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I'm going through elementary literature on measure theory from Shreve (Vol II) and having a hard interpreting the meaning of $\mathbb{E}[X|\mathcal{F(t)}]$ where $X$ is a random variable and $\mathcal{F(t)}$ is the sigma-algebra at time $t$.

Usually in non-measure theoretic definition of conditional probability $\mathbb{E}[X|Y]$, we assume that one needs to calculate the expected value of random variable $X$ on the condition that event $Y$ is already given. But in case of measure theoretic conditional expectation, we know the sigma-algebra $\mathcal{F(t)}$ even at the start of the process i.e. at $t=0$. Therefore, how does it makes any sense to say that one needs to calculate the expected value of $X$ given the set $\mathcal{F(t)}$. In other words, how have we gained more information by being provided with $\mathcal{F(t)}$ (which we already knew)?

Akshay Bansal
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  • Not sure about Shrevian notation, but $\mathcal{F}(t)$ would presumably be the distribution function of $X$. [see here](https://stats.stackexchange.com/questions/10159/find-expected-value-using-cdf). Sometimes the BB font-env is used to denote empirical operators, here that would be empirical expectation which would use a finite sample empirical distribution function yielding the sample average. That would require Lebesgue–Stieltjes integration. – AdamO Apr 28 '17 at 21:09
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    @AdamO $(\mathcal{F}(t))$ is a filtration of sigma-algebras; $\mathcal{F}(t)$ is the sigma-algebra at "time" $t$, as stated in the first paragraph of this question. Shreve uses what appears to be the most widespread notation: it appears in most of the textbooks on probability and measure theory that are oriented towards stochastic processes and financial applications. – whuber Apr 28 '17 at 21:17
  • @whuber okay that's totally new to me then. Thanks for the heads up. – AdamO Apr 28 '17 at 21:22

1 Answers1

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Informally, it is not what you actually know, it is what the symbols say you should "pretend" to know and not know.

So $E(X)$ says that you should calculate the expected value of $X$ in a situation where you don't know anything abut any event having happened or not having happened. $E(X\mid A)$ on the other hand says that you should calculate the expected value of $X$ in a situation where we assume that event $A$ has happened.

Even in a temporal setting, when looking at $E(X_{t})$ we mean "expected value of $X_{t}$ in a situation where we don't know anything", while with $E[X_t \mid \mathcal{F_{t-1})}]$ we mean "expected value of $X_{t}$ in a situation where we know what $\mathcal{F_{t-1}}$ can tell us.

WORKED OUT EXAMPLE
Following discussion in comments, the OP describes the following situation: Let an experiment where we throw two fair coins, perhaps sequentially but independently. Let $S_2$ represent the number of heads in the second time we do that. Let $S_2$ be stochastically dependent on what happened in the first period. And consider the information set $\mathcal{F_1} = \{ \{HH, HT \}, \{ TT, TH \} \}$. How will we obtain the conditional expectation function $E(S_2 \mid \mathcal{F_1})$?

Well, we have to determine the structure of the stochastic dependence first, and this may be many different things. And also, the Conditional Expectation Function should be completely determined, once we "feed" it with one of the events that we are considering.

So we assume that the two coins are indeed thrown sequentially in each period, and what matters for the outcome in the second period is heads in the first coin or not. This is binary, so we can map the information set given by defining the indicator function

$$I_1 \equiv I_1\{ (HH, HT )\}$$

So this random variable takes the value $1$ if we get heads in the first coin in the first period, and zero otherwise. We now assume the following:

If we get heads in the first coin in the first period, we will get $S_2 =2$. If we don't get heads in the first coin in the first period, we will get $S_2=0$. Then, we obtain

$$E(S_2 \mid \mathcal{F_1}) = 2I_1$$

which is a function, not a specific value. This should satisfy the defining property of the Conditional Expectation $E[E(S_2 \mid \mathcal{F_1})]=E(S_2)$. Does it?

We have

$$E(S_2) = 2\cdot 0.25 + 1\cdot 0.5 + 0\cdot 0.25 = 1$$

while

$$E[E(S_2 \mid \mathcal{F_1})] = E(2I_1) = 2E(I_1)= 2\cdot 0.5 =1$$

It does.

Note that we could use the usual shorthand and write $E(S_2 \mid I_1) = 2I_1$, because given how we assumed the dependence to be, conditioning on $\mathcal{F_1}$ is equivalent to conditioning on (the sigma algebra induced by) $I_1$.

I hope this helped. This answer may be relevant here.

Alecos Papadopoulos
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  • If we consider every element of sigma-algebra $\mathcal{F(t)}$ as an event, then exactly which event are we referring to when we need to calculate $\mathbb{E}[X|\mathcal{F(t)}]$? – Akshay Bansal Apr 22 '17 at 22:35
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    @akshay The Conditional Expectation is a function, not a value, so it is defined with respect to a sigma-algebra and not a specific event. – Alecos Papadopoulos Apr 23 '17 at 00:17
  • I still have doubt with this notion. Consider two different time points $t=1$ and $t=2$ for an event with sample space $\Omega = \{ HH, HT, TH, TT \} $. So $\mathcal{F_1} = \{ \{HH, HT \}, \{ TT, TH \} \}$ and $\mathcal{F_2} = \{HH, HT, TT, TH \}$ be two sigma algebras (augment the sets so it may satisfy the definition of sigma algebra). Let's say one needs to evaluate $\mathbb{E}[S_2 | \mathcal{F_1}]$. Could you please elaborate the procedure to evaluate this expectation step-wise so that this abstract definition becomes clear to me? – Akshay Bansal Apr 26 '17 at 18:59
  • @akshey What is $S_2$? What does it measure? Also, for the conditional expectation to be non-trivial, I guess the outcome in period 2 is assumed stochastically dependent on the outcome in period 1. – Alecos Papadopoulos Apr 26 '17 at 19:07
  • Let's say $S_2$ measures the no. of heads. Yes, the outcome in period 2 is stochastically dependent on the outcome on period 1. – Akshay Bansal Apr 26 '17 at 19:14
  • akshay, Shreve carries out these kinds of calculations explicitly, clearly, and carefully in Volume I: why not look there? – whuber Apr 28 '17 at 21:06
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    @amoeba I agree, done. – Alecos Papadopoulos Apr 30 '17 at 23:44