Density and distribution functions of X + Y

Question

Two numbers $x$ and $y$ are randomly and independently selected in the interval $[0,1]$. Find the density and distribution functions of the random variable $X+Y$.

I have doubts about how to approach this exercise, because I do not know if the variables satisfy a uniform distribution, although I have not until the moment very clear the concept of convolution to find what is required. Any contribution would be appreciated, thank you very much!

Not a duplicate. Uniform distribution is not mentioned in the case of this question

Answer 1

Let $W$ be the event, such as : $W = X + Y$. Since you are referring on a density function, I'll assume we are working over a continuous case.

Then, the Cumulative Distribution Function, will be given as :

$$F_{W}(w) = \text{Pr}\{S\leq w\} = \text{Pr}\{X+Y \leq w\} = \text{Pr}\{X \leq w -Y\}=\int\int_Df(x,y)dxdy$$

This means that :

$$F_W(w)=\int\bigg[\int f(x,y)dx\bigg]dy=\int_0^1\bigg[\int_0^{w-y}f(x,y)dx\bigg]dy$$

But, we also know that :

$$f_W(w)=\frac{dF_W(w)}{dw}=\frac{d}{dw}\int_0^1\bigg[\int_0^{w-y}f(x,y)dx\bigg]dy=\int_0^1\bigg[\frac{d}{dw}\int_0^{w-y}f(x,y)dx\bigg]dy$$

$$=$$

$$\int_0^1f(w-y,y)dy=\int_0^1f(x,w-x)dx$$

So, we get :

$$f_W(w)=\int_0^1f(w-y,y)dy=\int_0^1f(x,w-x)dx$$

Now since $X,Y$ are independent, then this means that :

$$f(x,y)=f_X(x)f_Y(y)$$

which, by applying to the result above, yields :

$$f_W(w) = \int_0^1 f_X(x)f_Y(w-x)dx=(f_x\cdot f_y)(w)$$

Answer 2

The idea of the convolution tells you that if Z=X+Y and $X$ and $Y$ are independent then $$ p_Z(z)= \int_\Omega dx p_X(x)p_Y(z-x)$$ which is sometimes written as $p_X*p_Y$ (convolution).
This formula is valid in general independently of the specifics of $p_X$ and $p_Y$. If they are uniform then for $z \in [0,2]$ $$ p_Z(z) = \int_0^1 dx \Theta(z-x)\Theta(1-z+x), $$ and you recover the well-known result that $p_Z(z)=z$ for $0 \leq z \leq 1$, and $p_Z(z)=2-z$ for $1 \leq z \leq 2$.