Conditional Expectation
Ordinary expectation
\mathbb{E}[X] collapses a random variable to a single number — the
best constant guess of X when you know nothing. But we
usually know something. Conditional expectation
\mathbb{E}[X \mid \mathcal{G}] is the best guess of
X given the information encoded in a sub-\sigma-algebra
\mathcal{G} \subseteq \mathcal{F} — and that "best guess" is itself a
random variable, not a number.
Formally, \mathbb{E}[X \mid \mathcal{G}] is the (almost surely unique)
random variable satisfying two requirements at once. The first is a
measurability clause and the second is an averaging clause:
(\text{i})\quad \mathbb{E}[X \mid \mathcal{G}] \text{ is } \mathcal{G}\text{-measurable},
(\text{ii})\quad \int_A \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_A X\, d\mathbb{P} \qquad \text{for every } A \in \mathcal{G}.
Take the two clauses one at a time, because each buys you something different.
Clause (i): measurability. Saying the estimate is
\mathcal{G}-measurable is saying it may be built from the
information in \mathcal{G} alone. You are forbidden from peeking at
anything finer than \mathcal{G} can resolve. If
\mathcal{G} only tells you which block of a partition you landed in, then
your estimate may only depend on that block — it cannot vary within a block, because nothing
in \mathcal{G} distinguishes two outcomes in the same block. This clause is
what makes \mathbb{E}[X \mid \mathcal{G}] an honest forecast rather than a
forbidden glance at X itself.
Clause (ii): averaging. Measurability alone is far too weak — the constant
0 is \mathcal{G}-measurable, and so is every
other constant. Clause (ii) is what selects the right
\mathcal{G}-measurable variable: it must carry the same total mass
as X over every event A you can resolve
with \mathcal{G}. Read the integral
\int_A X\, d\mathbb{P} as "the average value of
X on A, weighted by probability": clause (ii)
demands the estimate reproduce that quantity for all such A simultaneously.
No clairvoyance is needed to check it, only the sets in \mathcal{G}.
Together the two clauses pin the variable down uniquely (up to a null set): (i) restricts you to the
right shelf of candidates, and (ii) singles out exactly one of them.
We have described \mathbb{E}[X \mid \mathcal{G}] by two
properties, but description is not construction — why should a variable satisfying both exist? The
guarantee comes from the
Radon–Nikodym theorem.
Define a set function \nu(A) = \int_A X\, d\mathbb{P} on
A \in \mathcal{G}; it is a (signed) measure that is
absolutely continuous with respect to \mathbb{P} restricted to
\mathcal{G} (if \mathbb{P}(A) = 0 then
\nu(A) = 0). Radon–Nikodym then hands you a
\mathcal{G}-measurable density
\frac{d\nu}{d\mathbb{P}} — and that density is precisely
\mathbb{E}[X \mid \mathcal{G}]. Uniqueness up to a
\mathbb{P}-null set is part of the same theorem. So conditional
expectation is not a new axiom; it is a Radon–Nikodym derivative wearing a probabilist's hat.
The partition picture
The cleanest case: \mathcal{G} is generated by a partition
of \Omega into blocks B_1, B_2, \dots. Knowing
\mathcal{G} means knowing only which block you landed in — not
the exact outcome. So the best guess of X must be the same throughout a
block: \mathbb{E}[X \mid \mathcal{G}] is piecewise constant,
equal on block B to the average of X over
B,
\mathbb{E}[X \mid \mathcal{G}](\omega) = \frac{1}{\mathbb{P}(B)}\int_B X\, d\mathbb{P}, \qquad \omega \in B.
That formula is not a separate assumption — it falls straight out of the two defining clauses.
Suppose \mathcal{G} = \sigma(B_1, \dots, B_k) is generated by a finite
partition of \Omega into disjoint blocks
B_1, \dots, B_k (each of positive probability). We derive the value on a
single block B_j, step by step.
First, what does clause (i) force? The only \mathcal{G}-measurable functions
are those constant on each block — because the finest event
\mathcal{G} can resolve is "which block am I in?". So there is a number
c_j with
\mathbb{E}[X \mid \mathcal{G}](\omega) = c_j \qquad \text{for every } \omega \in B_j.
Now pin down c_j using clause (ii). Each block
B_j is itself a member of \mathcal{G}, so we are
entitled to take A = B_j in the averaging equation:
\int_{B_j} \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_{B_j} X\, d\mathbb{P}.
On the left, the integrand is the constant c_j throughout
B_j, so it pulls out of the integral, and what remains is just the measure
of the block:
\int_{B_j} \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_{B_j} c_j\, d\mathbb{P} = c_j \int_{B_j} 1\, d\mathbb{P} = c_j\,\mathbb{P}(B_j).
Equating the two sides of the averaging equation,
c_j\,\mathbb{P}(B_j) = \int_{B_j} X\, d\mathbb{P} = \mathbb{E}[X\,\mathbf{1}_{B_j}],
where \mathbf{1}_{B_j} is the indicator of the block (it equals
1 on B_j and 0
elsewhere, so \int_{B_j} X\, d\mathbb{P} = \mathbb{E}[X\,\mathbf{1}_{B_j}]).
Dividing by \mathbb{P}(B_j) > 0 gives the block value outright:
c_j = \frac{\mathbb{E}[X\,\mathbf{1}_{B_j}]}{\mathbb{P}(B_j)} = \frac{1}{\mathbb{P}(B_j)}\int_{B_j} X\, d\mathbb{P}.
So on block B_j the conditional expectation is exactly the
probability-weighted average of X over that block — the elementary
conditional expectation \mathbb{E}[X \mid B_j] you already know, now
assembled into one \mathcal{G}-measurable random variable that reads off the
right block average wherever \omega happens to land.
Step through the figure: first the raw values of X, then the flat level
the conditional expectation snaps each block to — its average.
The tower property, derived
The single most-used consequence is the tower property (also called the law of
iterated expectations): averaging the average recovers the plain average,
\mathbb{E}\!\big[\mathbb{E}[X \mid \mathcal{G}]\big] = \mathbb{E}[X].
This is not a new principle either — it is clause (ii) read at the single largest event. Recall
\Omega \in \mathcal{G} (the whole space sits in every
\sigma-algebra), so we are allowed to take
A = \Omega in the averaging equation:
\int_\Omega \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_\Omega X\, d\mathbb{P}.
Now use the basic identity \mathbb{E}[Z] = \int_\Omega Z\, d\mathbb{P} —
the expectation of any random variable is its integral over the whole space — applied to
each side. On the left, Z = \mathbb{E}[X \mid \mathcal{G}]; on the right,
Z = X:
\mathbb{E}\!\big[\mathbb{E}[X \mid \mathcal{G}]\big] = \int_\Omega \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_\Omega X\, d\mathbb{P} = \mathbb{E}[X].
Reading the outer two expressions, \mathbb{E}\big[\mathbb{E}[X \mid \mathcal{G}]\big] = \mathbb{E}[X].
The middle two steps are exactly clause (ii) at A = \Omega; the framing
steps are just the definition of expectation as an integral. That is the whole proof.
For an integrable random variable X and a sub-\sigma-algebra
\mathcal{G} \subseteq \mathcal{F}, the conditional expectation
\mathbb{E}[X \mid \mathcal{G}] is the almost-surely-unique random variable
that is
- (i) \mathcal{G}-measurable, and
- (ii) satisfies \displaystyle\int_A \mathbb{E}[X \mid \mathcal{G}]\, d\mathbb{P} = \int_A X\, d\mathbb{P} for every A \in \mathcal{G}.
Taking A = \Omega in (ii) yields the tower property
\mathbb{E}\big[\mathbb{E}[X \mid \mathcal{G}]\big] = \mathbb{E}[X]; taking
A = B_j for a generating block shows
\mathbb{E}[X \mid \mathcal{G}] is constant on each block of a finite
partition, equal there to \mathbb{E}[X\,\mathbf{1}_{B_j}] / \mathbb{P}(B_j).
A workhorse rule: if X is itself
\mathcal{G}-measurable, it behaves like a constant inside the
conditional expectation and may be pulled out,
\mathbb{E}[XY \mid \mathcal{G}] = X\,\mathbb{E}[Y \mid \mathcal{G}].
The intuition is exactly the partition picture: on each block of
\mathcal{G} a \mathcal{G}-measurable
X takes a single known value, so it is a constant for the purposes of
that block's average, and constants slide out of an average. To sketch the proof, one checks the
two defining clauses for the candidate X\,\mathbb{E}[Y \mid \mathcal{G}]:
it is a product of two \mathcal{G}-measurable variables, hence
\mathcal{G}-measurable (clause i); and for the averaging clause one
verifies \int_A X\,\mathbb{E}[Y \mid \mathcal{G}]\, d\mathbb{P} = \int_A XY\, d\mathbb{P}
first for X = \mathbf{1}_B an indicator (where it reduces to the defining
equation for Y over A \cap B \in \mathcal{G}),
then for simple functions by linearity, and finally for general
X by a limiting argument. In finance this is the rule that lets a
quantity already known today (a holding, a discount factor fixed at t)
come outside tomorrow's conditional expectation.
Jensen's inequality survives conditioning. If \varphi is a convex
function and X, \varphi(X) are integrable, then
\varphi\big(\mathbb{E}[X \mid \mathcal{G}]\big) \le \mathbb{E}\big[\varphi(X) \mid \mathcal{G}\big] \quad \text{a.s.}
Convexity says \varphi lies above each of its tangent lines:
\varphi(x) \ge \varphi(m) + \lambda\,(x - m) for a suitable slope
\lambda. Put m = \mathbb{E}[X \mid \mathcal{G}]
and x = X, then take
\mathbb{E}[\,\cdot \mid \mathcal{G}] of both sides: the linear term has
conditional mean zero (because m and
\lambda are \mathcal{G}-measurable and
\mathbb{E}[X - m \mid \mathcal{G}] = m - m = 0), leaving the inequality.
Taking \mathcal{G} trivial recovers the ordinary
\varphi(\mathbb{E}[X]) \le \mathbb{E}[\varphi(X)] — for example
\mathbb{E}[X \mid \mathcal{G}]^2 \le \mathbb{E}[X^2 \mid \mathcal{G}].
The other properties that make it work
Three further facts do almost all the remaining labour in finance, and each is a short consequence of
the defining property above (the first two are derived in the vignettes just above).
-
Independence. If X is
independent
of \mathcal{G}, the extra information is useless and
\mathbb{E}[X \mid \mathcal{G}] = \mathbb{E}[X].
-
Linearity.
\mathbb{E}[aX + bY \mid \mathcal{G}] = a\,\mathbb{E}[X \mid \mathcal{G}] + b\,\mathbb{E}[Y \mid \mathcal{G}].
These reappear the moment we meet
martingales,
where the whole "fair game" definition is a statement about a conditional expectation.