When reading another post, I was wondering about the definition of existence of expectation of a random variable.
-
From Kai Lai Chung,
We say a random variable $X$ has a
finite or infinite expectation (or
expected value) according as $E(X)$ is
a finite number or not. In the
expected case we shall say that the
expectation of X does not exist.I was wondering what it means by
"the expected case" in the last
sentence? Is this generally regarded
as the meaning of non-existence of
expectation? -
From Wikipedia:
Let X be a discrete random variable.
Then the expected value of this random
variable is the infinite sum
$$\operatorname{E}[X] = \sum_{i=1}^\infty x_i\, p_i, $$
provided that this series converges
absolutely (that is, the sum must
remain finite if we were to replace
all $x_i$'s with their absolute
values). If this series does not
converge absolutely, we say that the
expected value of $X$ does not exist.I was wondering
- if the meaning of nonexistence of expectation here is consistence with
the one by Kai Lai Chung, - if the meaning of nonexistence of expectation here is consistence with
the nonexistence of Lebesgue
integral in Rudin's book
where he says Legesgue integral
of a real-valued
Boreal-measurable function does not
exist if and only if the integrals
of the positive part and of the
negative part are both infinite, which allow the integral to exist when it is infinite. - if the expectation is infinite, then the expectation is regarded
as nonexistence?
- if the meaning of nonexistence of expectation here is consistence with
Best Answer
With usual notation, decompose $X$ as $X=X^+ - X^-$ (also note that $|X|=X^+ + X^-$). $X$ is said to have finite expectation (or to be integrable) if both ${\rm E}(X^+)$ and ${\rm E}(X^-)$ are finite. In this case ${\rm E}(X) = {\rm E}(X^+) - {\rm E}(X^-)$. Moreover, if ${\rm E}(X^+) = +\infty$ (respectively, ${\rm E}(X^-) = +\infty$) and ${\rm E}(X^-)<\infty$ (respectively, ${\rm E}(X^+)<\infty$), then ${\rm E}(X) = +\infty$ (respectively, ${\rm E}(X) = -\infty$). So, $X$ is allowed to have infinite expectation.
Whenever ${\rm E}(X)$ exists (finite or infinite), the strong law of large numbers holds. That is, if $X_1,X_2,\ldots$ is a sequence of i.i.d. random variables with finite or infinite expectation, letting $S_n = X_1+\cdots + X_n$, it holds $n^{-1}S_n \to {\rm E}(X_1)$ almost surely. The infinite expectation case follows from the finite case by the monotone convergence theorem.
If, on the other hand, ${\rm E}(X^+) = +\infty $ and ${\rm E}(X^-) = +\infty $, then $X$ does not admit an expectation. In this case, must of the following must occur (a result by Kesten, see Theorem 1 in the paper The strong law of large numbers when the mean is undefined, by K. Bruce Erickson): 1) Almost surely, $n^{-1}S_n \to +\infty$; 2) Almost surely, $n^{-1}S_n \to -\infty$; 3) Almost surely, $\lim \sup n^{ - 1} S_n = + \infty$ and $\lim \inf n^{ - 1} S_n = - \infty$.
EDIT: Since you mentioned the recent post "Are there any random variables so that ${\rm E}[X]$ and ${\rm E}[Y]$ exist but ${\rm E}[XY]$ doesn't?", it is worth stressing the difference between "$X$ has expectation" and "$X$ is integrable". By definition, $X$ is integrable if $|X|$ has finite expectation (recall that $|X|=X^+ + X^-$). So, for example, the random variable $X=1/U$, where $U \sim {\rm uniform}(0,1)$, is not integrable, yet has (infinite) expectation (indeed, $\int_0^1 {x^{ - 1} \,{\rm d}x} = \infty $). Further, it is worth noting the following. A random variable $X$ is integrable (i.e., ${\rm E}|X|<\infty$) if and only if $$ \int_\Omega {|X|\,{\rm dP}} = \int_{ - \infty }^\infty {|x|\,{\rm d}F(x)} < \infty . $$ A random variable has expectation if and only if $$ \int_\Omega {X^ + \,{\rm dP}} = \int_{ - \infty }^\infty {\max \{ x,0\} \,{\rm d}F(x)} = \int_0^\infty {x\,{\rm d}F(x)} < \infty $$ or $$ \int_\Omega {X^ - \,{\rm dP}} = \int_{ - \infty }^\infty {-\min \{ x,0\} \,{\rm d}F(x)} = \int_{ - \infty }^0 {|x|\,{\rm d}F(x)} < \infty. $$ In any of these cases, the expectation of $X$ is given by $$ {\rm E}(X) = \int_0^\infty {x\,{\rm d}F(x)} - \int_{ - \infty }^0 {|x|\,{\rm d}F(x)} \in [-\infty,\infty]. $$ Finally, $X$ does not admit an expectation if and only if both $\int_\Omega {X^ + \,{\rm dP}} = \int_0^\infty {x\,{\rm d}F(x)}$ and $\int_\Omega {X^ - \,{\rm dP}} = \int_{ - \infty }^0 {|x|\,{\rm d}F(x)} $ are infinite. Thus, for example, a Cauchy random variable with density function $f(x) = \frac{1}{{\pi (1 + x^2 )}}$, $x \in \mathbb{R}$, though symmetric, does not admit an expectation, since both $\int_0^\infty {xf(x)\,{\rm d}x}$ and $\int_{ - \infty }^0 {|x|f(x)\,{\rm d}x}$ are infinite.