Joint Density Function Boundaries

Question

I've been having trouble understanding the following exercise:

The joint density function of the random variables X and Y is:

$$ f(x, y) =\left\{ \begin{array}{ll} 6x,&0<x<1, &0<y<1-x \\ 0 &elsewhere \\ \end{array} \right. $$

a) Show that X and Y are not independent

b) Find P(X > 0.3|Y = 0.5)

I know that the answers are:

a)

and for Part b:

But what I don't get is how to get the integration boundaries. For example in Part b, I don't see why it should be integrated from 0.3 to 1 - .5, instead of integrating from 0.3 to 1.

How do I know when the boundaries of the integral are different?

Answer 1

Your domain on which $(X,Y)$ are defined is a triangle in $\mathbb{R}^2$ with vertices $(0,0),(1,0)$ and $(0,1)$. If you want to integrate $dx$ first, the boundaries would be $0 < y < 1$ and $0 < x< 1-y$.

This explains the choice of the upper bound of the integral in both parts of the problem.

Answer 2

The integration bounds are determined by the intersection of the condition and the support.

The support is $\{\langle x,y\rangle: 0\leqslant x\leqslant 1-y\leqslant 1\}$

Thus for the marginal density function for $Y$, we integrate wrt $x$ over its support relative to the value for $y$:- $$\begin{align}f_{\small Y}(y) &= \int_\Bbb R f_{\small X,Y}(x,y)\,\mathrm d x\\[1ex]&=\int_\Bbb R 6x\mathbf 1_{0\leqslant y\leqslant 1\cap 0\leqslant x\leqslant 1-y}\,\mathrm d x\\[1ex]&= 6\mathbf 1_{0\leqslant y\leqslant 1}\int_0^{1-y}x\,\mathrm d x\\[1ex]&=3(1-y)^2\mathbf 1_{0\leqslant y\leqslant 1}\\[4ex]f_{\small X\mid Y}(x\mid y)&=\dfrac{6x }{3(1-y)^2}\mathbf 1_{0\leqslant x\leqslant 1-y\leqslant 1}\\[2ex]\mathsf P(X>0.3\mid Y=0.5)&=\int_\Bbb R \mathbf 1_{x>0.3}f_{\small X\mid Y}(x\mid 0.5)\,\mathrm d x\\[1ex]&=\int_{\Bbb R}\dfrac{2x}{(1-0.5)^2}\mathbf 1_{0.3<x\leqslant 1-0.5}\,\mathrm d x\\[1ex]&=8\int_{0.3}^{0.5} x\,\mathrm d x\\[1ex]&=\dfrac{16}{25}\end{align}$$