I'm reading about conditional expectations from Achim Klenke's Probability Theory: A Comprehensive Course, which defines conditional expectations for integrable random variables $X$ with respect to a $\sigma$-algebra $\mathcal F$ as the unique (a.s.) $\mathcal F$-measurable random variable $Y$ for which $\mathbb E[X\mathbb 1_A] = \mathbb E[Y\mathbb 1_A]$ for all $A \in \mathcal F$. The following remark is made to generalize this definition to some non-integrable random variables:

Let $X : \Omega \to \mathbb R$ be a random variable such that $X^- \in \mathcal L^1(\mathbb P)$. We can define the conditional expectation as the monotone limit $$ \mathbb E[X\,|\,\mathcal F] = \lim_{n \to \infty} \mathbb E[X_n\,|\,\mathcal F] $$ where $-X^- \leq X_1$ and $X_n \uparrow X$. Due to the monotonicity of the conditional expectation, it is easy to show that the limit does not depend on the choice of the sequence $(X_n)$ and that it fulfills the conditions of [the definition of conditional convergence].

(Presumably we want $(X_n) \subset \mathcal L^1(\mathbb P)$ to avoid circular definitions.)

My question: Why do we need $-X^- \leq X_1$?

It's not hard to verify these claims. By monotonicity of conditional expectation, we get that $\mathbb E[X_n\,|\,\mathcal F] \uparrow \mathbb E[X\,|\,\mathcal F]$ (this limit may be infinite). So for any $A \in \mathcal F$, by the Beppo-Levi monotone convergence theorem, since the $\mathbb E[X_n\,|\,\mathcal F]$ are integrable, $$ \mathbb E[\mathbb E[X\,|\,\mathcal F]\mathbb 1_A] = \mathbb E\left[\lim_{n \to \infty} \mathbb E[X_n\,|\,\mathcal F]\mathbb 1_A\right] = \lim_{n \to \infty} \mathbb E\left[\mathbb E[X_n\,|\,\mathcal F]\mathbb 1_A\right] = \lim_{n \to \infty} \mathbb E[X_n\mathbb 1_A] = \mathbb E[X\mathbb 1_A]. $$ If $(\tilde X_n)$ is another sequence with $\tilde X_n \uparrow X$ and $-X^- \leq \tilde X_1$, letting $Y = \lim \mathbb E[X_n \,|\,\mathcal F]$ and $\tilde Y = \lim\mathbb E[\tilde X_n\,|\,\mathcal F]$, then both $Y$ and $\tilde Y$ are $\mathcal F$-measurable (as a limit of $\mathcal F$-measurable functions), and this calculation shows $\mathbb E[Y\mathbb 1_A] = \mathbb E[\tilde Y \mathbb 1_A]$ for every $A \in \mathcal F$. It follows that $Y = \tilde Y$.

I can see that we'd want $X^- \in \mathcal L^1(\mathbb P)$ because otherwise we can't have that both $(X_n) \in \mathcal L^1(\mathbb P)$ and $X_n \uparrow X$. But I don't see why we need $-X^- \leq X_1$. In fact this seems contradictory if we require $X_n \uparrow X$. Am I missing something?


2 Answers 2


Here is my understanding. If $-X^{-}\le X_1$, then we would have $X_n^-\uparrow X^-$ and $X_n^+\uparrow X^+$. Thus, $X_n\in\mathcal{L}^1$ iff $X_n^+\in \mathcal{L}^1$. We can take such sequence, say, $X_n=X\mathbb{I}_{\{|X|\le n \}}$. If instead $-X^- > X_1$, then we would have $X^-<X_1^-$ and $X_n^-\downarrow X^-$. It is not clear if one can take $(X_n)$ such that both $X_n^\pm\in \mathcal{L}^1$ since we only know $X^-\in \mathcal{L}^1$


Since your random variables may take negative values you need a generalized version of the monotone convergence theorem. Such a result exists which, in addition to $X_n\uparrow X$, assumes that $X_1^-$ is integrable. We ensure that this is satisfied by assuming $-X^-\leq X_1$.

Note that you can construct a sequence $(X_n)$ with the desired properties by taking $X_n=X1_{\{X\leq n\}}$.

  • $\begingroup$ The MCT you're talking about generalizing assumes the random variables are nonnegative, but not necessarily integrable. You shouldn't need to generalize the MCT in that way if the random variables $X_n$ are integrable (see Klenke, Theorem 4.20): if $N$ is the null event on which $X_n$ does not converge to $X$, then $X_n' = (X_n - X_1)\mathbb 1_{N^c}$ is nonnegative and $X-X_1$ a.s. By linearity and the MCT for nonnegative measurable random variables, $\mathbb E[X_n] \to \mathbb E[X]$. $\endgroup$
    – D Ford
    Commented Jan 19, 2023 at 13:42
  • $\begingroup$ I am just trying to guess why one would assume $-X^-\leq X_1$ and this is one possibility. But as you suggest it is sufficient that $X_1$ is integrable. However, it is my experience that people do not always care about assuming too much as long as it solves the problem. The assumptions on $(X_n)$ may be stricter than necessary but as long as such a sequence exists (and it does) it does not really matter. $\endgroup$
    – jakobdt
    Commented Jan 19, 2023 at 14:01

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