2
$\begingroup$

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?

  • 0
    Yes - as Sivaram Ambikasaran has said when you asked the same quaetion2012-03-12

2 Answers 2

2

$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx$

Note that both, not surprisingly, yield the same answer.

$\int_0^{\infty} \int_0^{y/2} \exp(x-y) dx dy = \int_0^{\infty} (\exp(-y/2) - \exp(-y)) dy = 1$

$\int_0^{\infty} \int_{2x}^{\infty} \exp(x-y) dy dx = \int_0^{\infty} \left. - \exp(x-y) \right|_{2x}^{\infty} dx = \int_0^{\infty} \exp(-x) dx = 1$

  • 0
    $\int_{2x}^{\infty} \exp(x-y)dy = \exp(-x)$ is the marginal distribution of $x$.2012-03-12
1

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.