Let X and Y have the joint probability density function given by:
f(x,y)=$\frac{1}{4}exp{\frac{-(x+y)}{2}}$, x>0 and y>0
(a) Find Pr(x<1,y>1)
(b) Find Pr(y<$x^2$)
This is how I tackled (a):
Pr(x<1,y>1)=$\int\int exp{\frac{-(x+y)}{2}}$ dydx where $0<x<1$ and $0<y<\infty$. My problem is with the interdependency of the two variables. Are they dependent or independent?
For (b), I had no clue at all. Someone should please help me out.
Both are just double integrals. (a) should be $$\int_0^1 \int_1^\infty \frac{1}{4}e^{-(x+y)/2} \, dy \, dx$$ Along similar line of reasoning, (b) should be $$\int_0^\infty \int_0^{x^2} \frac{1}{4}e^{-(x+y)/2} \, dy \, dx$$
We address your dependence/independence problem, but point out that one does not need to do it to set up and evaluate the integrals.
To find the marginal density of $X$, "integrate out" $y$. We get $f_X(x)$, where $f_X(x)=0$ for $x\le 0$, and $f_X(x)=\frac{1}{2}e^{-x/2}$ for $x\gt 0$.
Similarly, we find the marginal density $f_Y(y)$ of $Y$.
Note that the given joint density function is $f_X(x)f_Y(y)$. So $X$ and $Y$ are independent.
Remark: In general, if the joint density of $X$ and $Y$ factors as a product of a density function $g(x)$ and a density function $h(y)$, then $X$ and $Y$ are independent. We have to be careful about applying this, for the factorization must apply to full joint density, including the parts where the joint density is $0$.
So for example the function $f(x,y)$ which is $8xy$ for $0\le x\le y\le 1$ and $0$ elsewhere is not the joint density function of a pair of independent random variables. The problem is that the joint density "lives" on a triangle, not a rectangle. The full joint density does not factor.
