# Finding marginal pdf

I'm given a joint pdf with the following bounds

f(x,y)= \begin{cases} x^2 & \text{ for } 0 \leq y \leq 1-x^2 \\ 0 & \text{ otherwise} \end{cases}

I am trying to find the Marginal pdf of y. I thought the integral would be from -1 to 1 since that's what it seems to be bound by on the graph, but that's wrong. What am I doing wrong?

Draw the downward-facing parabola $y=1-x^2$. To find the (marginal) density of the random variable $y$, we "integrate out" $x$.
But first we need to make a fix. The "density" is not a density, since the double integral of our "density" over the region below the parabola and above the $x$-axis is $\frac{4}{15}$. To make it a density we need to use $\frac{15}{4}x^2$, not $x^2$.
It is clear that the density is $0$ outside the interval $[0,1]$. Inside the interval, we go from $x=-\sqrt{1-y}$ (the left branch of the parabola) to $x=\sqrt{1-y}$ (the right branch). So inside the interval we want $$\int_{x=-\sqrt{1-y}}^{\sqrt{1-y}} \frac{15}{4} x^2\,dx.$$ Integrate. We get $\frac{5}{2}(1-y)^{3/2}$ on the interval $[0,1]$.
$$f_Y(y)=\int_{-\sqrt{1-y}}^{\sqrt{1-y}}x^2dx=\dfrac{2(1-y)^{3/2}}{3}$$ for $0<y<1$.
• @risdel from the support: $0< y < 1- x^2 < 1$. This means that $1>1-y>x^2>0$, which in turn means that ${-}\sqrt{1-y}<x<\sqrt{1-y}$ and $0<y<1$. Jul 16 '15 at 3:38
• @risdel Yes. It is algebraically reasonable to do so. Plot $\,y= 1-x^2\,$. The area below this curve and above the x-axis is the support of the joint distribution. Draw a series of horizontal line segments across this area (particularly at $y=0.36$ and $y=0.64$). From this we can see from the plot we have $\,x\in\big({-}\sqrt{1-y\,}..{+}\sqrt{1-y\,}\big)\,$ for each $\,y\in\big(0..1\big)\,$. Jul 16 '15 at 4:20