the 'joins' at $z=0$ and $z=2$. – Paul Aljabar Feb 11 at 15:27

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 $$:$$

$$f_Z(z) = \begin{cases} \frac 1 3(z+1) & \text{for -1 \leq z < 0} \\ \frac 1 3 & \text{for 0 \leq z < 2} \\ \frac 1 3 (3-z) & \text{for}\ 2 \leq z \leq 3 \\ 0 & \text{elsewhere} \end{cases}$$

First notice that for the inside integral to contribute something you must have:

$$0 \leq t-y\leq 1$$

Which means that:

$$\int_{-1}^{2}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy= \int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy$$

But you should notice that:

$$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy\neq \int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$$

But instead: $$t\leq -1$$ then $$\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$$, and if $$t>2$$ $$\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$$. else:

$$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=\begin{cases}0 & ,t<-1 \\ 1 & ,t>2 \\ \int_{(t-1)\vee -1}^{t}\int_{0}^{t-y} \frac{1}{3} dxdy & ,\text{else} \end{cases}$$

Assuming that you know that the sum of two independent random variables is their convolution, we have $$F_z(t)= \int_{-\infty}^{\infty}f_X(t-\tau)f_Y(\tau) \ d \tau = \frac{1}{3} \int_{-1}^{2}f_X(t-\tau) \ d \tau = \frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx$$ Now, we have cases according to the integral boundaries and the PDF of $$X$$, namely:

1) If $$t+1<0$$, i.e. $$t<-1$$, the integral evaluates to zero.

2)If $$t-2>1$$, i.e. $$t>3$$, the integral evaluates to zero.

3) If $$t-2<0$$ ($$t<2$$) given that $$t+1<1$$, i.e. $$t<0$$ So for $$-1 < t < 0$$ $$F_z(t)=\frac{1}{3}\int_{0}^{t+1}f_X(x) \ dx = \frac{1}{3}(t+1)$$

4) If $$t-2 > 0$$ ($$t > 2$$). So for $$2

$$F_z(t)=\frac{1}{3}\int_{t-2}^{1}f_X(x) \ dx = \frac{1}{3}(3-t)$$

5) Now $$0 < t < 2$$

$$F_z(t)=\frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx =\frac{1}{3}\int_{0}^{1} 1 \ dx =\frac{1}{3}$$