Conditional expectation: Worked examples — Maths deep dive

Before anything else, three reminders so every symbol on this page is earned:

Recall What the notation means (from the parent)

$E [X]$ ::: your best single-number guess for $X$ before any info — a plain number. $E [X ∣ Y = y]$ ::: your guess after learning $Y$ took the specific value $y$ — still a plain number, one per $y$ . $E [X ∣ Y]$ (capital $Y$ ) ::: leave $Y$ unknown → the guess is itself a random variable, the function $g (Y) = E [X ∣ Y = y]$ fed the random $Y$ . Tower Rule ::: $E [E [X ∣ Y]] = E [X]$ — average the group-averages, weighted by group sizes.

The scenario matrix

Every conditional-expectation problem lands in one of these cells. The examples below are labelled by cell.

Cell	Case class	What makes it tricky	Example
A	Discrete, condition on an event $B$	must renormalise by $P (B)$	Ex 1
B	Discrete, condition on a variable $Y$ ; use Tower	must weight by $P (Y = y)$	Ex 2
C	Continuous, $E [X ∣ Y] = g (Y)$ then integrate	inner density $f_{X ∣ Y}$	Ex 3
D	Degenerate: $X ⊥ Y$	knowing $Y$ changes nothing	Ex 4
E	Take-out-what-is-known with $g (Y) X$	don't fuse rules	Ex 5
F	Real-world word problem (random count)	Wald-style sum	Ex 6
G	Exam twist: recover a hidden mean by tower	reverse the tower	Ex 7
H	Limiting/edge: $P (B) \to 0$ , or $Y$ near boundary	check it stays sane	Ex 8

Cell A — condition on an event

Worked example Ex 1 (Cell A). Roll one fair die. Given the outcome is

at least 3, find $E [X ∣ B]$ . Here $X$ = the die face, $B = {X \geq 3}$ .

Forecast: guess first — the unconditional mean is $3.5$ . Chopping off ${1, 2}$ pushes the average up. Above $3.5$ ? Below? Make a bet.

Find $P (B)$ . The favourable faces are ${3, 4, 5, 6}$ , so $P (B) = 4/6 = 2/3$ . Why this step? Cell-A problems condition on an event, and the definition $E [X ∣ B] = \sum_{x} x P (X = x, B) / P (B)$ divides by $P (B)$ — we must know it first.
Renormalise. For each surviving face $x \in {3, 4, 5, 6}$ , $P (X = x ∣ B) = \frac{1/6}{2/3} = \frac{1}{4}$ . Why? Conditioning on $B$ keeps only these outcomes and stretches their probabilities so they sum to 1.
Average. $E [X ∣ B] = \frac{1}{4} (3 + 4 + 5 + 6) = \frac{1}{4} \cdot 18 = 4.5$ . Why? This is just ordinary expectation under the renormalised distribution.

Verify: $4.5 > 3.5$ ✓ (removing the two smallest faces must raise the mean). The four kept faces are symmetric about $4.5$ , so their equal-weight average sits exactly in the middle — sanity confirmed.

Cell B — condition on a variable, then Tower

Worked example Ex 2 (Cell B). Roll two fair dice;

$X$ = sum, $Y$ = first die. Find $E [X]$ via the Tower Rule, being careful about weights.

Forecast: two dice, each mean $3.5$ , so you probably already smell $E [X] = 7$ . Good — but let's derive it through the tower so the weighting habit sticks.

Conditional mean. $E [X ∣ Y = y] = y + E [second die] = y + 3.5$ . Why this step? The second die is independent of $Y$ ; conditioning fixes only the first term, leaving the second's mean untouched.
Weight by $P (Y = y)$ . Each $y \in {1, \dots, 6}$ has $P (Y = y) = 1/6$ : $E [X] = \sum_{y = 1}^{6} (y + 3.5) \cdot \frac{1}{6} .$ Why? The Tower Rule is $E [X] = \sum_{y} E [X ∣ Y = y] P (Y = y)$ — a weighted average of group means, not a plain one.
Compute. $\sum_{y = 1}^{6} y = 21$ , so $E [X] = \frac{1}{6} (21) + 3.5 = 3.5 + 3.5 = 7$ . Why this step? Distributing the sum, $\frac{1}{6} \sum_{y} (y + 3.5) = \frac{1}{6} \sum_{y} y + \frac{1}{6} \cdot 6 \cdot 3.5$ ; the constant $3.5$ averaged over six equal weights is just $3.5$ , and $\frac{1}{6} \cdot 21 = 3.5$ , so the two halves add to $7$ .

Verify: matches $E [X] = E [die_{1}] + E [die_{2}] = 3.5 + 3.5 = 7$ by linearity (Expectation and its linearity) ✓.

Twist inside Cell B — condition on an even first die (this reuses Cell A renormalisation):

Renormalise the first die. Among ${1, \dots, 6}$ the even faces are ${2, 4, 6}$ , so $P (Y even) = 3/6 = 1/2$ , and for each even $y$ , $P (Y = y ∣ Y even) = \frac{1/6}{1/2} = \frac{1}{3}$ . Why this step? Just like Cell A, conditioning on the event ${Y even}$ keeps only ${2, 4, 6}$ and stretches their probabilities to sum to 1.
Weighted average of $y$ . $E [Y ∣ Y even] = \frac{1}{3} (2 + 4 + 6) = \frac{1}{3} \cdot 12 = 4$ . Why this step? With the three even faces now equally likely at $1/3$ each, $E [Y ∣ Y even] = \sum y P (y ∣ even)$ is their equal-weight average — the midpoint $4$ of ${2, 4, 6}$ .
Add the second die. $E [X ∣ Y even] = 4 + 3.5 = 7.5$ (matches parent Example 1) ✓. Why this step? By linearity $E [X ∣ Y even] = E [Y ∣ Y even] + E [second die]$ ; the second die is independent so its mean stays $3.5$ , giving $4 + 3.5$ .

Cell C — continuous

Worked example Ex 3 (Cell C).

$Y \sim U (0, 1)$ and, given $Y = y$ , $X \sim U (0, y)$ . Find $E [X]$ and $E [X^{2}]$ .

Forecast: $X$ lives inside $[0, y]$ and $Y < 1$ , so $X$ is squeezed toward small values. Guess $E [X]$ is well below $0.5$ .

Inner mean. For a uniform on $[0, y]$ , $E [X ∣ Y = y] = \frac{0 + y}{2} = \frac{y}{2}$ . Why this step? A uniform's mean is its midpoint. This gives the function $g (y) = y /2$ , hence the random variable $E [X ∣ Y] = Y /2$ .
Tower for the mean. $E [X] = E [Y /2] = \frac{1}{2} E [Y] = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} .$ Why? Replaces the hard double integral over the joint density with two easy one-liners.
Second moment via tower. For $U (0, y)$ , $E [X^{2} ∣ Y = y] = y^{2} /3$ . Why? $\int_{0}^{y} x^{2} \cdot \frac{1}{y} d x = \frac{y ^{2}}{3}$ — the variance-plus-mean-squared of a uniform, but we just integrate directly. Then $E [X^{2}] = E [\frac{Y ^{2}}{3}] = \frac{1}{3} E [Y^{2}] = \frac{1}{3} \cdot \frac{1}{3} = \frac{1}{9} .$ (Used $E [Y^{2}] = \int_{0}^{1} y^{2} d y = 1/3$ .)

Verify: $E [X] = 1/4 = 0.25 < 0.5$ ✓ (squeezed low, as forecast). As a second sanity check we use the variance $Var (X)$ , which is defined as the average squared distance from the mean and can be computed as $Var (X) = E [X^{2}] - E [X]^{2}$ . Why use it as a check? Because a variance is an average of squares, it can never be negative — so if our $E [X^{2}]$ and $E [X]$ were mutually inconsistent, this would come out negative and flag the error. Here $Var (X) = \frac{1}{9} - \frac{1}{16} = \frac{7}{144} > 0$ ✓ (safely positive — the tools didn't break). This continuous case connects to Joint and marginal distributions and Conditional variance and Eve's law.

Cell D — degenerate: independence

Worked example Ex 4 (Cell D).

$X$ = number of heads in 3 fair-coin flips; $Y$ = a completely separate fair die. Find $E [X ∣ Y]$ and $E [X ∣ Y = 6]$ .

Forecast: the die has nothing to do with the coins. Does learning $Y$ change your guess for $X$ ? Predict before reading.

Check independence. The coin flips and the die are physically unrelated, so $X ⊥ Y$ . Why this step? The independence-collapse rule $E [X ∣ Y] = E [X]$ only fires when $X$ and $Y$ are independent — we must justify it, not assume it. See Conditional probability.
Collapse. $E [X ∣ Y] = E [X]$ , a constant random variable. Why? Knowing $Y$ gives zero new information about $X$ , so the updated guess equals the original.
Value. $X \sim Binomial (3, \frac{1}{2})$ , so $E [X] = 3 \cdot \frac{1}{2} = 1.5$ . Hence $E [X ∣ Y = 6] = 1.5$ too — same for every $y$ . Why this step? A Binomial $(n, p)$ counts successes in $n$ independent trials, and by linearity its mean is $n p$ ; here $n = 3, p = \frac{1}{2}$ give $n p = 1.5$ . Since the collapse made $E [X ∣ Y]$ constant, plugging in any $y$ (including $6$ ) returns that same $1.5$ .

Verify: Tower must still hold: $E [E [X ∣ Y]] = E [1.5] = 1.5 = E [X]$ ✓. A constant's expectation is itself, so the tower trivially closes — the degenerate case is consistent.

Cell E — take out what is known (don't fuse rules)

Worked example Ex 5 (Cell E).

$Y \sim U (0, 1)$ , and given $Y = y$ , $X \sim U (0, y)$ (same setup as Ex 3). Find $E [X Y]$ .

Forecast: it's tempting to write $E [X Y ∣ Y] = Y E [X]$ . That's the classic trap. Watch.

Pull out the known factor. Given $Y$ , the value $Y$ is a constant, so $E [X Y ∣ Y] = Y E [X ∣ Y] .$ Why this step? "Take out what is known": once $Y$ is fixed, $Y$ is a constant and leaves the inner expectation. This is always valid.
Do NOT replace $E [X ∣ Y]$ by $E [X]$ . Here $E [X ∣ Y] = Y /2$ (from Ex 3), which depends on $Y$ — $X$ and $Y$ are not independent. Why? Swapping in $E [X]$ needs independence (Cell D); fusing the two rules gives the wrong answer $Y \cdot \frac{1}{4}$ . Keep them separate.
Assemble and take the outer expectation. $E [X Y ∣ Y] = Y \cdot \frac{Y}{2} = \frac{Y ^{2}}{2}, E [X Y] = E [\frac{Y ^{2}}{2}] = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} .$ Why this step? $E [X Y ∣ Y]$ is a random variable (a function of $Y$ ); the Tower Rule says its outer expectation $E [E [X Y ∣ Y]]$ equals $E [X Y]$ . Pulling the constant $\frac{1}{2}$ out and using $E [Y^{2}] = \frac{1}{3}$ from Ex 3 finishes it.

Verify: Contrast the wrong route: $Y E [X] = Y \cdot \frac{1}{4} \Rightarrow E [X Y] = ? \frac{1}{4} E [Y] = \frac{1}{8}$ . Since $\frac{1}{6} \neq = \frac{1}{8}$ , fusing the rules genuinely breaks — our careful answer $\frac{1}{6}$ is the correct one ✓.

Cell F — real-world word problem (random count)

Worked example Ex 6 (Cell F). A call centre gets

$N \sim Poisson (λ = 10)$ calls an hour. Each call independently lasts $L$ minutes with $E [L] = 4$ . Let $S = \sum_{i = 1}^{N} L_{i}$ be the total minutes. Find $E [S]$ .

Forecast: roughly "10 calls times 4 minutes" ⇒ 40 minutes. Let's prove that the tower makes this exact even though $N$ is random.

Condition on the count $N$ . Given $N = n$ , $S$ is a sum of $n$ fixed-mean pieces: $E [S ∣ N = n] = \sum_{i = 1}^{n} E [L_{i}] = n \cdot 4.$ Why this step? With $N$ fixed the sum has a known number of terms, so linearity of expectation applies directly (Expectation and its linearity).
Package as a random variable. $E [S ∣ N] = 4 N$ . Why? Every value $n$ maps to $4 n$ , so the function $g (N) = 4 N$ is the conditional expectation.
Tower. $E [S] = E [4 N] = 4 E [N] = 4 \cdot 10 = 40$ minutes. Why? Averaging the conditional means, weighted by $P (N = n)$ — exactly the Tower Rule — collapses to $4 E [N]$ .

Verify: This is Wald's identity: $E [S] = E [N] E [L] = 10 \cdot 4 = 40$ ✓. Units: (calls) $\times$ (minutes/call) = minutes ✓.

Cell G — exam twist: run the tower backwards

Worked example Ex 7 (Cell G). An exam has two versions. A student gets version

$A$ with prob $0.3$ and version $B$ with prob $0.7$ . The overall mean score is known to be $E [X] = 68$ . Version- $B$ takers average $E [X ∣ B] = 72$ . Find the version- $A$ mean $E [X ∣ A]$ .

Forecast: overall is 68, and B (the majority) averages 72 (above overall). So A must drag the mean down — guess A's mean is below 68.

Write the tower as a weighted sum. $E [X] = P (A) E [X ∣ A] + P (B) E [X ∣ B] .$ Why this step? Conditioning on which version is a two-value partition; the tower says the overall mean is the crowd-weighted blend of the two group means (Law of total probability).
Plug in knowns. $68 = 0.3 E [X ∣ A] + 0.7 \cdot 72 = 0.3 E [X ∣ A] + 50.4.$ Why? We know three of the four quantities; the tower gives one equation in the one unknown.
Solve. $0.3 E [X ∣ A] = 68 - 50.4 = 17.6 \Rightarrow E [X ∣ A] = \frac{17.6}{0.3} = \frac{176}{3} = 58. \overline{6} \approx 58.67.$ Why this step? It is a single linear equation $68 = 0.3 m + 50.4$ in the unknown $m = E [X ∣ A]$ ; subtracting the known $50.4$ isolates the $0.3 m$ term, and dividing by the coefficient $0.3$ frees $m$ .

Verify: $0.3 \cdot \frac{176}{3} + 0.7 \cdot 72 = 17.6 + 50.4 = 68$ ✓, and $\frac{176}{3} < 68$ ✓ (A drags the mean down, as forecast). The trap here is forgetting the weights — a naive $\frac{68 + 72}{2}$ reasoning would be nonsense.

Cell H — limiting / edge behaviour

Worked example Ex 8 (Cell H). Two edge checks so the reader never hits an unshown case.

(H1) A shrinking event. For one fair die, condition on $B_{k} = {X \geq k}$ and let $k \to 6$ .

Forecast: as the event narrows to just the face 6, the conditional mean should slide up to 6.

$E [X ∣ X \geq k] = \frac{k + ( k + 1 ) + \dots + 6}{7 - k}$ = midpoint of the kept faces $= \frac{k + 6}{2}$ . Why this step? Kept faces ${k, \dots, 6}$ are equally spaced and equally likely, so their mean is the midpoint.
At $k = 6$ : $E [X ∣ X \geq 6] = \frac{6 + 6}{2} = 6$ , and $P (B_{6}) = 1/6 > 0$ so the formula is still legal. Why? Conditional expectation needs only $P (B) > 0$ ; here that holds right up to the singleton ${6}$ .

Verify: $k = 1$ gives $\frac{1 + 6}{2} = 3.5 = E [X]$ ✓ (conditioning on the certain event changes nothing); $k = 6$ gives $6$ ✓. Monotone increase from 3.5 to 6 as the event shrinks — exactly as forecast. The forbidden case is $P (B) = 0$ (e.g. "given $X = 7$ "), where $E [X ∣ B]$ is undefined, not zero.

(H2) A boundary continuous value. In Ex 3, let $Y \to 0^{+}$ . Then $X ∣ Y = y \sim U (0, y)$ collapses toward the point $0$ .

Read off the conditional mean near the boundary. From Ex 3, $E [X ∣ Y = y] = y /2$ , so as $y \to 0^{+}$ we get $E [X ∣ Y = y] = y /2 \to 0$ . Why this step? The edge case is handled by the same formula we already derived — we just track what it does at the boundary rather than inventing a new rule.
Sanity of the collapse. A uniform on the shrinking interval $[0, y]$ has all its mass sliding to $0$ , so its mean must go to $0$ too — and crucially the formula $y /2$ never divides by anything that vanishes, so nothing blows up. Why this step? We must confirm the limit is finite and sensible: degenerate inputs are exactly where careless formulas explode, and here they don't ✓.

Figure — Left panel (Cells A & H): the blue faded bars are the fair die's original $P = \frac{1}{6}$ ; conditioning on ${X \geq 3}$ crosses out faces 1 and 2 (yellow X's) and the pink bars show the survivors renormalised to $P = \frac{1}{4}$ , whose balance point (yellow dashed line) sits at $E [X ∣ B] = 4.5$ . Right panel (Cells B & G): two bars whose heights are the group means ( $E [X ∣ A] \approx 58.7$ pink, $E [X ∣ B] = 72$ blue) and whose widths are the crowd weights ( $0.3$ and $0.7$ ); the yellow dashed line is the width-weighted blend $E [X] = 68$ — the tower is literally "average the heights, weighted by the widths".

The chalkboard above shows the two workhorse pictures: left, Cell A/H — conditioning as slicing off outcomes and renormalising; right, Cell B/G — the tower as weighted bars whose blended height is the overall mean.

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

Scenario Map

The flowchart below is not a restatement of the matrix table — it is a decision procedure: read it top-to-bottom to pick which technique a fresh problem needs (divide by $P (B)$ ? weight by $P (Y = y)$ ? integrate? collapse?), then jump to the labelled example. The table classifies; this diagram routes.