marginal probability density function for variables with uneven range

Question

I'm working through some mock exam questions in preparation for an upcoming exam, and I'm stumped on this one.

given that $ f(x,y) = \frac 3 4 y, \quad x \in ]-1,1[, \quad 0 \le y \le x+1$, I'm to prove that the density functions are:

$f(x)=\frac 3 8 (1+x)^2$ if $x \in ]-1,1[$,

$f(y)=\frac 3 4 y(2-y)$ if $y \in ]0,2[$.

To find $x$, I integrated for $y$

$$f(x) = \int_0^{1+x} \frac 3 4 y\ dy = \frac 3 8 y^2 \Big| _0^{(1+x)} = \frac 3 8 (1+x)^2 .$$

So far so good, but when I try to integrate for $x$ I get lost.

$$f(y) = \int_{-1}^1 \frac 3 4 y\ dx = \frac 3 4 xy \Big| _{-1}^1 = - \frac 3 4 y + \frac 3 4 y=0 .$$

Can anyone explain what I'm doing wrong?

Answer 1

Two issues to consider:

• $\frac 3 4 xy \Big| _{-1}^1 = +\frac 3 4 y + \frac 3 4 y = \frac 3 2 y$, though ...
• ... for given $y$ from $0$ to $2$, you might have $y-1 \le x \lt 1$ as the limits of integration