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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?

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    Because the way you have interchanged the integral is incorrect.2012-03-12
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    The domain of integration on the right-hand side is $$ D = \{ (x, y) \in \mathbb{R}^2 \ : \ 0 \leq x \leq y/2 \}.$$ You must carefully reinterpret this domain when interchanging the order of integration. In this example, you will have $0 \leq 2x \leq y$ and hence $$\int_{0}^{\infty} \int_{2x}^{\infty} e^{x-y} \; dy dx.$$ Limits can be deceiving, but the domain not.2012-03-12
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    The LHS has $y$ integrated away initially and then reappearing as a limit in the integral over $x$. This is confusing at best, but more likely an error.2012-03-12
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    @Henry: So $\int_{2x}^{\infty} \exp(x-y) \ dy$ be would be $f_{X}(x)$?2012-03-12
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    Yes - as Sivaram Ambikasaran has said when you asked the same quaetion2012-03-12

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

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

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

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

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    So $\int_{2x}^{\infty} \exp(x-y) \ dy$ be would be $f_{X}(x)$?2012-03-12
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    $\int_{2x}^{\infty} \exp(x-y)dy = \exp(-x)$ is the marginal distribution of $x$.2012-03-12
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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.