find $P(X>Y)$ in $f(x; y) = \frac67(x^2 − y^2) $; $x > 0$; $ y < 1$

Question

find $P(X>Y)$ in $f(x; y) = \frac67(x^2 − y^2) $; $x > 0$; $ y < 1$

I am aware this can be solved with a double integral, such that:

$$ P(X>Y)=\int\int\frac67(x^2 − y^2) dydx$$

However, I do not know how to set the boundaries?

$$ P(X>Y)=\int_\infty^0\int_{\infty}^x\frac67(x^2 − y^2) dydx$$

What is the solution to $P(X>Y)$ ?

Answer 1

You have to see how the set $\{ (x,y): x>y \}$ intersects with $\{(x,y): x>0, y<1 \}$. The intersection can be written as:

$D_1\sqcup D_2$, where $D_1:= \{ (x,y): x>y, 0<x<1, y<1\}$ and $D_2:=\{ (x,y): x\geq 1, y<1 \}$.

By additivity you then have to calculate:

$\int_{D_1}f(x,y)dydx+ \int_{D_2}f(x,y)dydx$

Notice that for a rectangle $D=\{ (x,y) \in \mathbb{R}^2: a\leq x\leq b, c\leq y \leq d \}$, we have:

$\int_D f(x,y)dx dy= \int_a^b \Big( \int_c^d f(x,y) dy \Big) dx= \int_c^d \Big( \int_a^b f(x,y) dx \Big) dy$

In our case:

$\int_{D_2} f(x,y)dx dy= \int_1^\infty \Big( \int_{-\infty}^1 f(x,y) dy \Big) dx$

And for a set $C=\{ (x,y)\in \mathbb{R}^2: a\leq x\leq b, g_1(x) \leq y\leq g_2(x) \}$, where $g_1,g_2$ are functions of $x$, we have:

$\int_Cf(x,y)dxdy= \int_a^b \Big( \int_{g_1(x)}^{g_2(x)} f(x,y)dy \Big)dx$

In our case:

$\int_{D_1}f(x,y)dydx = \int_0^1 \Big( \int_{-\infty}^x f(x,y) dy \Big)dx$

I decomposed the intersection to sets such that the upper bound is a constant and explicit function on them. Were you to impose more conditions, say for example $Y<-X+3$ and $Y>-100$, then I would have to consider the set:

$\{ (x,y): x<y, x>0 , y<1, y<-x+3, y>-100 \}$

And decompose them into sets where the upper bound and lower bounds are constant functions. Which observing this set in $\mathbb{R}^2$ would give me a natural decomposition looking at the set with respect to the '$x$' axis:

$E_1:= \{ (x,y): 0<x<1, -100<y<x \}$, $E_2:=\{ (x,y): 1 \leq x \leq 2, -100<y<1 \}$ and $E_3:=\{ (x,y): 2<x<103, -100<y<-x+3 \}$

And summed the integrals on each set.

Answer 2

Note that $x > 0$ and $y < 1$ can only satisfy $x > y$ on the interval $(0, 1)$.

We can choose $x$ arbitrarily on the interval $(0, 1)$.

We must have $y$ varying from $0$ to $x$ (so that the inequality $x > y$ is satisfied!).

Thus,

$$P(X > Y) = \int_{0}^{x}\int_{0}^{1} f(x, y) \mathop{dx} \mathop{dy}. $$

Alternatively, we can have $y$ vary arbitrarily, and we can have $x$ vary from $y$ to $1$.

Therefore,

$$P(X >Y) = \int_{y}^{1} \int_{0}^{1} f(x, y) \mathop{dy} \mathop{dx} $$