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Is there some sort of intuition or a good ilustrative example for random variables being $\sigma$-algebra measurable? I understand the definition, but when looking at martingales, the meaning of random variables being measurable eludes me. So my question is mainly aimed on the case of martingales where sequence of random variables is adapted to some filtration.

In Interpretation of sigma algebra, the asker asks (among many others) a similar question, but I don't think it contains an actual answer to this question.

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Maybe this can help you to understand the concept of conditional expectation, behind your question.

Suppose you have a probability space $(\Omega, \mathcal P (\Omega), \mathbb{P})$, where $\mathcal P (\Omega)$ denotes the set of all possible subsets of $\Omega$ (evidently, a $\sigma$-algebra), and $\mathbb{P}$ is a probability measure (in this case, a function from $\mathcal P (\Omega)$ to [0,1]).

Suppose you have a random variable (measurable function) $X:(\Omega, \mathcal P (\Omega)) \to (\mathbb{R}, \mathcal B (\mathbb R ))$, where $\mathcal B (\mathbb R )$ is the usual Borel $\sigma$-algebra.

Take as a sub-$\sigma$-algebra the trivial one, $\mathcal F = \{\emptyset, \Omega\}$. Suppose we only know the conditional expectation $\mathbb E(X | \mathcal F)$, but not $X$ itself. How much do we know about X? Well, $Y = \mathbb E(X | \mathcal F)$ is a random variable, $\mathcal F$/ $\mathcal B (\mathbb R )$- measurable. From Y, we can only determine ONE thing (think about this!): $$\mathbb E(Y) = \mathbb E(\mathbb E(X | \mathcal F)) = \mathbb E X.$$ So, what is $\mathbb{E}(X | \mathcal F)$? It is the most simplified knowledge that we can have; we arrive at this if we determine the expectation of the random variable but know nothing about its values in particular events (in $\mathcal P (\Omega)$).

(In fact, $Y$ is constant... otherwise, it would not be measurable.)

Suppose now that we enlarge this $\sigma$-algebra, say to $\mathcal F' = \{\emptyset, A, A^c, \Omega\}$, for some non-trivial set $A$. Again, suppose that we only know $\mathbb{E}(X | \mathcal F')$, not X. Then, we can determine three things about the variable: $$\mathbb E(X 1_A), \, \mathbb E(X 1_{A^c}) \text{ and } \mathbb E (X).$$ Conclusion: a bigger $\sigma$ algebra implies more knowledge about the random variable X (we are interested in that one)!

Check that in the extreme case, when $\mathcal F'' =\mathcal P (\Omega)$, the knowledge of $\mathbb E (X|\mathcal F'')$ allows us to determine all the expected values $\mathbb E(X 1_{\{X=x\}})= x\mathbb P (X=x)$, because the events $\{X=x\}$ are contained in $\mathcal F''$ (like every other subset). If $X$ only take a finite number of different values (for instance, when $\Omega$ is finite), these expectations are enough to determine the probability of all the events $\{X=x\}$. (When $X$ is continuous, the above reasoning is not very useful, for the subsets $\{X=x\}$ have probability zero and the expectations above are zero too. Anyway, by the general properties of the conditional expectation, $\mathbb E(X|\mathcal F'') = X$, because $X$ is $F''$-measurable. In this sense, we can say that the variable is recovered from its conditional expectation.)

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    $\begingroup$ Additionally, you could think that knowing $\mathbb E(X1_A),\mathbb E(X1_{A^c})$ and $\mathbb E(X)$ is equivalent to knowing the average value of $X$ in the sets $A,A^c$ and $\Omega$ (just divide by the probability of the set). I like to think about conditional expectations as "averaged out" versions of the initial variables. Of course, averaging will lose some information, and the bigger the $\sigma$-algebra, the less information you lose by the averaging. $\sigma(X)$ is the smallest sigma algebra that won't result in information loss. $\endgroup$ – dafinguzman Feb 28 '14 at 2:34
  • $\begingroup$ @jpvigneaux Can you please justify why we could determine $\mathbb{P}(X=x)$ when we are given complete information? $\endgroup$ – wayne May 18 '18 at 20:24
  • $\begingroup$ I updated the answer! I wrote this several years ago, so I do not remember what I had in mind. I hope this helps. $\endgroup$ – jpvigneaux May 20 '18 at 14:41
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The random variable $X$ is measurable with respect to $\sigma$-algebra $\mathfrak F$ if $X=\mathbb E(X\mid\mathfrak F)$.

One can understand this in a few steps:

  1. $\mathbb E(X\mid A)$, where $A$ is an event, is the expected value of $X$ given that $A$ occurs;
  2. $\mathbb E(X\mid Y)$, where $Y$ is a random variable, is a random variable whose value at $\omega\in\Omega$ is $\mathbb E(X\mid A)$ where $A$ is the event $\{Y=y\}$ and $y=Y(\omega)$;
  3. $\mathbb E(X\mid \mathbf 1_A)$ is the case $Y=\mathbf 1_A$, and $\mathbf 1_A(\omega)$ is 1 if $\omega\in A$, and 0 otherwise. This is the random variable that returns $\mathbb E(X\mid A)$ if $\omega\in A$, and $\mathbb E(X\mid A^c)$ if $\omega\not\in A$;

  4. $\mathbb E(X\mid \mathfrak F)$, where $\mathfrak F=\{\varnothing, \Omega, A, A^c\}$, is the same as $\mathbb E(X\mid 1_A)$;

  5. $\mathbb E(X\mid \mathfrak F)$, where $\mathfrak F=\{\varnothing, \Omega, A, A^c, B, B^c, A\cup B, A\cup B^c,\dots\}$ ($2^{2^2}=16$ elements), is something we could call $\mathbb E(X\mid 1_A, 1_B)$ and which returns $\mathbb E(X\mid A\cap B^c)$, or $\mathbb E(X\mid A^c\cap B)$, or $\mathbb E(X\mid A\cap B)$, or $\mathbb E(X\mid A^c\cap B^c)$; it is sort of superfluous to list $A\cup B$ etc. in $\mathfrak F$, it would suffice to list a generating set, but the generating set may not be unique so it is best to list all of $\mathfrak F$;

  6. $\mathbb E(X\mid \mathfrak F)$, where $\mathfrak F=\mathfrak F_t$ is a $\sigma$-algebra corresponding to what's known at time $t$, is an infinite version of (5). It's a random variable that returns our best estimate of $X$, given answers to all the questions "$\omega\in A$?" for $A\in\mathcal F$. If that answer is always just the same as $X$, then $X$ is $\mathfrak F$-measurable or "known at time $t$".

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  • $\begingroup$ Just like the standard machine for Lebesgue integration. Brilliant! $\endgroup$ – BCLC May 23 '18 at 14:01
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    $\begingroup$ Thanks, @BCLC!! $\endgroup$ – Bjørn Kjos-Hanssen May 23 '18 at 22:55
  • $\begingroup$ Bjørn Kjos-Hanssen, in between 5 and 6, could we perhaps have a countably infinite definition (then 6 would be uncountable) ? $\endgroup$ – BCLC May 31 '18 at 9:17

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