# Distribution function method for transformation.

Let $x_1, \ x_2$ be independent random variables.

The probability density function $f(x_i)= 2x \ \ \ \ for \ \ \ \ 0<x_i < 1$ and $f(x_i)=0 \ \ \ \text{otherwise}$

Then, the density function $f(x_1,x_2)=4x_1x_2$ for $0< x_1, x_2 < 1$

And Y is defined as $Y=X_1+X_2$

What is the distribution function $G(Y)$? Solve by using distribution function method(graph method)

What I did is as follows:

$G(Y)=P(Y \le y)= P( X_1+X_2 \le y)=\int \int f(x_1,x_2) dx_2 dx_1$

And I draw the plot of $Y=X_1+X_2$. But I'm stacking here. Please show me.

Comment: Notice that $X_i \sim Beta(2,1).$ The following R code simulates 100,000 realizations of $Y$, and makes their 'empirical distribution function'. The CDF of $Y$ obtained from the Answers by @BCLC or @probablyme should very closely match the ECDF.

 m = 10^5;  x1 = rbeta(m, 2, 1);  x2 = rbeta(m, 2, 1);  y = x1 + x2
plot.ecdf(y, pch=20) Wolfram Alpha gives us bounds. Try changing $y=1$ with different numbers here.

So we have

$$F_Y(y) = P(X_1 + X_2 \le y) = \int_0^{y} \int_0^{y - x_1} 4x_1x_2 dx_2dx_1$$

Keep going, you just need the bounds. Draw a picture for yourself to the see regions of interest. Then, if you want to use a double integral, then when $0\leq y\leq 1$, \begin{align*} P(X_1+X_2\leq y)&= P(X_2 \leq -X_1 +y)\\ &= \int_0^y \int_0^{-x_2+y} f(x_1,x_2)\,dx_1dx_2 \end{align*}

When $1<y\leq 2$, it is easier to do \begin{align*} P(X_1+X_2\leq y)&= 1 - P(X_2 > -X_1 +y)\\ &= 1 - \int_y^2 \int_{-x_2+y}^2 f(x_1,x_2)\,dx_1dx_2 \end{align*}