PS 1.1 — Conditional Probability & LIE/LTV Drills

Course: ASSIP 2026 · 8-Week Empirical Finance Track · Week 1 · Problem Set 1.1 Covers: Ch 1.1 (Joint, Conditional, and the Two Laws That Run Everything). Methods allowed: only what is built through Ch 1.1 — joint/marginal/conditional distributions, the multiplication rule, independence, conditional expectation as a random variable, the Law of Iterated Expectations (LIE), and the Law of Total Variance (LTV). You will not need covariance algebra (Ch 1.2), sampling theory (Ch 1.3), or anything later. Where a problem says “explain,” a clear sentence or two beats a wall of symbols.

Total: 100 points. Point values are stated per problem. Show your reasoning — a correct final number with no work earns roughly half credit, because the reasoning is the skill we are grading. Solutions are in Appendix E (E-w1-ps1.1-solutions.md); try every part before you look.

A note on the difficulty curve: Problem 1 is a warm-up you should finish in five minutes; Problems 4 and 5 are genuinely hard and are where the learning lives. Budget your time accordingly.

Problem 1 — Reading a joint table (14 points)

Priya is studying flood-insurance claims. For each insured home in a coastal county she records two things in a single year:

Y = the home’s flood-zone rating, coded H for “high risk” and L for “low risk.”
X = whether the home filed a claim that year, coded 1 for “filed a claim” and 0 for “no claim.”

Her data give the following joint distribution — each cell is the joint probability \Pr(X = x,\, Y = y) that a randomly drawn insured home falls in that combination:

	X = 0 (no claim)	X = 1 (claim)
Y = L (low risk)	0.55	0.05
Y = H (high risk)	0.25	0.15

(a) (3 pts) Verify this is a valid joint distribution, and compute the two marginal distributions p_Y(y) and p_X(x).

(b) (4 pts) Compute the conditional distribution of X given Y = H — that is, \Pr(X = 0 \mid Y = H) and \Pr(X = 1 \mid Y = H). Confirm the two numbers sum to 1, and state in one sentence what the claim rate is among high-risk homes.

(c) (4 pts) Compute \Pr(X = 1 \mid Y = L), the claim rate among low-risk homes. By what factor is a high-risk home more likely to file a claim than a low-risk home? (This ratio is exactly the kind of number an insurer uses to set a premium differential.)

(d) (3 pts) Are X and Y independent? Show the single check that settles it, and say in one sentence why the answer is what an insurer would expect. (Reveal-the-trick reminder: independence is an all-cells claim, but disproving it takes only one cell — you do not have to check the whole table. Find the cheapest cell that breaks the product rule and stop there.)

Problem 2 — Maya, reverse conditioning, and a fair-lending audit (18 points)

Maya is auditing a bank’s small-business loan model for fair lending. The model assigns each applicant a binary flag F: F = 1 means “flagged as high default risk” (loan likely denied), F = 0 means “not flagged.” Separately, each applicant truly either defaults or does not over the loan term; let D = 1 if the applicant would default and D = 0 if they would repay. From a large validation sample the bank reports:

The true default rate in the applicant pool is \Pr(D = 1) = 0.10.
Among applicants who would default, the model flags \Pr(F = 1 \mid D = 1) = 0.80 (it catches 80\% of true defaulters).
Among applicants who would repay, the model still flags \Pr(F = 1 \mid D = 0) = 0.15 (a 15\% false-flag rate on good borrowers).

(a) (4 pts) Find the overall fraction of applicants the model flags, \Pr(F = 1). (Hint: a flagged applicant is either a true defaulter who got caught or a good borrower who got falsely flagged. Use the multiplication rule on each piece, then add — this is the marginalization-by-conditioning move from §1–§2 of the chapter.)

(b) (6 pts) Maya’s real question is the reverse of what the bank reported. The bank told her \Pr(F = 1 \mid D = 1) — flag given default. Maya wants \Pr(D = 1 \mid F = 1) — given that an applicant was flagged, what is the probability they truly would have defaulted? Compute it. Show the reverse-conditioning step explicitly (joint over the relevant marginal).

(c) (4 pts) Interpret the gap. The model “catches 80\% of defaulters,” yet the number you found in (b) is much smaller than 80\%. Explain in two or three sentences why these two numbers are so different, and which one a denied applicant should care about.

(d) (4 pts) Now compute \Pr(D = 1 \mid F = 0) — the true default rate among applicants the model did not flag. In one sentence, what does comparing this to the pool’s base rate of 0.10 tell Maya about whether the flag carries real information?

Problem 3 — The Law of Iterated Expectations, proved then applied (16 points)

(a) (8 pts) Prove the Law of Iterated Expectations for discrete random variables: for any X with a finite mean and any discrete Y, \mathbb{E}\big[\mathbb{E}[X \mid Y]\big] = \mathbb{E}[X]. Write the proof in your own steps. You may use, without re-deriving them, the definition of conditional expectation \mathbb{E}[X \mid Y = y] = \sum_x x\, p_{X\mid Y}(x \mid y), the multiplication rule p(x,y) = p_{X\mid Y}(x\mid y)\, p_Y(y), and marginalization \sum_y p(x,y) = p_X(x). State clearly at each step which of these you are invoking. The whole proof is three or four honest lines; the grading is on whether each line names the rule that justifies it, not on length. Finally, state in one sentence where the assumption “X has a finite mean” actually gets used — there is exactly one step that silently relies on it.

(b) (8 pts) Now apply it. Devon tracks the daily return X (in percent) on a crypto token. The return depends on whether the day sees a major exchange-listing announcement, a rare event. Let Y = 1 if there is an announcement that day and Y = 0 otherwise, with \Pr(Y = 1) = 0.04. Devon estimates the conditional means \mathbb{E}[X \mid Y = 1] = 12.0\%, \qquad \mathbb{E}[X \mid Y = 0] = -0.25\%. Use LIE to find the unconditional expected daily return \mathbb{E}[X]. Then explain in one sentence why the unconditional mean is so close to the boring non-announcement number despite the huge announcement-day return — i.e., name the mechanism in the LIE formula that produces this.

Problem 4 — Sam’s regimes and the Law of Total Variance (22 points)

Sam refines the momentum strategy from the chapter and re-estimates the joint distribution of the daily portfolio return X (in percent) and the market regime Y. The new table has three return values and two regimes:

	X = -3\%	X = 0\%	X = +3\%	row total
Y = 1 (calm)	0.04	0.16	0.40	0.60
Y = 2 (stormy)	0.24	0.10	0.06	0.40

(a) (4 pts) Compute the conditional mean return in each regime, \mathbb{E}[X \mid Y = 1] and \mathbb{E}[X \mid Y = 2]. Then write out the conditional-expectation random variable \mathbb{E}[X \mid Y] explicitly (which value it takes, with what probability).

(b) (5 pts) Compute the within-regime variances \operatorname{Var}(X \mid Y = 1) and \operatorname{Var}(X \mid Y = 2), and from them the within-regime piece \mathbb{E}\big[\operatorname{Var}(X \mid Y)\big].

(c) (4 pts) Compute the across-regime piece \operatorname{Var}\big(\mathbb{E}[X \mid Y]\big).

(d) (4 pts) Add the two pieces, then verify your answer by computing \operatorname{Var}(X) directly from the marginal distribution of X. The two routes must agree exactly; if they do not, find your arithmetic error before reading on.

(e) (5 pts) Report the fraction of total variance that is across-regime. Compare it to the chapter’s strategy, where the across-regime share was about 19\%. In two or three sentences, tell Sam what this new number means for whether regime forecasting (predicting calm vs. stormy in advance) is worth the effort for this strategy versus the chapter’s.

Problem 5 — Uncorrelated but dependent (18 points)

The chapter warns (§3) that “uncorrelated” is strictly weaker than “independent”: zero covariance only rules out linear association, not all association. This problem makes you build the counterexample yourself and see the trap from both directions. You may use the fact that \operatorname{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y]; everything else uses only Ch 1.1 tools.

Devon models a “volatility-harvesting” position whose daily P&L X depends on the size of the market move but not its direction. Let Y be the signed market move, taking the three values -1, 0, +1 each with probability \tfrac{1}{3}, and let the position’s payoff be X = Y^2 (so X = 1 on a big move either way, X = 0 on a flat day).

(a) (3 pts) Write down the conditional distribution of X given each value of Y. Use it to argue in one sentence that X and Y are dependent — i.e., that knowing Y changes what you know about X. (You should not need any arithmetic beyond reading off values.)

(b) (6 pts) Now show they are uncorrelated: compute \mathbb{E}[Y], \mathbb{E}[XY], and hence \operatorname{Cov}(X, Y), and confirm it is zero. (Hint: XY = Y^3. Use LIE or a direct sum — your choice — but show the expectation calculation.)

(c) (5 pts) This is the part most students get wrong, so think before writing. Reconsider the claim “X and Y are uncorrelated.” Compute \mathbb{E}[X \mid Y = y] for each y and write out the random variable \mathbb{E}[X \mid Y]. Is \mathbb{E}[X \mid Y] equal to the constant \mathbb{E}[X]? What does your answer say about whether “\operatorname{Cov}(X,Y) = 0” and “\mathbb{E}[X \mid Y] = \mathbb{E}[X]” are the same condition? (They are not — explain the difference in plain words.)

(d) (4 pts) Tie it to Sam. In the chapter’s regime example, \operatorname{Cov}(\text{return}, \text{regime}) happened to be nonzero. Suppose a different strategy had \operatorname{Cov}(X, Y) = 0 between return and a regime indicator. Using what you just learned, is Sam safe to conclude “the regime is useless for forecasting my return”? Answer yes/no and give the one-sentence reason, referencing the difference between parts (b) and (c).

Problem 6 — Simulation design (pointing to nb1.1) (12 points)

You will not write working code here — you will design the experiment that notebooks/week-01/nb1.1 carries out, in enough detail that a classmate could implement it. Use Sam’s three-value table from Problem 4 as the universe. The point of this problem is that a simulation is only trustworthy if you can state, in advance, exactly what it should produce; here you have an analytic answer (Problem 4) to check the code against, which is the gold standard for debugging a sampler.

(a) (5 pts) Describe, step by step in plain English (pseudocode is fine), how to draw N independent samples of the pair (X, Y) from Problem 4’s joint distribution using only a uniform random-number generator. Be explicit about how you turn the six joint probabilities into a sampling rule. (Hint: flatten the table into six outcomes with probabilities summing to 1, take the cumulative sums, draw a uniform u \in [0,1), and pick the first outcome whose cumulative probability exceeds u.)

(b) (4 pts) Given a simulated sample of N pairs, describe how you would estimate the within-regime piece \mathbb{E}[\operatorname{Var}(X\mid Y)] and the across-regime piece \operatorname{Var}(\mathbb{E}[X\mid Y]) from the data alone (i.e., grouping by the realized regime), and what single equation you would check to confirm the Law of Total Variance held in your sample.

(c) (3 pts) As N grows from 10^2 to 10^6, what do you expect to happen to the gap between (i) your simulated within-plus-across sum and (ii) the exact \operatorname{Var}(X) you computed in Problem 4(d)? Will it ever be exactly zero at finite N? Answer in two sentences, and name the qualitative reason (you do not yet have the formal tool — that is Ch 1.3 — but state the intuition).

# PS 1.1 — Conditional Probability & LIE/LTV Drills **Course:** ASSIP 2026 · 8-Week Empirical Finance Track · Week 1 · Problem Set 1.1 **Covers:** Ch 1.1 (Joint, Conditional, and the Two Laws That Run Everything). **Methods allowed:** only what is built through Ch 1.1 — joint/marginal/conditional distributions, the multiplication rule, independence, conditional expectation as a random variable, the Law of Iterated Expectations (LIE), and the Law of Total Variance (LTV). You will *not* need covariance algebra (Ch 1.2), sampling theory (Ch 1.3), or anything later. Where a problem says "explain," a clear sentence or two beats a wall of symbols. **Total: 100 points.** Point values are stated per problem. Show your reasoning — a correct final number with no work earns roughly half credit, because the reasoning *is* the skill we are grading. Solutions are in Appendix E (`E-w1-ps1.1-solutions.md`); try every part before you look. A note on the difficulty curve: Problem 1 is a warm-up you should finish in five minutes; Problems 4 and 5 are genuinely hard and are where the learning lives. Budget your time accordingly. --- ## Problem 1 — Reading a joint table (14 points) Priya is studying flood-insurance claims. For each insured home in a coastal county she records two things in a single year: - $Y$ = the home's flood-zone rating, coded $H$ for "high risk" and $L$ for "low risk." - $X$ = whether the home filed a claim that year, coded $1$ for "filed a claim" and $0$ for "no claim." Her data give the following joint distribution — each cell is the joint probability $\Pr(X = x,\, Y = y)$ that a randomly drawn insured home falls in that combination: | | $X = 0$ (no claim) | $X = 1$ (claim) | |---|---|---| | **$Y = L$ (low risk)** | $0.55$ | $0.05$ | | **$Y = H$ (high risk)** | $0.25$ | $0.15$ | **(a) (3 pts)** Verify this is a valid joint distribution, and compute the two marginal distributions $p_Y(y)$ and $p_X(x)$. **(b) (4 pts)** Compute the conditional distribution of $X$ given $Y = H$ — that is, $\Pr(X = 0 \mid Y = H)$ and $\Pr(X = 1 \mid Y = H)$. Confirm the two numbers sum to $1$, and state in one sentence what the claim rate is among high-risk homes. **(c) (4 pts)** Compute $\Pr(X = 1 \mid Y = L)$, the claim rate among low-risk homes. By what factor is a high-risk home more likely to file a claim than a low-risk home? (This ratio is exactly the kind of number an insurer uses to set a premium differential.) **(d) (3 pts)** Are $X$ and $Y$ independent? Show the single check that settles it, and say in one sentence why the answer is what an insurer would *expect*. (Reveal-the-trick reminder: independence is an *all-cells* claim, but *dis*proving it takes only one cell — you do not have to check the whole table. Find the cheapest cell that breaks the product rule and stop there.) --- ## Problem 2 — Maya, reverse conditioning, and a fair-lending audit (18 points) Maya is auditing a bank's small-business loan model for fair lending. The model assigns each applicant a binary flag $F$: $F = 1$ means "flagged as high default risk" (loan likely denied), $F = 0$ means "not flagged." Separately, each applicant truly either defaults or does not over the loan term; let $D = 1$ if the applicant *would* default and $D = 0$ if they would repay. From a large validation sample the bank reports: - The true default rate in the applicant pool is $\Pr(D = 1) = 0.10$. - Among applicants who *would* default, the model flags $\Pr(F = 1 \mid D = 1) = 0.80$ (it catches $80\%$ of true defaulters). - Among applicants who *would* repay, the model still flags $\Pr(F = 1 \mid D = 0) = 0.15$ (a $15\%$ false-flag rate on good borrowers). **(a) (4 pts)** Find the overall fraction of applicants the model flags, $\Pr(F = 1)$. (Hint: a flagged applicant is *either* a true defaulter who got caught *or* a good borrower who got falsely flagged. Use the multiplication rule on each piece, then add — this is the marginalization-by-conditioning move from §1–§2 of the chapter.) **(b) (6 pts)** Maya's real question is the *reverse* of what the bank reported. The bank told her $\Pr(F = 1 \mid D = 1)$ — flag given default. Maya wants $\Pr(D = 1 \mid F = 1)$ — given that an applicant was flagged, what is the probability they truly would have defaulted? Compute it. Show the reverse-conditioning step explicitly (joint over the relevant marginal). **(c) (4 pts)** Interpret the gap. The model "catches $80\%$ of defaulters," yet the number you found in (b) is much smaller than $80\%$. Explain in two or three sentences *why* these two numbers are so different, and which one a denied applicant should care about. **(d) (4 pts)** Now compute $\Pr(D = 1 \mid F = 0)$ — the true default rate among applicants the model did *not* flag. In one sentence, what does comparing this to the pool's base rate of $0.10$ tell Maya about whether the flag carries real information? --- ## Problem 3 — The Law of Iterated Expectations, proved then applied (16 points) **(a) (8 pts)** Prove the Law of Iterated Expectations for discrete random variables: for any $X$ with a finite mean and any discrete $Y$, $$ \mathbb{E}\big[\mathbb{E}[X \mid Y]\big] = \mathbb{E}[X]. $$ Write the proof in your own steps. You may use, without re-deriving them, the definition of conditional expectation $\mathbb{E}[X \mid Y = y] = \sum_x x\, p_{X\mid Y}(x \mid y)$, the multiplication rule $p(x,y) = p_{X\mid Y}(x\mid y)\, p_Y(y)$, and marginalization $\sum_y p(x,y) = p_X(x)$. State clearly at each step which of these you are invoking. The whole proof is three or four honest lines; the grading is on whether each line *names* the rule that justifies it, not on length. Finally, state in one sentence where the assumption "$X$ has a finite mean" actually gets used — there is exactly one step that silently relies on it. **(b) (8 pts)** Now *apply* it. Devon tracks the daily return $X$ (in percent) on a crypto token. The return depends on whether the day sees a major exchange-listing announcement, a rare event. Let $Y = 1$ if there is an announcement that day and $Y = 0$ otherwise, with $\Pr(Y = 1) = 0.04$. Devon estimates the conditional means $$ \mathbb{E}[X \mid Y = 1] = 12.0\%, \qquad \mathbb{E}[X \mid Y = 0] = -0.25\%. $$ Use LIE to find the unconditional expected daily return $\mathbb{E}[X]$. Then explain in one sentence why the unconditional mean is so close to the boring non-announcement number despite the huge announcement-day return — i.e., name the mechanism in the LIE formula that produces this. --- ## Problem 4 — Sam's regimes and the Law of Total Variance (22 points) Sam refines the momentum strategy from the chapter and re-estimates the joint distribution of the daily portfolio return $X$ (in percent) and the market regime $Y$. The new table has **three** return values and **two** regimes: | | $X = -3\%$ | $X = 0\%$ | $X = +3\%$ | row total | |---|---|---|---|---| | **$Y = 1$ (calm)** | $0.04$ | $0.16$ | $0.40$ | $0.60$ | | **$Y = 2$ (stormy)** | $0.24$ | $0.10$ | $0.06$ | $0.40$ | **(a) (4 pts)** Compute the conditional mean return in each regime, $\mathbb{E}[X \mid Y = 1]$ and $\mathbb{E}[X \mid Y = 2]$. Then write out the conditional-expectation random variable $\mathbb{E}[X \mid Y]$ explicitly (which value it takes, with what probability). **(b) (5 pts)** Compute the within-regime variances $\operatorname{Var}(X \mid Y = 1)$ and $\operatorname{Var}(X \mid Y = 2)$, and from them the within-regime piece $\mathbb{E}\big[\operatorname{Var}(X \mid Y)\big]$. **(c) (4 pts)** Compute the across-regime piece $\operatorname{Var}\big(\mathbb{E}[X \mid Y]\big)$. **(d) (4 pts)** Add the two pieces, then verify your answer by computing $\operatorname{Var}(X)$ directly from the marginal distribution of $X$. The two routes must agree exactly; if they do not, find your arithmetic error before reading on. **(e) (5 pts)** Report the fraction of total variance that is *across-regime*. Compare it to the chapter's strategy, where the across-regime share was about $19\%$. In two or three sentences, tell Sam what this new number means for whether **regime forecasting** (predicting calm vs. stormy in advance) is worth the effort for *this* strategy versus the chapter's. --- ## Problem 5 — Uncorrelated but dependent (18 points) The chapter warns (§3) that "uncorrelated" is strictly weaker than "independent": zero covariance only rules out *linear* association, not all association. This problem makes you build the counterexample yourself and see the trap from both directions. You may use the fact that $\operatorname{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y]$; everything else uses only Ch 1.1 tools. Devon models a "volatility-harvesting" position whose daily P&L $X$ depends on the size of the market move but not its direction. Let $Y$ be the signed market move, taking the three values $-1$, $0$, $+1$ each with probability $\tfrac{1}{3}$, and let the position's payoff be $X = Y^2$ (so $X = 1$ on a big move either way, $X = 0$ on a flat day). **(a) (3 pts)** Write down the conditional distribution of $X$ given each value of $Y$. Use it to argue in one sentence that $X$ and $Y$ are **dependent** — i.e., that knowing $Y$ changes what you know about $X$. (You should not need any arithmetic beyond reading off values.) **(b) (6 pts)** Now show they are **uncorrelated**: compute $\mathbb{E}[Y]$, $\mathbb{E}[XY]$, and hence $\operatorname{Cov}(X, Y)$, and confirm it is zero. (Hint: $XY = Y^3$. Use LIE or a direct sum — your choice — but show the expectation calculation.) **(c) (5 pts)** This is the part most students get wrong, so think before writing. Reconsider the claim "$X$ and $Y$ are uncorrelated." Compute $\mathbb{E}[X \mid Y = y]$ for each $y$ and write out the random variable $\mathbb{E}[X \mid Y]$. Is $\mathbb{E}[X \mid Y]$ equal to the constant $\mathbb{E}[X]$? What does your answer say about whether "$\operatorname{Cov}(X,Y) = 0$" and "$\mathbb{E}[X \mid Y] = \mathbb{E}[X]$" are the same condition? (They are not — explain the difference in plain words.) **(d) (4 pts)** Tie it to Sam. In the chapter's regime example, $\operatorname{Cov}(\text{return}, \text{regime})$ happened to be nonzero. Suppose a *different* strategy had $\operatorname{Cov}(X, Y) = 0$ between return and a regime indicator. Using what you just learned, is Sam safe to conclude "the regime is useless for forecasting my return"? Answer yes/no and give the one-sentence reason, referencing the difference between parts (b) and (c). --- ## Problem 6 — Simulation design (pointing to nb1.1) (12 points) You will *not* write working code here — you will *design* the experiment that `notebooks/week-01/nb1.1` carries out, in enough detail that a classmate could implement it. Use Sam's three-value table from Problem 4 as the universe. The point of this problem is that a simulation is only trustworthy if you can state, in advance, exactly what it should produce; here you have an analytic answer (Problem 4) to check the code against, which is the gold standard for debugging a sampler. **(a) (5 pts)** Describe, step by step in plain English (pseudocode is fine), how to draw $N$ independent samples of the pair $(X, Y)$ from Problem 4's joint distribution using only a uniform random-number generator. Be explicit about how you turn the six joint probabilities into a sampling rule. (Hint: flatten the table into six outcomes with probabilities summing to $1$, take the cumulative sums, draw a uniform $u \in [0,1)$, and pick the first outcome whose cumulative probability exceeds $u$.) **(b) (4 pts)** Given a simulated sample of $N$ pairs, describe how you would *estimate* the within-regime piece $\mathbb{E}[\operatorname{Var}(X\mid Y)]$ and the across-regime piece $\operatorname{Var}(\mathbb{E}[X\mid Y])$ from the data alone (i.e., grouping by the realized regime), and what single equation you would check to confirm the Law of Total Variance held in your sample. **(c) (3 pts)** As $N$ grows from $10^2$ to $10^6$, what do you expect to happen to the gap between (i) your simulated within-plus-across sum and (ii) the exact $\operatorname{Var}(X)$ you computed in Problem 4(d)? Will it ever be *exactly* zero at finite $N$? Answer in two sentences, and name the qualitative reason (you do not yet have the formal tool — that is Ch 1.3 — but state the intuition).