I am having difficulty finding the posterior distribution of a Bayesian model with two parameters, which involves evaluating a double integral over an unbounded region. I prefer not to post the model for now, since this question is not strictly related to this problem in particular, but I probably will end up looking for help in another question.


I have been taught that Fubini's Theorem holds if the region of integration is bounded, but all the books in Statistical Inference I own and my teachers always evaluate such multiple improper integrals (expectations, densities, and so forth) as iterated single improper integrals, and most of the times those are Gamma, Beta or density functions.

I recall from multivariable calculus that if a function $f(x,y)$ is positive and $D = [a,\infty)\times[b,\infty)$ then, if the limit exists:

$$\int_D f = \lim_{(c,d) \to (\infty,\infty)}\int_b^d\int_a^c f(x,y) \, dx dy = \lim_{(c,d) \to (\infty,\infty)}\int_a^c\int_b^d f(x,y) \, dy dx $$

Which I think it is not quite "the product" of two single improper integrals, so there must be something I am missing. Whenever I have applied this theorem in the context of multivariable calculus (homework, exams) $f$ was always a carefully chosen function with anti-derivatives. However, as I have mentioned before, the improper double integrals I have come across in Statistical Inference are not that easy.

So, for instance, would it be correct to do something like this?

Let $f(x,y) = g(x)h(y)$, $D = [0,\infty)\times[b,\infty)$ and $g(x) = x^{\alpha-1}e^{-x}$:

$$\int_D f = \int_b^\infty\int_0^\infty f(x,y) \, dx dy = \Gamma(\alpha) \int_b^\infty h(y) \, dy $$

If so, how can it be justified?

Sorry for my imprecision and thanks in advance!

  • 2
    $\begingroup$ The statement of Fubini's theorem at Wikipedia does not require the region to be bounded. $\endgroup$ – hmakholm left over Monica Oct 31 '11 at 13:04
  • 1
    $\begingroup$ If $g(x)$ is a constant, then the area under it from zero to infinity ought to be infinite, no? $\endgroup$ – J. M. is a poor mathematician Oct 31 '11 at 13:05
  • $\begingroup$ @J.M. Oops, sorry! I've just edited my question. Hope it makes sense now. $\endgroup$ – user14559 Oct 31 '11 at 13:14
  • $\begingroup$ @HenningMakholm Thank you! I don't know anyhing about measure theory, but at least now I know I'm not doing things wrong. $\endgroup$ – user14559 Oct 31 '11 at 13:14

Two important points:

  • The hypotheses of Fubini's theorem should say nothing about boundedness or unboundedness of the regions over which you integrate.
  • You shouldn't call an integral "improper" merely because it's over an unbounded region.

Fubini's theorem makes this assumption about a double integral: $\displaystyle\int_{A\times B} |f(x,y)|\;d(x,y)<\infty$

It concludes that two iterated integrals are equal to the double integral: $$ \int_A \left(\int_B f(x,y)\;dx\right) \;dy = \int_{A\times B} f(x,y)\; d(x,y)= \int_B \left(\int_A f(x,y)\;dy\right) \;dx. $$

An iterated integral is not the same thing as a double integral. A double integral is an integral with respect to product measure on the same $A\times B$. An iterated integral involves integrating with respect to a measure on one space, getting a function of one variable (thus $\int_B f(x,y)\;dx$ is a function of $y$), then integrating that with respect to a measure on the other space.

The integral $\displaystyle\int_{(0,\infty)} x^{\alpha-1} e^{-x}\; dx$ is not an improper integral. It is defined like any Lebesgue integral of a nonnegative function: it's the smallest number that's not too small to be the integral (and if all numbers are too small, then it's $\infty$), and "too small" is defined by using integrals of simple functions that are dominated by the function being integrated.

An improper integral, on the other hand, is defined as a limit as one of the bounds of integration approaches something. For example $$ \lim_{x\to\infty}\int_0^x f(w)\;dw\text{ and }\lim_{x\to0+} \int_x^1 f(w)\;dw $$ are improper integrals. This must be done in cases where the integrals of the positive and negative parts of the function are both infinite. For example, that can be shown to happen with $\displaystyle\int_0^\infty \frac{\sin x}{x}\;dx$, so that integral is defined as $\displaystyle\lim_{b\to\infty}\int_0^b \frac{\sin x}{x}\;dx$. That's an improper integral.

  • $\begingroup$ Thank you for your answer. My theoretical background in integration is weak to say the least and my ignorance leads to questions like this one. I wrote "improper" in the sense of improper (Riemann) integrals of the first kind, which doesn't make sense at all in this context. $\endgroup$ – user14559 Oct 31 '11 at 15:38
  • $\begingroup$ Maybe I should add that even when an integral like $\int_0^\infty \frac{dx}{1+x^2}$ is not really "improper" in the sense that it's defined as $\lim\limits_{b\to\infty}\int_0^b \frac{dx}{1+x^2}$, nonetheless finding that limit is often the best way to actually compute the integral. $\endgroup$ – Michael Hardy Oct 31 '11 at 17:26

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