interchanging integrals Why does $$\int_0^{y/2} \int_0^\infty e^{x-y} \ dy \ dx \neq \int_0^\infty \int_0^{y/2} e^{x-y} \ dx \ dy$$
The RHS is 1 and the LHS side is not. Would this still be a legitimate joint pdf even if Fubini's Theorem does not hold?
 A: $$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx$$
Note that both, not surprisingly, yield the same answer.
$$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} (\exp(-y/2) - \exp(-y)) dy = 1$$
$$\int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx = \int_0^{\infty} \left. - \exp(x-y) \right|_{2x}^{\infty} dx = \int_0^{\infty} \exp(-x) dx = 1$$
A: The right side,
$$\int_0^\infty \int_0^{y/2} e^{x-y} \ dx \ dy,$$
refers to something that exists.  The left side, as you've written it, does not.  Look at the outer integral:
$$
\int_0^\infty \cdots\cdots\; dy.
$$
The variable $y$ goes from $0$ to $\infty$.  For any particular value of $y$ between $0$ and $\infty$, the integral $\displaystyle \int_0^{y/2} e^{x-y}\;dx$ is something that depends on the value of $y$.
The integral $\displaystyle \int_0^\infty \cdots\cdots dy$ does not depend on anything called $y$.
But when you write $\displaystyle \int _0^{y/2} \int_\text{?}^\text{?} \cdots \cdots$ then that has to depend on something called $y$.  What is this $y$?  On the inside you've got $\displaystyle\int_0^\infty e^{x-y}\;dy$.  Something like that does not depend on anything called $y$, but does depend on $x$.  It's like what happens when you write
$$
\sum_{k=1}^4 k^2.
$$
What that means is
$$
1^2 + 2^2 + 3^2 + 4^2
$$
and there's nothing called "$k$" that it could depend on.
