Suppose that the random variable $X$ is $\mid X \mid<M$ almost surely, for some constant $M<\infty.$ Then can we say that $E(X)<C$ for some constant $C<\infty$?

If the expectation is not bounded, is there any additional (other than bounded a.s.) conditions on $X$ that would give us a finite expectation?

My confusion is there could be $0\cdot \infty$ terms where the random variable is infinite with probability $0$. And these terms are not defined well. I believe this post disproves the converse. This is not homework just self studying. Thank you.

  • $\begingroup$ Just for the sake of nitpicking: is $X$ non-negative? Otherwise, you may need absolute values around $X$ in your assumption of upper bounded a.s. $\endgroup$ – Clement C. Feb 14 '16 at 0:34
  • $\begingroup$ @ClementC. Thanks, Edited as you suggested. $\endgroup$ – seek Feb 14 '16 at 0:35
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    $\begingroup$ In measure theory, the $0 \cdot \infty$ terms are zero by convention. $\endgroup$ – fourierwho Feb 14 '16 at 0:51
  • $\begingroup$ @fourierwho Thanks, does that mean the statement is correct? $\endgroup$ – seek Feb 14 '16 at 6:17

The definition of almost-sure boundedness, or essential boundedness is the following (see also this definition):

$Y:(\Omega, \mathcal{F}, \mathrm{P})\to \mathbb{R}$ is an essentially bounded random variable if there is a $M\geq 0$ so that $\mathrm{P}[|X|>M]=0$, i.e., the set $\{\omega: |X(\omega)|>M\}$ has measure zero.

Remark An essentially bounded function is not necessarily bounded. You can find relevant examples here.

A very easy way to give an affirmative answer to your question is the following: Let $X:(\Omega, \mathcal{F}, \mathrm{P})\to [0,+\infty)$ be a finite-valued non-negative continuous random variable. In this case we may use the following representation of the expectation (see this Wikipedia article):

$$\mathbb{E}[X]=\int_{\Omega}X(\omega)\mathrm{P}(\mathrm{d}\omega)=\int_{0}^{\infty}\mathrm{P}[X\geq x]\mathrm{d}x,$$

where $f(x) = \mathrm{P}[X\geq x]$ has finite support because there is $M>0$ so that $\mathrm{P}[X\geq x]=0$ for all $x>M$ because of our hypothesis that $X$ is non-negative and essentially bounded (almost-surely bounded). As a result, $\mathbb{E}[X]$ will be finite.

We also have the following result (without assuming continuity):

Let $X\in \mathcal{L}^{\infty}(\Omega, \mathcal{F}, \mathrm{P})$ and $Z\in \mathcal{L}^{1}(\Omega, \mathcal{F}, \mathrm{P})$. Then, $XZ$ is measurable and $XZ$ is integrable and

$$ \|XZ\|_1 \leq \|X\|_{\infty}\|Z\|_{1}, $$

take $Z=1$ and we get

$$ \|X\|_1 \equiv \mathbb{E}[|X|] \leq \|X\|_{\infty} < \infty, $$

Therefore $\mathbb{E}[|X|]$ is finite, but by Jensen's inequality

$$ |\mathbb{E}[X]| \leq \mathbb{E}[|X|], $$

so, $\mathbb{E}[X]$ is also finite. As a result, essentially bounded random variables have finite expectation.

With some little abuse of notation we can write

$$ \mathcal{L}^{\infty}(\Omega, \mathcal{F}, \mathrm{P}) \subseteq \mathcal{L}^{1}(\Omega, \mathcal{F}, \mathrm{P}) $$

In general, we can also show that

$$ \mathcal{L}^{p}(\Omega, \mathcal{F}, \mathrm{P}) \subseteq \mathcal{L}^{p'}(\Omega, \mathcal{F}, \mathrm{P}), \text{ whenever } p' \leq p, $$

but the converse is not true. A random variable with finite expectation does not have finite second moment. A random variable with all moments up to $p>1$ may not have finite moments of order $p'>p$ and, further, may not be essentially bounded.

A necessary clarification: A typical example of an essentially bounded function that is not bounded is the following: enter image description here This is given by (here $\Omega=\mathbb{R}$) $$ X(\omega)=\begin{cases} 1, &\text{if} x\in \mathbb{R}\setminus \mathbb{N},\\ x, &\text{otherwise} \end{cases} $$ Indeed, notice that $\mathrm{P}[X>1]=0$.

The notion you mentioned in your comment below is not essential boundedness: for $X:\Omega\to \mathbb{R}\cup\{+\infty\}$ the property $\mathrm{P}[X=\infty]=0$ (equivalently $\mathrm{P}[X<\infty]=1$) is that $X$ is almost-everywhere finite or almost-surely finite-valued. No way, should this be confused with essential boundedness. Take for instance the following very simple function

$$ X(\omega) = \omega, $$

where $\Omega = \mathbb{R}$. Notice that $\{\omega, X(\omega) = +\infty\}=\varnothing$, but $X$ is not essentially bounded and is not bounded.

If a random variable is almost-everywhere finite, it does not necessarily have finite expectation. Take for instance the counterexample provided in this answer. Another example, if you prefer an example not in discrete space, is the Pareto distribution with parameter $\alpha=1$. There are numerous other examples.

  • $\begingroup$ Almost surely bounded, or, essentially bounded, means that there is a $M$ so that $\mathrm{P}[|X|>M]=0$. For instance the function $f(x)=0$ when $x$ is not an integer and $f(n)=n$ for n: integer is essentially bounded (wrt Lebesgue measure), but not surely bounded and it is always finite-valued (in contrast to extended-real-valued functions which may take the value $+\infty$). Continuity is used only in the first part of the answer - I'll update my answer. $\endgroup$ – Pantelis Sopasakis Feb 14 '16 at 13:58
  • $\begingroup$ Thank you. Could you please explain why you excluded infinity as $[0,+\infty)$. Doesn't this mean you have taken that the random variable is sure bounded. In my case I only have that $X=+\infty$ with zero probability. What do you mean by continuous random variable, is it in the sense take values in a continuous range? Also how is continuity and non-negativity used here? Finally does this mean the statement is valid only in some special cases and that it is not valid in general? Can you please give a counter example? Apologies, had to make a change so this comment should proceed your comment. $\endgroup$ – seek Feb 14 '16 at 14:14
  • $\begingroup$ The way you have stated your questions, $X$ is almost surely bounded. This does not mean that it is finite-valued with probability one. If $X:(\Omega, \mathcal{F}, \mathrm{P})\to \mathbb{R}\cup \{+\infty\}$ and $\mathrm{P}[X=\infty]=0$, then this doesn't mean that $X$ is essentially bounded and the expectation may be infinite. Would you like me to give a counterexample? $\endgroup$ – Pantelis Sopasakis Feb 14 '16 at 14:47
  • $\begingroup$ What I meant by almost surely bounded is $P[\mid X\mid < \infty]=1$, which I belive is equal to $P[\mid X\mid = \infty]=0.$ So as I understand almost sure bounded and this condition are not the same? Could you please explain how they are different? Thank you. $\endgroup$ – seek Feb 14 '16 at 14:52
  • $\begingroup$ Yes, these two are not the same. I'll update my answer in a while with a figure. $\endgroup$ – Pantelis Sopasakis Feb 14 '16 at 14:59

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