Appendix w — Solutions — PS 1.1 (Conditional Probability & LIE/LTV Drills)

Problem set: book/weeks/week-01/ps1.1.md (PS 1.1, Week 1). Chapter: Ch 1.1 — Joint, Conditional, and the Two Laws That Run Everything.

These are full worked solutions, not just answers. Notation follows CONVENTIONS.md: \mathbb{E}[\cdot] for expectation, \operatorname{Var}(\cdot) for variance, \operatorname{Cov}(\cdot,\cdot) for covariance, p(x,y) for a joint pmf, p_{X\mid Y}(x\mid y) for a conditional pmf. Rounding: intermediate conditional probabilities are kept exact as fractions where it helps, and final numbers are reported to three decimals.

Problem 1 — Reading a joint table (14 points)

(a) (3 pts) A valid joint pmf needs every entry \ge 0 (true here) and all entries summing to 1: 0.55 + 0.05 + 0.25 + 0.15 = 1.00. \checkmark Marginals come from summing across rows (for Y) and down columns (for X): p_Y(L) = 0.55 + 0.05 = 0.60, \qquad p_Y(H) = 0.25 + 0.15 = 0.40, p_X(0) = 0.55 + 0.25 = 0.80, \qquad p_X(1) = 0.05 + 0.15 = 0.20. So 60\% of insured homes are low-risk, and 20\% of all homes file a claim in a given year.

(b) (4 pts) Condition on Y = H by dividing the high-risk row by its total p_Y(H) = 0.40: \Pr(X = 0 \mid Y = H) = \frac{0.25}{0.40} = 0.625, \qquad \Pr(X = 1 \mid Y = H) = \frac{0.15}{0.40} = 0.375. These sum to 0.625 + 0.375 = 1.000. \checkmark The claim rate among high-risk homes is 37.5\%.

(c) (4 pts) Condition on Y = L by dividing the low-risk row by p_Y(L) = 0.60: \Pr(X = 1 \mid Y = L) = \frac{0.05}{0.60} = 0.0833\ldots \approx 0.083. The factor by which a high-risk home is more likely to claim is the ratio of the two conditional claim rates: \frac{\Pr(X=1\mid Y=H)}{\Pr(X=1\mid Y=L)} = \frac{0.375}{0.0833} = 4.5. A high-risk home is 4.5\times as likely to file a claim as a low-risk home.

(d) (3 pts) X and Y are independent iff p(x,y) = p_X(x)\,p_Y(y) for every cell. One failing cell settles it. Test the (X=1, Y=H) cell: p_X(1)\,p_Y(H) = 0.20 \times 0.40 = 0.08, \quad\text{but the actual joint is } 0.15 \neq 0.08. So X and Y are dependent. An insurer expects exactly this — if claims were independent of flood-zone rating, the rating would carry no information about risk and there would be nothing to price. Equivalently, the conditional claim rates (37.5\% vs. 8.3\%) differ from the marginal rate (20\%), which is the definition of dependence.

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

Set up the given quantities. Let D be true default (D=1) and F the model’s flag (F=1). We are told the base rate \Pr(D=1) = 0.10 (so \Pr(D=0) = 0.90) and two forward conditionals — flag given truth: \Pr(F=1 \mid D=1) = 0.80, \qquad \Pr(F=1 \mid D=0) = 0.15.

(a) (4 pts) A flagged applicant is either a caught defaulter or a falsely-flagged good borrower. By the multiplication rule, \Pr(F=1, D=d) = \Pr(F=1\mid D=d)\,\Pr(D=d), and summing over the two values of D marginalizes D away: \Pr(F=1) = \Pr(F=1\mid D=1)\Pr(D=1) + \Pr(F=1\mid D=0)\Pr(D=0) = (0.80)(0.10) + (0.15)(0.90) = 0.080 + 0.135 = \boxed{0.215}. The model flags 21.5\% of applicants.

(b) (6 pts) Maya wants the reverse conditional \Pr(D=1\mid F=1). Reverse conditioning routes through the joint and divides by the other margin (here \Pr(F=1), which we just found): \Pr(D=1 \mid F=1) = \frac{\Pr(F=1, D=1)}{\Pr(F=1)} = \frac{\Pr(F=1\mid D=1)\Pr(D=1)}{\Pr(F=1)} = \frac{(0.80)(0.10)}{0.215} = \frac{0.080}{0.215} = \boxed{0.372}. So only about 37.2\% of flagged applicants are true defaulters. (This is exactly Bayes’ rule in table form, §2 of the chapter: the numerator is one joint cell, the denominator is the column total.)

(c) (4 pts) The two numbers answer different questions. \Pr(F=1\mid D=1) = 0.80 asks “among true defaulters, how many get flagged?” — it conditions on the small group (10\% of the pool). \Pr(D=1\mid F=1) = 0.372 asks “among the flagged, how many truly would default?” Because defaulters are rare (10\%) and the false-flag rate (15\%) is applied to the large pool of good borrowers, the flagged group is dominated by false positives: 0.135 of the 0.215 flagged are good borrowers. The base rate dilutes the flag’s meaning. A denied applicant should care about \Pr(D=1\mid F=1) = 37.2\% — given that they were flagged, the chance they would actually have defaulted is well under half, so the flag is far from a verdict.

(d) (4 pts) The true default rate among the un-flagged is, by the same reverse-conditioning move, \Pr(D=1\mid F=0) = \frac{\Pr(F=0\mid D=1)\Pr(D=1)}{\Pr(F=0)} = \frac{(0.20)(0.10)}{0.785} = \frac{0.020}{0.785} = \boxed{0.025}, where \Pr(F=0) = 1 - 0.215 = 0.785 and \Pr(F=0\mid D=1) = 1 - 0.80 = 0.20. Comparing 0.025 among the un-flagged to the pool’s 0.10 base rate, the flag clearly carries real information: un-flagged applicants default at one-quarter the base rate, and flagged applicants (37.2\%) default at nearly four times it. The model separates the two groups even though, per (c), being flagged is not a near-certain sentence.

Problem 3 — LIE proved then applied (16 points)

(a) (8 pts) We prove \mathbb{E}\big[\mathbb{E}[X\mid Y]\big] = \mathbb{E}[X]. The outer expectation averages the random variable \mathbb{E}[X\mid Y] over the distribution of Y: \mathbb{E}\big[\mathbb{E}[X\mid Y]\big] = \sum_y \mathbb{E}[X\mid Y=y]\, p_Y(y). \tag{outer expectation over $Y$} Substitute the definition of conditional expectation: = \sum_y \left(\sum_x x\, p_{X\mid Y}(x\mid y)\right) p_Y(y). \tag{def. of $\mathbb{E}[X\mid Y=y]$} Push p_Y(y) inside the inner sum (it does not depend on x) and apply the multiplication rule p_{X\mid Y}(x\mid y)\,p_Y(y) = p(x,y): = \sum_y \sum_x x\, p_{X\mid Y}(x\mid y)\, p_Y(y) = \sum_y \sum_x x\, p(x,y). \tag{multiplication rule} Swap the order of summation (a finite double sum, so this is legal) and sum the joint over y to recover the marginal of X: = \sum_x x \sum_y p(x,y) = \sum_x x\, p_X(x). \tag{marginalization} The last expression is the definition of \mathbb{E}[X]: = \mathbb{E}[X]. \qquad \blacksquare Where finiteness is used: the assumption that \mathbb{E}[X] exists (the sums converge absolutely) is what licenses swapping the order of summation in the marginalization step — for a conditionally convergent double sum, reordering can change the value, so without a finite mean the manipulation is not valid. With a finite outcome set, as in every problem on this set, absolute convergence is automatic.

(b) (8 pts) Apply LIE with Y the announcement indicator. By the same formula, \mathbb{E}[X] = \mathbb{E}[X\mid Y=1]\,\Pr(Y=1) + \mathbb{E}[X\mid Y=0]\,\Pr(Y=0) = (12.0)(0.04) + (-0.25)(0.96) = 0.480 - 0.240 = \boxed{0.24\%}. Why the unconditional mean sits so close to the boring -0.25\%: in the LIE formula each conditional mean is weighted by how often its condition occurs. The spectacular +12\% announcement day is real, but it is multiplied by its tiny probability 0.04, so it contributes only +0.48 to the average. The non-announcement mean, multiplied by its near-certain weight 0.96, dominates the blend. The mechanism is the probability weighting p_Y(y) inside LIE: rare states, however dramatic, move the unconditional mean only in proportion to their frequency.

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

The table (return X \in \{-3, 0, +3\}, regime Y \in \{1,2\}):

	-3	0	+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

Marginals: p_Y = (0.60, 0.40); column totals give p_X(-3)=0.28, p_X(0)=0.26, p_X(+3)=0.46.

(a) (4 pts) Conditional distributions divide each row by its total.

Calm (\div 0.60): \big(\tfrac{0.04}{0.60}, \tfrac{0.16}{0.60}, \tfrac{0.40}{0.60}\big) = (0.0667, 0.2667, 0.6667). \mathbb{E}[X\mid Y=1] = (-3)(0.0667) + (0)(0.2667) + (3)(0.6667) = -0.200 + 2.000 = 1.800. Stormy (\div 0.40): \big(\tfrac{0.24}{0.40}, \tfrac{0.10}{0.40}, \tfrac{0.06}{0.40}\big) = (0.60, 0.25, 0.15). \mathbb{E}[X\mid Y=2] = (-3)(0.60) + (0)(0.25) + (3)(0.15) = -1.800 + 0.450 = -1.350. The conditional-expectation random variable: \mathbb{E}[X\mid Y] = \begin{cases} 1.800\% & \text{w.p. } 0.60 \ (Y=1),\\[2pt] -1.350\% & \text{w.p. } 0.40 \ (Y=2). \end{cases}

(b) (5 pts) Within-regime variances via \operatorname{Var}(X\mid Y=y) = \mathbb{E}[X^2\mid Y=y] - (\mathbb{E}[X\mid Y=y])^2. Note X^2 \in \{9, 0, 9\}.

Calm: \mathbb{E}[X^2\mid Y=1] = (9)(0.0667) + (0)(0.2667) + (9)(0.6667) = 0.600 + 6.000 = 6.600, so \operatorname{Var}(X\mid Y=1) = 6.600 - (1.800)^2 = 6.600 - 3.240 = 3.360. Stormy: \mathbb{E}[X^2\mid Y=2] = (9)(0.60) + (0)(0.25) + (9)(0.15) = 5.400 + 1.350 = 6.750, so \operatorname{Var}(X\mid Y=2) = 6.750 - (-1.350)^2 = 6.750 - 1.8225 = 4.9275. The within-regime piece is the regime-weighted average: \mathbb{E}\big[\operatorname{Var}(X\mid Y)\big] = (3.360)(0.60) + (4.9275)(0.40) = 2.016 + 1.971 = \boxed{3.987}.

(c) (4 pts) First the overall mean, by LIE: \mathbb{E}[X] = (1.800)(0.60) + (-1.350)(0.40) = 1.080 - 0.540 = 0.540. Then \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = (1.800 - 0.540)^2(0.60) + (-1.350 - 0.540)^2(0.40). Compute each term: (1.260)^2(0.60) = 1.5876 \times 0.60 = 0.95256 and (-1.890)^2(0.40) = 3.5721 \times 0.40 = 1.42884. So \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = 0.95256 + 1.42884 = \boxed{2.3814}.

(d) (4 pts) Add the pieces: \mathbb{E}\big[\operatorname{Var}(X\mid Y)\big] + \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = 3.987 + 2.3814 = 6.3684. Direct check from the marginal of X (probabilities (0.28, 0.26, 0.46), mean 0.540): \mathbb{E}[X^2] = (9)(0.28) + (0)(0.26) + (9)(0.46) = 2.520 + 4.140 = 6.660, \operatorname{Var}(X) = 6.660 - (0.540)^2 = 6.660 - 0.2916 = 6.3684. \checkmark The two routes agree exactly: \boxed{6.3684 = 3.987 + 2.3814}.

(e) (5 pts) The across-regime share is \frac{\operatorname{Var}(\mathbb{E}[X\mid Y])}{\operatorname{Var}(X)} = \frac{2.3814}{6.3684} = 0.374 \approx 37\%. In the chapter’s strategy the across-regime share was only about 19\%; here it is nearly double, about 37\%. What this tells Sam: for this strategy, a much larger chunk of total risk comes purely from which regime you are in — the conditional means (+1.8\% vs. -1.35\%) are far apart and the regimes are close to evenly weighted. So regime forecasting is meaningfully more valuable here: perfectly anticipating calm vs. stormy would erase about 37\% of the variance, versus only \sim19\% for the chapter’s strategy. It is still not a magic bullet — the majority (63\%) of the wobble remains within regimes — but the payoff to a good calm/stormy timing signal is clearly higher for this strategy than for the chapter’s.

Problem 5 — Uncorrelated but dependent (18 points)

Setup: Y \in \{-1, 0, +1\} each w.p. \tfrac13, and X = Y^2 \in \{1, 0, 1\}.

(a) (3 pts) The conditional distribution of X given Y is degenerate — X is a function of Y, so once Y is known, X is certain: X\mid (Y=-1): X=1 \text{ w.p. } 1; \quad X\mid(Y=0): X=0 \text{ w.p. }1; \quad X\mid(Y=+1): X=1 \text{ w.p. }1. These conditional distributions are not all the same (given Y=0, X is surely 0; given Y=\pm1, X is surely 1), and they differ from the marginal of X. Knowing Y changes — in fact completely determines — X, so X and Y are dependent.

(b) (6 pts) By symmetry \mathbb{E}[Y] = \tfrac13(-1) + \tfrac13(0) + \tfrac13(1) = 0. For \mathbb{E}[XY], note XY = Y^2 \cdot Y = Y^3, and Y^3 \in \{-1, 0, 1\} at the three points: \mathbb{E}[XY] = \mathbb{E}[Y^3] = \tfrac13(-1)^3 + \tfrac13(0)^3 + \tfrac13(1)^3 = \tfrac13(-1 + 0 + 1) = 0. Therefore \operatorname{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y] = 0 - (\mathbb{E}[X])(0) = \boxed{0}. X and Y are uncorrelated, regardless of the value of \mathbb{E}[X] (which is \tfrac23, but it multiplies \mathbb{E}[Y]=0 and drops out).

(c) (5 pts) From part (a), X is determined by Y, so the conditional expectation just reads off the certain value: \mathbb{E}[X\mid Y=-1] = 1, \quad \mathbb{E}[X\mid Y=0] = 0, \quad \mathbb{E}[X\mid Y=+1] = 1, i.e. \mathbb{E}[X\mid Y] = Y^2, which takes the value 1 when Y = \pm1 and 0 when Y = 0. The marginal mean is \mathbb{E}[X] = \tfrac13(1) + \tfrac13(0) + \tfrac13(1) = \tfrac23, a constant. So \mathbb{E}[X\mid Y] = Y^2 \neq \tfrac23 = \mathbb{E}[X] — the conditional expectation is not the constant marginal mean. Hence “\operatorname{Cov}(X,Y)=0” and “\mathbb{E}[X\mid Y]=\mathbb{E}[X]” are not the same condition. In plain words: zero covariance only says there is no linear (straight-line) relationship — Y does not push X consistently up or down on average. But \mathbb{E}[X\mid Y] here bends (it is a parabola in Y): X’s mean genuinely depends on Y, just in a U-shape that a single covariance number averages out to zero. The conditional mean carries information that covariance cannot see.

(d) (4 pts) No. Even if a strategy had \operatorname{Cov}(X, Y) = 0 between return and a regime indicator, Sam cannot conclude the regime is useless. Zero covariance rules out only a linear relationship; as part (c) shows, \mathbb{E}[X\mid Y] can still vary with Y (a nonlinear dependence), in which case the regime does help forecast the return even though the covariance is zero. Forecasting power lives in \mathbb{E}[X\mid Y], not in \operatorname{Cov}(X,Y). (For a binary regime indicator the two conditions happen to coincide, but Sam should not generalize the binary-case coincidence into a rule — the safe habit is to check the conditional means directly.)

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

(a) (5 pts) Flatten Problem 4’s table into six labeled outcomes with their probabilities (summing to 1):

outcome k	(X, Y)	prob q_k	cumulative Q_k
1	(-3, \text{calm})	0.04	0.04
2	(0, \text{calm})	0.16	0.20
3	(+3, \text{calm})	0.40	0.60
4	(-3, \text{stormy})	0.24	0.84
5	(0, \text{stormy})	0.10	0.94
6	(+3, \text{stormy})	0.06	1.00

Sampling rule (inverse-CDF / “roulette wheel”), repeated N times independently: 1. Draw u \sim \text{Uniform}[0,1). 2. Find the smallest k such that u < Q_k (the first cumulative bucket that u falls into). 3. Record that outcome’s (X, Y) pair as one draw.

Because outcome k is selected exactly when Q_{k-1} \le u < Q_k, an interval of width q_k, outcome k is drawn with probability q_k — reproducing the joint distribution. Repeat for N independent draws to get a sample \{(X_i, Y_i)\}_{i=1}^N. (In NumPy this is one call to rng.choice(6, size=N, p=q) plus a lookup table, but the uniform-draw logic above is what it does under the hood.)

(b) (4 pts) From the simulated sample, estimate by grouping on the realized regime Y_i: - For each regime g \in \{\text{calm, stormy}\}, take the subsample of days with Y_i = g. Compute its sample mean \bar X_g (estimate of \mathbb{E}[X\mid Y=g]) and sample variance \widehat{\operatorname{Var}}_g (estimate of \operatorname{Var}(X\mid Y=g)). Let \hat p_g be the fraction of the sample in group g. - Within piece: \widehat{\mathbb{E}[\operatorname{Var}(X\mid Y)]} = \sum_g \hat p_g\, \widehat{\operatorname{Var}}_g (group-size-weighted average of the within-group variances). - Across piece: \widehat{\operatorname{Var}(\mathbb{E}[X\mid Y])} = \sum_g \hat p_g\,(\bar X_g - \bar X)^2, where \bar X is the overall sample mean (the size-weighted average of the \bar X_g).

The single equation to check is the Law of Total Variance on the sample: \widehat{\mathbb{E}[\operatorname{Var}(X\mid Y)]} + \widehat{\operatorname{Var}(\mathbb{E}[X\mid Y])} \;\approx\; \widehat{\operatorname{Var}}(X), where the right side is the overall sample variance of all X_i ignoring regime. (Use the same variance convention — dividing by N — on both sides so the identity holds exactly in-sample; with an N-1 divisor the two sides match only up to the usual finite-sample factor.)

(c) (3 pts) As N grows from 10^2 to 10^6, the gap between the simulated within-plus-across sum and the exact \operatorname{Var}(X) = 6.3684 should shrink toward zero, because the sample frequencies and sample moments converge to the true probabilities and moments as the sample grows. It will essentially never be exactly zero at finite N: any finite sample carries sampling noise, so the estimates wobble around the truth rather than hitting it. The intuition is that more data averages out the randomness — the formal statement (the Law of Large Numbers, and the 1/\sqrt{N} rate at which the error shrinks) is the subject of Ch 1.3–1.4.

# Solutions — PS 1.1 (Conditional Probability & LIE/LTV Drills) **Problem set:** `book/weeks/week-01/ps1.1.md` (PS 1.1, Week 1). **Chapter:** Ch 1.1 — Joint, Conditional, and the Two Laws That Run Everything. These are full worked solutions, not just answers. Notation follows `CONVENTIONS.md`: $\mathbb{E}[\cdot]$ for expectation, $\operatorname{Var}(\cdot)$ for variance, $\operatorname{Cov}(\cdot,\cdot)$ for covariance, $p(x,y)$ for a joint pmf, $p_{X\mid Y}(x\mid y)$ for a conditional pmf. Rounding: intermediate conditional probabilities are kept exact as fractions where it helps, and final numbers are reported to three decimals. --- ## Problem 1 — Reading a joint table (14 points) **(a) (3 pts)** A valid joint pmf needs every entry $\ge 0$ (true here) and all entries summing to $1$: $$ 0.55 + 0.05 + 0.25 + 0.15 = 1.00. \checkmark $$ Marginals come from summing across rows (for $Y$) and down columns (for $X$): $$ p_Y(L) = 0.55 + 0.05 = 0.60, \qquad p_Y(H) = 0.25 + 0.15 = 0.40, $$ $$ p_X(0) = 0.55 + 0.25 = 0.80, \qquad p_X(1) = 0.05 + 0.15 = 0.20. $$ So $60\%$ of insured homes are low-risk, and $20\%$ of all homes file a claim in a given year. **(b) (4 pts)** Condition on $Y = H$ by dividing the high-risk row by its total $p_Y(H) = 0.40$: $$ \Pr(X = 0 \mid Y = H) = \frac{0.25}{0.40} = 0.625, \qquad \Pr(X = 1 \mid Y = H) = \frac{0.15}{0.40} = 0.375. $$ These sum to $0.625 + 0.375 = 1.000$. $\checkmark$ **The claim rate among high-risk homes is $37.5\%$.** **(c) (4 pts)** Condition on $Y = L$ by dividing the low-risk row by $p_Y(L) = 0.60$: $$ \Pr(X = 1 \mid Y = L) = \frac{0.05}{0.60} = 0.0833\ldots \approx 0.083. $$ The factor by which a high-risk home is more likely to claim is the ratio of the two conditional claim rates: $$ \frac{\Pr(X=1\mid Y=H)}{\Pr(X=1\mid Y=L)} = \frac{0.375}{0.0833} = 4.5. $$ **A high-risk home is $4.5\times$ as likely to file a claim as a low-risk home.** **(d) (3 pts)** $X$ and $Y$ are independent iff $p(x,y) = p_X(x)\,p_Y(y)$ for *every* cell. One failing cell settles it. Test the $(X=1, Y=H)$ cell: $$ p_X(1)\,p_Y(H) = 0.20 \times 0.40 = 0.08, \quad\text{but the actual joint is } 0.15 \neq 0.08. $$ So **$X$ and $Y$ are dependent.** An insurer expects exactly this — if claims were independent of flood-zone rating, the rating would carry no information about risk and there would be nothing to price. Equivalently, the conditional claim rates ($37.5\%$ vs. $8.3\%$) differ from the marginal rate ($20\%$), which is the definition of dependence. --- ## Problem 2 — Maya, reverse conditioning, and a fair-lending audit (18 points) Set up the given quantities. Let $D$ be true default ($D=1$) and $F$ the model's flag ($F=1$). We are told the *base rate* $\Pr(D=1) = 0.10$ (so $\Pr(D=0) = 0.90$) and two *forward* conditionals — flag given truth: $$ \Pr(F=1 \mid D=1) = 0.80, \qquad \Pr(F=1 \mid D=0) = 0.15. $$ **(a) (4 pts)** A flagged applicant is either a caught defaulter or a falsely-flagged good borrower. By the multiplication rule, $\Pr(F=1, D=d) = \Pr(F=1\mid D=d)\,\Pr(D=d)$, and summing over the two values of $D$ marginalizes $D$ away: $$ \Pr(F=1) = \Pr(F=1\mid D=1)\Pr(D=1) + \Pr(F=1\mid D=0)\Pr(D=0) $$ $$ = (0.80)(0.10) + (0.15)(0.90) = 0.080 + 0.135 = \boxed{0.215}. $$ The model flags $21.5\%$ of applicants. **(b) (6 pts)** Maya wants the *reverse* conditional $\Pr(D=1\mid F=1)$. Reverse conditioning routes through the joint and divides by the *other* margin (here $\Pr(F=1)$, which we just found): $$ \Pr(D=1 \mid F=1) = \frac{\Pr(F=1, D=1)}{\Pr(F=1)} = \frac{\Pr(F=1\mid D=1)\Pr(D=1)}{\Pr(F=1)} = \frac{(0.80)(0.10)}{0.215} = \frac{0.080}{0.215} = \boxed{0.372}. $$ So only about **$37.2\%$ of flagged applicants are true defaulters.** (This is exactly Bayes' rule in table form, §2 of the chapter: the numerator is one joint cell, the denominator is the column total.) **(c) (4 pts)** The two numbers answer different questions. $\Pr(F=1\mid D=1) = 0.80$ asks "*among true defaulters*, how many get flagged?" — it conditions on the small group ($10\%$ of the pool). $\Pr(D=1\mid F=1) = 0.372$ asks "*among the flagged*, how many truly would default?" Because defaulters are rare ($10\%$) and the false-flag rate ($15\%$) is applied to the large pool of good borrowers, the flagged group is dominated by false positives: $0.135$ of the $0.215$ flagged are good borrowers. The base rate dilutes the flag's meaning. **A denied applicant should care about $\Pr(D=1\mid F=1) = 37.2\%$** — given that *they* were flagged, the chance they would actually have defaulted is well under half, so the flag is far from a verdict. **(d) (4 pts)** The true default rate among the *un*-flagged is, by the same reverse-conditioning move, $$ \Pr(D=1\mid F=0) = \frac{\Pr(F=0\mid D=1)\Pr(D=1)}{\Pr(F=0)} = \frac{(0.20)(0.10)}{0.785} = \frac{0.020}{0.785} = \boxed{0.025}, $$ where $\Pr(F=0) = 1 - 0.215 = 0.785$ and $\Pr(F=0\mid D=1) = 1 - 0.80 = 0.20$. **Comparing $0.025$ among the un-flagged to the pool's $0.10$ base rate, the flag clearly carries real information:** un-flagged applicants default at one-quarter the base rate, and flagged applicants ($37.2\%$) default at nearly four times it. The model separates the two groups even though, per (c), being flagged is not a near-certain sentence. --- ## Problem 3 — LIE proved then applied (16 points) **(a) (8 pts)** We prove $\mathbb{E}\big[\mathbb{E}[X\mid Y]\big] = \mathbb{E}[X]$. The outer expectation averages the random variable $\mathbb{E}[X\mid Y]$ over the distribution of $Y$: $$ \mathbb{E}\big[\mathbb{E}[X\mid Y]\big] = \sum_y \mathbb{E}[X\mid Y=y]\, p_Y(y). \tag{outer expectation over $Y$} $$ Substitute the definition of conditional expectation: $$ = \sum_y \left(\sum_x x\, p_{X\mid Y}(x\mid y)\right) p_Y(y). \tag{def. of $\mathbb{E}[X\mid Y=y]$} $$ Push $p_Y(y)$ inside the inner sum (it does not depend on $x$) and apply the multiplication rule $p_{X\mid Y}(x\mid y)\,p_Y(y) = p(x,y)$: $$ = \sum_y \sum_x x\, p_{X\mid Y}(x\mid y)\, p_Y(y) = \sum_y \sum_x x\, p(x,y). \tag{multiplication rule} $$ Swap the order of summation (a finite double sum, so this is legal) and sum the joint over $y$ to recover the marginal of $X$: $$ = \sum_x x \sum_y p(x,y) = \sum_x x\, p_X(x). \tag{marginalization} $$ The last expression is the definition of $\mathbb{E}[X]$: $$ = \mathbb{E}[X]. \qquad \blacksquare $$ **Where finiteness is used:** the assumption that $\mathbb{E}[X]$ exists (the sums converge absolutely) is what licenses *swapping the order of summation* in the marginalization step — for a conditionally convergent double sum, reordering can change the value, so without a finite mean the manipulation is not valid. With a finite outcome set, as in every problem on this set, absolute convergence is automatic. **(b) (8 pts)** Apply LIE with $Y$ the announcement indicator. By the same formula, $$ \mathbb{E}[X] = \mathbb{E}[X\mid Y=1]\,\Pr(Y=1) + \mathbb{E}[X\mid Y=0]\,\Pr(Y=0) $$ $$ = (12.0)(0.04) + (-0.25)(0.96) = 0.480 - 0.240 = \boxed{0.24\%}. $$ **Why the unconditional mean sits so close to the boring $-0.25\%$:** in the LIE formula each conditional mean is *weighted by how often its condition occurs*. The spectacular $+12\%$ announcement day is real, but it is multiplied by its tiny probability $0.04$, so it contributes only $+0.48$ to the average. The non-announcement mean, multiplied by its near-certain weight $0.96$, dominates the blend. The mechanism is the probability weighting $p_Y(y)$ inside LIE: rare states, however dramatic, move the unconditional mean only in proportion to their frequency. --- ## Problem 4 — Sam's regimes and the Law of Total Variance (22 points) The table (return $X \in \{-3, 0, +3\}$, regime $Y \in \{1,2\}$): | | $-3$ | $0$ | $+3$ | row total | |---|---|---|---|---| | $Y=1$ (calm) | $0.04$ | $0.16$ | $0.40$ | $0.60$ | | $Y=2$ (stormy) | $0.24$ | $0.10$ | $0.06$ | $0.40$ | Marginals: $p_Y = (0.60, 0.40)$; column totals give $p_X(-3)=0.28$, $p_X(0)=0.26$, $p_X(+3)=0.46$. **(a) (4 pts)** Conditional distributions divide each row by its total. *Calm* ($\div 0.60$): $\big(\tfrac{0.04}{0.60}, \tfrac{0.16}{0.60}, \tfrac{0.40}{0.60}\big) = (0.0667, 0.2667, 0.6667)$. $$ \mathbb{E}[X\mid Y=1] = (-3)(0.0667) + (0)(0.2667) + (3)(0.6667) = -0.200 + 2.000 = 1.800. $$ *Stormy* ($\div 0.40$): $\big(\tfrac{0.24}{0.40}, \tfrac{0.10}{0.40}, \tfrac{0.06}{0.40}\big) = (0.60, 0.25, 0.15)$. $$ \mathbb{E}[X\mid Y=2] = (-3)(0.60) + (0)(0.25) + (3)(0.15) = -1.800 + 0.450 = -1.350. $$ The conditional-expectation random variable: $$ \mathbb{E}[X\mid Y] = \begin{cases} 1.800\% & \text{w.p. } 0.60 \ (Y=1),\\[2pt] -1.350\% & \text{w.p. } 0.40 \ (Y=2). \end{cases} $$ **(b) (5 pts)** Within-regime variances via $\operatorname{Var}(X\mid Y=y) = \mathbb{E}[X^2\mid Y=y] - (\mathbb{E}[X\mid Y=y])^2$. Note $X^2 \in \{9, 0, 9\}$. *Calm:* $\mathbb{E}[X^2\mid Y=1] = (9)(0.0667) + (0)(0.2667) + (9)(0.6667) = 0.600 + 6.000 = 6.600$, so $$ \operatorname{Var}(X\mid Y=1) = 6.600 - (1.800)^2 = 6.600 - 3.240 = 3.360. $$ *Stormy:* $\mathbb{E}[X^2\mid Y=2] = (9)(0.60) + (0)(0.25) + (9)(0.15) = 5.400 + 1.350 = 6.750$, so $$ \operatorname{Var}(X\mid Y=2) = 6.750 - (-1.350)^2 = 6.750 - 1.8225 = 4.9275. $$ The within-regime piece is the regime-weighted average: $$ \mathbb{E}\big[\operatorname{Var}(X\mid Y)\big] = (3.360)(0.60) + (4.9275)(0.40) = 2.016 + 1.971 = \boxed{3.987}. $$ **(c) (4 pts)** First the overall mean, by LIE: $\mathbb{E}[X] = (1.800)(0.60) + (-1.350)(0.40) = 1.080 - 0.540 = 0.540$. Then $$ \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = (1.800 - 0.540)^2(0.60) + (-1.350 - 0.540)^2(0.40). $$ Compute each term: $(1.260)^2(0.60) = 1.5876 \times 0.60 = 0.95256$ and $(-1.890)^2(0.40) = 3.5721 \times 0.40 = 1.42884$. So $$ \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = 0.95256 + 1.42884 = \boxed{2.3814}. $$ **(d) (4 pts)** Add the pieces: $$ \mathbb{E}\big[\operatorname{Var}(X\mid Y)\big] + \operatorname{Var}\big(\mathbb{E}[X\mid Y]\big) = 3.987 + 2.3814 = 6.3684. $$ Direct check from the marginal of $X$ (probabilities $(0.28, 0.26, 0.46)$, mean $0.540$): $$ \mathbb{E}[X^2] = (9)(0.28) + (0)(0.26) + (9)(0.46) = 2.520 + 4.140 = 6.660, $$ $$ \operatorname{Var}(X) = 6.660 - (0.540)^2 = 6.660 - 0.2916 = 6.3684. \checkmark $$ The two routes agree exactly: $\boxed{6.3684 = 3.987 + 2.3814}$. **(e) (5 pts)** The across-regime share is $$ \frac{\operatorname{Var}(\mathbb{E}[X\mid Y])}{\operatorname{Var}(X)} = \frac{2.3814}{6.3684} = 0.374 \approx 37\%. $$ In the chapter's strategy the across-regime share was only about $19\%$; here it is nearly double, about $37\%$. **What this tells Sam:** for *this* strategy, a much larger chunk of total risk comes purely from *which regime you are in* — the conditional means ($+1.8\%$ vs. $-1.35\%$) are far apart and the regimes are close to evenly weighted. So regime forecasting is meaningfully more valuable here: perfectly anticipating calm vs. stormy would erase about $37\%$ of the variance, versus only $\sim19\%$ for the chapter's strategy. It is still not a magic bullet — the majority ($63\%$) of the wobble remains *within* regimes — but the payoff to a good calm/stormy timing signal is clearly higher for this strategy than for the chapter's. --- ## Problem 5 — Uncorrelated but dependent (18 points) Setup: $Y \in \{-1, 0, +1\}$ each w.p. $\tfrac13$, and $X = Y^2 \in \{1, 0, 1\}$. **(a) (3 pts)** The conditional distribution of $X$ given $Y$ is *degenerate* — $X$ is a function of $Y$, so once $Y$ is known, $X$ is certain: $$ X\mid (Y=-1): X=1 \text{ w.p. } 1; \quad X\mid(Y=0): X=0 \text{ w.p. }1; \quad X\mid(Y=+1): X=1 \text{ w.p. }1. $$ These conditional distributions are *not all the same* (given $Y=0$, $X$ is surely $0$; given $Y=\pm1$, $X$ is surely $1$), and they differ from the marginal of $X$. **Knowing $Y$ changes — in fact completely determines — $X$, so $X$ and $Y$ are dependent.** **(b) (6 pts)** By symmetry $\mathbb{E}[Y] = \tfrac13(-1) + \tfrac13(0) + \tfrac13(1) = 0$. For $\mathbb{E}[XY]$, note $XY = Y^2 \cdot Y = Y^3$, and $Y^3 \in \{-1, 0, 1\}$ at the three points: $$ \mathbb{E}[XY] = \mathbb{E}[Y^3] = \tfrac13(-1)^3 + \tfrac13(0)^3 + \tfrac13(1)^3 = \tfrac13(-1 + 0 + 1) = 0. $$ Therefore $$ \operatorname{Cov}(X,Y) = \mathbb{E}[XY] - \mathbb{E}[X]\,\mathbb{E}[Y] = 0 - (\mathbb{E}[X])(0) = \boxed{0}. $$ $X$ and $Y$ are **uncorrelated**, regardless of the value of $\mathbb{E}[X]$ (which is $\tfrac23$, but it multiplies $\mathbb{E}[Y]=0$ and drops out). **(c) (5 pts)** From part (a), $X$ is determined by $Y$, so the conditional expectation just reads off the certain value: $$ \mathbb{E}[X\mid Y=-1] = 1, \quad \mathbb{E}[X\mid Y=0] = 0, \quad \mathbb{E}[X\mid Y=+1] = 1, $$ i.e. $\mathbb{E}[X\mid Y] = Y^2$, which takes the value $1$ when $Y = \pm1$ and $0$ when $Y = 0$. The marginal mean is $\mathbb{E}[X] = \tfrac13(1) + \tfrac13(0) + \tfrac13(1) = \tfrac23$, a constant. **So $\mathbb{E}[X\mid Y] = Y^2 \neq \tfrac23 = \mathbb{E}[X]$ — the conditional expectation is *not* the constant marginal mean.** Hence "$\operatorname{Cov}(X,Y)=0$" and "$\mathbb{E}[X\mid Y]=\mathbb{E}[X]$" are *not* the same condition. In plain words: zero covariance only says there is no *linear* (straight-line) relationship — $Y$ does not push $X$ consistently up or down on average. But $\mathbb{E}[X\mid Y]$ here bends (it is a parabola in $Y$): $X$'s mean genuinely depends on $Y$, just in a U-shape that a single covariance number averages out to zero. The conditional mean carries information that covariance cannot see. **(d) (4 pts)** **No.** Even if a strategy had $\operatorname{Cov}(X, Y) = 0$ between return and a regime indicator, Sam cannot conclude the regime is useless. Zero covariance rules out only a *linear* relationship; as part (c) shows, $\mathbb{E}[X\mid Y]$ can still vary with $Y$ (a nonlinear dependence), in which case the regime *does* help forecast the return even though the covariance is zero. Forecasting power lives in $\mathbb{E}[X\mid Y]$, not in $\operatorname{Cov}(X,Y)$. (For a binary regime indicator the two conditions happen to coincide, but Sam should not generalize the binary-case coincidence into a rule — the safe habit is to check the conditional means directly.) --- ## Problem 6 — Simulation design (pointing to nb1.1) (12 points) **(a) (5 pts)** Flatten Problem 4's table into six labeled outcomes with their probabilities (summing to $1$): | outcome $k$ | $(X, Y)$ | prob $q_k$ | cumulative $Q_k$ | |---|---|---|---| | 1 | $(-3, \text{calm})$ | $0.04$ | $0.04$ | | 2 | $(0, \text{calm})$ | $0.16$ | $0.20$ | | 3 | $(+3, \text{calm})$ | $0.40$ | $0.60$ | | 4 | $(-3, \text{stormy})$ | $0.24$ | $0.84$ | | 5 | $(0, \text{stormy})$ | $0.10$ | $0.94$ | | 6 | $(+3, \text{stormy})$ | $0.06$ | $1.00$ | Sampling rule (inverse-CDF / "roulette wheel"), repeated $N$ times independently: 1. Draw $u \sim \text{Uniform}[0,1)$. 2. Find the smallest $k$ such that $u < Q_k$ (the first cumulative bucket that $u$ falls into). 3. Record that outcome's $(X, Y)$ pair as one draw. Because outcome $k$ is selected exactly when $Q_{k-1} \le u < Q_k$, an interval of width $q_k$, outcome $k$ is drawn with probability $q_k$ — reproducing the joint distribution. Repeat for $N$ independent draws to get a sample $\{(X_i, Y_i)\}_{i=1}^N$. (In NumPy this is one call to `rng.choice(6, size=N, p=q)` plus a lookup table, but the uniform-draw logic above is what it does under the hood.) **(b) (4 pts)** From the simulated sample, estimate by *grouping on the realized regime* $Y_i$: - For each regime $g \in \{\text{calm, stormy}\}$, take the subsample of days with $Y_i = g$. Compute its sample mean $\bar X_g$ (estimate of $\mathbb{E}[X\mid Y=g]$) and sample variance $\widehat{\operatorname{Var}}_g$ (estimate of $\operatorname{Var}(X\mid Y=g)$). Let $\hat p_g$ be the fraction of the sample in group $g$. - **Within piece:** $\widehat{\mathbb{E}[\operatorname{Var}(X\mid Y)]} = \sum_g \hat p_g\, \widehat{\operatorname{Var}}_g$ (group-size-weighted average of the within-group variances). - **Across piece:** $\widehat{\operatorname{Var}(\mathbb{E}[X\mid Y])} = \sum_g \hat p_g\,(\bar X_g - \bar X)^2$, where $\bar X$ is the overall sample mean (the size-weighted average of the $\bar X_g$). The single equation to check is the Law of Total Variance on the sample: $$ \widehat{\mathbb{E}[\operatorname{Var}(X\mid Y)]} + \widehat{\operatorname{Var}(\mathbb{E}[X\mid Y])} \;\approx\; \widehat{\operatorname{Var}}(X), $$ where the right side is the overall sample variance of all $X_i$ ignoring regime. (Use the same variance convention — dividing by $N$ — on both sides so the identity holds exactly in-sample; with an $N-1$ divisor the two sides match only up to the usual finite-sample factor.) **(c) (3 pts)** As $N$ grows from $10^2$ to $10^6$, the gap between the simulated within-plus-across sum and the exact $\operatorname{Var}(X) = 6.3684$ should *shrink toward zero*, because the sample frequencies and sample moments converge to the true probabilities and moments as the sample grows. It will essentially never be *exactly* zero at finite $N$: any finite sample carries sampling noise, so the estimates wobble around the truth rather than hitting it. The intuition is that more data averages out the randomness — the formal statement (the Law of Large Numbers, and the $1/\sqrt{N}$ rate at which the error shrinks) is the subject of Ch 1.3–1.4.