Probability distribution for sum of RV's

Question

I know my answer is incorrect because I end up with a pdf that doesn't integrate to $1$.

Given that $X$ and $Y$ are continuous RV's with joint density function:

$$f_{X,Y}(x,y) = \begin{cases}24xy \text{} \begin{cases} 0 < x < 1 \\ 0< y < 1 \\ 0 < x + y < 1\end{cases} \\ 0 \text{ otherwise}\end{cases}$$

Find the probability density of $Z = X + Y$.

My attempt:

Define $W = Y$ so that:

$$\begin{matrix} x = z - w, \ \ \ y = w \\z = x + y,\ \ \ w=y \ \end{matrix}$$

That means:

$$f_{Z,W}(z,w) = \begin{cases}24(z-w)w \left\vert{\dfrac{\partial(x,y)}{\partial(w,z)}}\right\vert \begin{cases} 0 < w < 1 \\ 0< z < 1 \end{cases} \\ 0 \text{ otherwise}\end{cases}$$

Calculating the determinant of the Jacobian gives

$$\begin{vmatrix} x_z \ x_w \\ y_z \ y_w \end{vmatrix} = \begin{vmatrix} 1 \ -\!1 \\ 0 \ \ \ \ 1 \end{vmatrix} = 2$$

Meaning

$$f_{Z,W}(z,w) = 48(zw-w^2)$$

So I should get the p.d.f. by integrating this over $w$

$$f_Z(z) = \int_0^1 48(zw-w^2) \, \mathrm dw = 48\left.\left({\frac {zw^2}{2}-\frac{w^3}{3} }\right)\right \vert_{w=0}^{w=1} = 24z - \frac {48} 3$$

but this is not a probability density function :(

Answer 1

Corrections: Determinant of jacobian is $1$ and limits of the integral should be $0$ to $z$. You didn't use $0 < x < 1 \implies 0 < z-w<1$.

Answer 2

Or you can do this without Jacobians:

Let $Z = X+Y$. Then $$ P(Z \leq z) = \int_{x=0}^z \int_{y=0}^{z-x}24xy\,dx\,dy = \int_{x=0}^z 12x(z-x)^2 dx= \int_{x=0}^z (12z^2x - 24 zx^2+12x^3) dx\\ = 6z^4 -8z^4 + 3z^4 = z^4 $$ (and of course, $P(Z \leq z) = 1$ for $z \geq 1$).

Then $$f_Z(z) = \frac{d(P(Z \leq z) )}{dz} = \frac14z^3$$ on $0\leq z \leq 1$ and zero elsewhere.