This isn't one specific homework question, but a concept I'm having trouble with in class. We were asked on a couple of questions recently on homework dealing with the probability density function of two random variables, $f_{X,Y}(x,y)$. We were given a joint probability density function and asked to find a given probability, like $P[X + Y > 5]$. One solid example being:
Given a Joint PDF $f_{X,Y}(x,y)$ let A be the event that $X + Y \leq 1$. Now, I know that $$P[A] = \int_A\int f_{X,Y}(x,y)dxdy$$ (at least, that's how my book writes it). My question is, how do I determine the bounds of those two integrals? In the solution manual for this particular problem, it is stated that the solution is $$P[X+Y\leq 1] = \int_0^1\int_0^{1-x}f_{X,Y}(x,y)dxdy$$
I can see pretty clearly that $x + (1-x)$ will always be less than or equal to one for $0 \leq x \leq 1$, so I understand why those bounds work. I'm just not sure I could have come up with them. Is there a process for figuring this out that I am missing? A similar question posed to us was given another joint PDF, let A be the event that $ X+Y>5$. I have no idea what the bounds would be for this one. Can someone show me how to find the integral bounds for questions like this?
Best Answer
In your example, it seems the density $f_{X,Y}(x,y)$ is zero when $x\leqslant0$ or when $y\leqslant0$ hence, to compute $\mathbb P(X+Y\leqslant1)$, the domain of integration is defined by the inequalities $$ x\geqslant0,\qquad y\geqslant0,\qquad x+y\leqslant1. $$ This is indeed equivalent to $$ 0\leqslant x\leqslant1,\qquad 0\leqslant y\leqslant 1-x, $$ but I know no systematic way of deducing the latter from the former, except drawing a rough sketch of the domain of interest.