Conditional expectation: Question bank — Maths deep dive

Two objects you must keep apart at all times:

$E [X ∣ Y = y]$ — a number, one per fixed value $y$ .
$E [X ∣ Y]$ — a random variable, the function $g (y) = E [X ∣ Y = y]$ evaluated at the random $Y$ .

True or false — justify

E [X ∣ Y]

is always a single number.

False — it is a random variable

g (Y)

. Only after you fix

Y = y

does it collapse to the number

E [X ∣ Y = y]

E [E [X ∣ Y]] = E [X]

requires

X

and

Y

to be independent.

False — the Tower Rule holds for any

X, Y

(with finite mean). Independence is never needed to average the conditional averages back to the whole.

X ⊥ Y

then

E [X ∣ Y] = E [X]

True — knowing

Y

tells you nothing new about

X

, so your updated guess equals the prior guess

E [X]

, a constant. See Conditional probability.

E [X ∣ Y] = E [X]

implies

X

and

Y

are independent.

False — it only says

Y

carries no linear/mean information about

X

. Independence is stronger; e.g.

Y

could still change the variance of

X

E [g (Y) ∣ Y] = g (Y)

for any function

g

True — once

Y

is known,

g (Y)

is a fixed constant, and the expectation of a constant is itself.

E [X Y ∣ Y] = Y E [X]

always.

False — pulling out gives

E [X Y ∣ Y] = Y E [X ∣ Y]

. You may only replace

E [X ∣ Y]

E [X]

when

X ⊥ Y

The Tower Rule is a plain average of the conditional means

E [X ∣ Y = y]

False — it is a weighted average, weighted by

P (Y = y)

. Bigger groups count more; equal weighting only coincides when all

y

are equally likely.

E [X ∣ Y]

can take values outside the range of

X

False — each

E [X ∣ Y = y]

is an average of possible

X

values, so it stays within the min–max range of

X

Var (E [X ∣ Y]) \leq Var (X)

True — conditioning averages away part of the fluctuation, so the "explained" variance can't exceed the total. This is Conditional variance and Eve's law.

E [X ∣ Y] = c

(a constant), then

c = E [X]

True — apply the Tower:

E [X] = E [c] = c

, so the constant must be the unconditional mean.

Spot the error

E [X ∣ Y = y] = \sum_{x} x P (X = x, Y = y)

Missing the divide by

P (Y = y)

— you must use the conditional distribution

P (X = x ∣ Y = y) = P (X = x, Y = y) / P (Y = y)

, not the joint.

"By the Tower,

E [X] = \frac{1}{n} \sum_{y} E [X ∣ Y = y]

for the

n

values of

Y

Wrong weights — the correct sum is

\sum_{y} E [X ∣ Y = y] P (Y = y)

. Averaging with

1/ n

silently assumes every

Y = y

is equally likely.

E [X + Z ∣ Y] = E [X ∣ Y] + E [Z]

The second term should be

E [Z ∣ Y]

, not

E [Z]

. Conditional expectation is linear while keeping the same conditioning — you can't drop

Y

from one term.

"Since

E [X ∣ Y]

is random,

E [E [X ∣ Y]]

is also random."

The outer

E [\cdot]

averages over

Y

, producing the constant

E [X]

. Applying

E

turns the random variable into a fixed number.

E [X^{2} ∣ Y] = (E [X ∣ Y])^{2}

False in general — this is Jensen's inequality territory;

E [X^{2} ∣ Y] \geq (E [X ∣ Y])^{2}

with a gap equal to the conditional variance

Var (X ∣ Y)

"For

X ∣ Y = y \sim U (0, y)

, we have

E [X ∣ Y] = 1/2

The mean of

U (0, y)

y /2

, so

E [X ∣ Y] = Y /2

— a random variable, not the constant

1/2

. Only after taking

E [\cdot]

do we get

E [X] = 1/4

P (A ∣ Y)

and

E [1_{A} ∣ Y]

are different things."

They are the same object — the conditional probability of an event is the conditional expectation of its indicator

1_{A}

. See Law of total probability.

Why questions

Why is

E [X ∣ Y]

defined as a function of

Y

rather than just a table of numbers?

Packaging every answer "

E [X ∣ Y = y]

for all

y

" into one function

g (Y)

lets us feed the random

Y

back in, so we can take its expectation, variance, and use it inside Martingales.

Why does the inner marginalising sum

\sum_{y} P (X = x, Y = y)

collapse to

P (X = x)

Summing the joint over all values of

Y

accounts for every way

X = x

can happen, which is exactly the marginal

P (X = x)

. See Joint and marginal distributions.

Why can we "take out what is known" —

E [g (Y) X ∣ Y] = g (Y) E [X ∣ Y]

Conditioning on

Y

treats

Y

as fixed, so

g (Y)

becomes a constant multiplier, and constants factor out of any expectation by Expectation and its linearity.

Why must the tower weight by

P (Y = y)

and not treat all groups equally?

Because more probable

y

contribute more sample mass to

E [X]

; ignoring the weights would over-count rare groups and distort the overall mean.

Why is the Tower Rule "split then recombine can't change the mean" true intuitively?

Averaging within each group and then blending by group size re-touches every original outcome exactly once with its true weight, so it must reproduce the global average.

Why does Wald's identity rely on conditional expectation?

It conditions the sum

S_{N} = \sum_{i = 1}^{N} X_{i}

on the random count

N

, computes

E [S_{N} ∣ N] = N E [X]

, then applies the Tower — see Wald's identity.

Why is

E [X ∣ X] = X

Once

X

is the very thing you condition on, it is fully known, so the best guess for

X

X

itself with no averaging left to do.

Edge cases

What is

E [X ∣ Y = y]

P (Y = y) = 0

Undefined by the ratio formula — you cannot condition on a probability-zero event in the discrete definition; the continuous case handles this via densities instead.

What does

E [X ∣ Y]

equal when

Y

is a constant

c

(degenerate, no information)?

It equals

E [X]

— conditioning on something that never varies gives you no new information, so the guess stays at the prior mean.

X

has no finite mean (e.g. Cauchy), is

E [X ∣ Y]

defined?

No — conditional expectation inherits the finite-mean requirement; without a well-defined

E [X]

the averages don't exist.

What is

E [X ∣ Y]

when

Y = X

(perfect information)?

It is

X

itself — knowing

Y = X

pins down

X

exactly, leaving nothing to average over.

For continuous

X, Y

, what replaces

P (X = x ∣ Y = y)

, which is

0/0

The conditional density

f_{X ∣ Y} (x ∣ y) = f_{X, Y} (x, y) / f_{Y} (y)

, and

E [X ∣ Y = y] = \int x f_{X ∣ Y} (x ∣ y) d x

Y

takes only one possible value with probability 1, what is

Var (E [X ∣ Y])

Zero —

E [X ∣ Y]

is then the constant

E [X]

, and constants have no variance, matching the intuition that

Y

explained nothing.

What happens to the tower when

X

is a constant

a

E [X ∣ Y] = a

for every

y

, and

E [a] = a = E [X]

— the rule holds trivially, since no group's average can differ from

a

Connections

Conditional expectation (parent)
Expectation and its linearity
Conditional probability
Law of total probability
Conditional variance and Eve's law
Joint and marginal distributions
Wald's identity
Martingales