Finding $P(X+Y<1)$ given $f(x,y) = 12xy^3$ when $0 < x < y < 1$

Question

So none of the student-assistants are able to figure this one out, but maybe you will. $$f(x,y) = 12xy^3\quad \text{for}\quad 0 < x < y < 1$$ If you draw out the figures you get that in this question $0<x<0.5$ and $0.5<y<1$, furthermore if you write $P(Y<1-X)$ you get $0.5 < y < 1-x$. Setting up the double integral for $x$ and $y$, I arrive at:

Integrate[Integrate[12*x*Power[y,3],{y,0.5,1-x}],{x,0,0.5}]=0.065625

This is not the right answer?! So, what am I doing wrong?

So the correct way of doing it was only slightly different as in the first integration i forget that $x < y $

Integrate[Integrate[12*x*Power[y,3],{y,x,1-x}],{x,0,0.5}]

Answer 1

I think you are trying to go too fast, trying to integrate everything at once. Whereas you have to split your reasoning into two cases.

From the definition of your joint probability density, $0 < x < y < 1$.

And you want to enforce the additional condition $X + Y < 1$.

So all in all, you have the two conditions: $\begin{cases} x < y \\ x + y < 1 \end{cases} \iff \begin{cases} x < y \\x < 1 - y \end{cases}$

From now on, you have to think about two cases:

• if $y < 1 - y$ (which is the same as $y < 0.5$), then $P(X+Y < 1) = \displaystyle\int_0^{0.5} \int_0^y 12xy^3 \mathrm{d}x\mathrm{d}y$
• if $1-y < y$ ($\iff y > 0.5$), then $P(X + Y < 1) = \displaystyle\int_{0.5}^1 \int_0^{1-y} 12xy^3 \mathrm{d}x\mathrm{d}y$^

To get the actual probability, all you have to do now is sum these two cases :) (since they are disjoint events)

---

Note: in your question, the integral you had was $\displaystyle\int_0^{0.5} \int_{0.5}^{1-x} 12xy^3 \mathrm{d}y \mathrm{d}x$

This didn't work because, not only is it just one of the two integrals that you needed, but also, in choosing the bounds of the inner integral to be $0.5$ and $1-x$, you are implicitly assuming that $y > 0.5$, and $y < 1-x < x$.

But $1 - x < x \iff x > 0.5$, and your integral for $x$ is between $0$ and $0.5$. So the bounds don't work.