# Help me with this question. This is the question 5.7 of the book Hoel

Let (u, v)  be chosen uniformly in the square  0<=u<=1 and 0<=v<=1. Let x be the random variable  that associate the number  u+v to the point (u, v). Get the Distribution function of x.

Let $f_X$ be the full density function of$X$, and let $f_Y$ be the full density function of $Y$. Let $Z=X+Y$. Then the pdf of $Z$ is given by
$$f_Z(z)=\int_{-\infty}^\infty f_X(x)f_Y(z-x)\,dx. (*)$$

Note that $f_x =1$ for $x\in(0,1)$ and  $f_y =1$ for $y\in(0,1)$. Using (*) we can easily see that  $f_Z(z)=0$ for $z\lt 0$, and also for $z\ge 2$. Next we compute $f_Z$ on $(0,2)$. We break it down into two cases (i) $0\lt z\le 1$ and (ii) $1\lt z\lt 2$.

(i) The product $f_X(x)f_Y(z-x)$ is ither $0$ or $1$. We have $f_Y(z-x)=1$, if $z-x\ge 0$, that is, $x\le z$. Hence in this case
$$f_Z(z)= \int_0^z 1\,dx=z, 0\lt z\le 1.$$

(ii) Suppose that $1\lt z\lt 2$. Then $f_Y(z-x)$ to be $1$, if we have $z-x\le 1$, that is  $x\ge z-1$. Thus
$$f_Z(z)=\int_{z-1}^1 1\,dx=2-z, 1\lt z\lt 2.$$

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