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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

Hendrik Matamoros
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2 Answers2

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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 :

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

Rebellos
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  • The information is very complete, thanks for that, but I can not see clearly what is the explicit form of the density and distribution functions that I am asked to find? – Hendrik Matamoros Feb 15 '18 at 23:37
  • @HendrikMatamoros If you are not given a specific $f(x,y)$, you cannot yield a specific example of course. It depends on the form of $f(x,y)$. The question you're being asked is expected to be answered exactly like that. The explicit form is the expressions given and elaborated by integration from $F_W(w)$ and $f_W(w)$. – Rebellos Feb 15 '18 at 23:38
  • If they tell me that both $X$ and $Y$ have a uniform distribution in $[0,1]$, they are no longer implicitly giving me the form of the variables $X$ and $Y$, like that: $f_X(x)=f_Y(x)=1$ if $0\leq x\leq1$ and $0$ otherwise. And then PDF is $f(w) = \begin{cases} w & \text{for $0 < w < 1$} \ 2-w & \text{for $1 \le w < 2$} \ 0 & \text{otherwise.} \end{cases}$ – Hendrik Matamoros Feb 15 '18 at 23:47
  • @HendrikMatamoros Yes, but you haven't been given that $X,Y$ are uniformly distributed in your question form. This is a statement from the link that was posted in your question, which is slightly something different, as the information for the uniform distribution opens a path to known forms for the functions describing the events, which means you can figure out a specific result. – Rebellos Feb 15 '18 at 23:50
  • Sorry, I thought that was implicit in the statement, but I see that apparently not. In the same way, thank you very much for the support, it has been very useful! – Hendrik Matamoros Feb 15 '18 at 23:57
  • @HendrikMatamoros I'm glad to have helped ! And yes, uniform distribution means that the distribution of the events have some specific characteristics, which lead to the result you mentioned. Your specific case is different though, as you haven't been given any type of distribution, so this means you have to work around a general case! – Rebellos Feb 15 '18 at 23:59
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    You're absolutely right, thank you very much for everything!. Greetings from Venezuela. – Hendrik Matamoros Feb 16 '18 at 00:01
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    Greetings from Greece ! – Rebellos Feb 16 '18 at 00:03
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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 $1 \leq z \leq 2$.

Jonathan D
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