Given the following joint density function

$$ f_{X,Y}(x,y) = \begin{cases} 3x, & \text{if $0\lt y \lt x \lt 1$} \\ 0, & \text{otherwise} \end{cases}$$.

We're then asked to find the marginal probability density function of Y.

[ so for $y\not\in (0,1)$ the marginal density of Y:

$$f_Y(y)=\int_{-\infty}^{\infty} 0 dx =0$$

then, for $y\in(0,1)$ and $0\lt y \lt x \lt 1$, the joint pdf is: $f_{X,Y}(x,y)=3x$ and $0$ otherwise.

Thus $$f_Y(y)=\int^{1}_{y} 3x dx =\frac{3}{2}-\frac{3}{2}y^2$$.]

and finally we end up with: $$f_{Y}(y) = \begin{cases} \frac{3}{2} - \frac{3}{2}y^2, & \text{if $0<y \lt 1$} \\ 0, & \text{otherwise} \end{cases}$$.

My questions is: Why exactly is it "if $0 \lt y \lt 1$"?

Is it because this is the range of the random variable Y such that the joint pdf is not zero and hence the marginal pdf of y is not zero?


1 Answer 1


$X$ and $Y$ are constrained by $0<Y<X<1$. For a given value of $X$, $Y$ is restricted to $(0,X)$ but as $X$ varies the only restriction you have on $Y$ is $0<Y<1$. So the density of $Y$ is supported by $(0,1)$.


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