Appendix — Solutions — Problem Set 2.5 (OVB Sign Predictions and the Measurement-Error Simulation)

Full worked solutions to book/weeks/week-02/ps2.5.md, covering Chapter 2.5. Notation follows the Conventions: the true (long) model is y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon with \mathbb{E}[\varepsilon \mid x,z] = 0; the short model is y = \tilde\beta_0 + \tilde\beta_1 x + u; the auxiliary slope is \delta_1 = \operatorname{Cov}(x,z)/\operatorname{Var}(x); for measurement error x = x^\* + m and \lambda = \sigma_{x^\*}^2/(\sigma_{x^\*}^2 + \sigma_m^2). The recurring theme of the key: bias lives in the probability limit and moves the center of the sampling distribution; precision lives in the standard error and controls only the spread. No amount of data, and no standard-error flavor, fixes a biased center — only a research design does.

Problem 1 — Derive the OVB formula, then read the bias term (14 pts)

(a) [6 pts] By definition the short-regression slope is \hat{\tilde\beta}_1 = \frac{\widehat{\operatorname{Cov}}(x,y)}{\widehat{\operatorname{Var}}(x)}. Substitute the true model y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon into the covariance. Covariance is linear in each argument and ignores additive constants (\operatorname{Cov}(x,\beta_0)=0), so \operatorname{Cov}(x,y) = \operatorname{Cov}(x,\ \beta_0 + \beta_1 x + \beta_2 z + \varepsilon) = \beta_1 \operatorname{Var}(x) + \beta_2 \operatorname{Cov}(x,z) + \operatorname{Cov}(x,\varepsilon). The last term is zero because the long model is well-specified: \mathbb{E}[\varepsilon\mid x,z]=0 implies \operatorname{Cov}(x,\varepsilon)=0. Divide through by \operatorname{Var}(x): \frac{\operatorname{Cov}(x,y)}{\operatorname{Var}(x)} = \beta_1 + \beta_2 \cdot \frac{\operatorname{Cov}(x,z)}{\operatorname{Var}(x)} = \beta_1 + \beta_2\,\delta_1, recognizing the fraction as the slope \delta_1 from the auxiliary regression of the omitted z on the included x. The Law of Large Numbers (Chapter 1.4) guarantees the sample covariance and variance converge in probability to their population counterparts, so the sample slope inherits the limit: \hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\,\delta_1.

(b) [4 pts] The bias is the product \beta_2\,\delta_1; a product is zero whenever either factor is zero. So OVB requires the omitted variable to both genuinely affect the outcome (\beta_2 \neq 0) and be correlated with the included regressor (\delta_1 \neq 0). Example of harmless omission: a variable that drives y but is uncorrelated with x — say, omitting a firm’s headquarters time zone from a return regression on momentum, when the time zone affects returns through some channel but is unrelated to the momentum signal. Here \delta_1 = 0, so the bias term vanishes and \hat{\tilde\beta}_1 stays consistent. The cost is not bias but precision: the omitted relevant variable sits in the error, inflating \operatorname{Var}(u) and therefore the standard error of \hat{\tilde\beta}_1.

(c) [4 pts] The short-model error is u = \beta_2 z + \varepsilon. Then \operatorname{Cov}(x,u) = \beta_2 \operatorname{Cov}(x,z) + \operatorname{Cov}(x,\varepsilon) = \beta_2 \operatorname{Cov}(x,z), which is nonzero whenever \beta_2 \neq 0 and \operatorname{Cov}(x,z)\neq 0 — i.e. whenever \delta_1 \neq 0 (and z matters). A nonzero \operatorname{Cov}(x,u) means \mathbb{E}[u\mid x]\neq 0: x carries information about the error. This is exactly the failure of the Chapter 2.2 zero-conditional-mean assumption — OVB is endogeneity caused by banishing a relevant, correlated variable into the error term, and it is the canonical example of why \mathbb{E}[\varepsilon\mid x]=0 breaks.

Problem 2 — The two-sign rule in three finance scenarios (20 pts)

The rule: sign of bias = \operatorname{sign}(\beta_2)\times\operatorname{sign}(\delta_1). “Same signs push up, opposite signs push down.” A downward bias makes the coefficient too negative; an upward bias makes it too positive — relative to the included regressor’s own true effect.

(a) [7 pts] Maya. (i) Omitted variable: creditworthiness C. (ii) \beta_2 > 0 — more creditworthy applicants are approved more often. (iii) \delta_1 < 0 — regressing C on D gives a negative slope because the D=1 group has lower average measured creditworthiness in this sample. (iv) Bias = (+)(-) = -: biased downward, too negative. As an estimate of D’s own effect on approval, the short coefficient is too negative — it overstates the approval penalty against the D=1 group, because part of the lower approval is really the lender responding to lower measured C, which the regression mistakenly loads onto D. (v) Honest conclusion: “The raw D coefficient is an upper bound (in magnitude) on the disadvantage attributable to D itself; it is contaminated by omitted creditworthiness, and I cannot separate discrimination from creditworthiness without controlling for C — or, better, a design (Week 4) that compares applicants identical on C.” Note we have not shown there is no discrimination; we have shown the number is biased.

(b) [7 pts] Sam. (i) Omitted variable: firm size, entered as “smallness.” (ii) \beta_2 > 0 — smaller firms earn higher returns (smallness raises return), by stipulation. (iii) \delta_1 > 0 — momentum scores are higher for smaller firms, so regressing smallness on momentum gives a positive slope. (iv) Bias = (+)(+) = +: biased upward, too positive. Sam’s momentum coefficient overstates momentum’s own profitability — a structural cause of the kind of mirage the multiple-testing discussion warned about. (v) Honest conclusion: “The momentum coefficient is biased upward by omitted size; some of the apparent momentum premium is the small-firm premium traveling with the momentum signal. I should control for size (a bet) or use a design that isolates momentum variation orthogonal to size.” Caveat: the empirical sign of the size–momentum relationship must be verified in the actual sample before asserting it in print — the mechanics of the two-sign rule hold regardless of which sign is true, but the direction of the conclusion depends on the real \delta_1.

(c) [6 pts] Priya. Omitted variable: physical climate exposure. \beta_2 > 0 (higher exposure raises premiums). \delta_1 > 0 (high-ESG firms tend to have higher physical exposure in her sample). Bias = (+)(+) = +: the ESG coefficient is biased upward. If the true ESG effect on premiums is negative or zero, the upward bias pulls the estimate toward positive, which undercuts a naive “ESG lowers premiums” claim — the omitted exposure is masking (or even reversing) the apparent ESG effect. Priya cannot claim ESG reduces premiums until physical exposure is held fixed (control, or a design that does so).

Problem 3 — OVB does not shrink with N (the precisely-wrong number) (16 pts)

(a) [4 pts] By the OVB formula, \hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\,\delta_1 = 0 + (0.30)(-0.50) = -0.15. Maya reads a -0.15 coefficient on D — apparently a 15-percentage-point approval penalty against the D=1 group — in a world with literally zero discrimination (\beta_1 = 0). The entire “effect” is borrowed from creditworthiness through the correlation between C and D.

(b) [5 pts] (i) The point estimate \hat{\tilde\beta}_1 stays centered on -0.15 at both sample sizes. Bias is a property of the probability limit; it does not depend on N. At N = 5{,}000{,}000 the estimate is, if anything, more reliably -0.15 — the wrong number, measured precisely. (ii) The standard error shrinks like 1/\sqrt{N}. Multiplying N by 1000 divides the standard error by \sqrt{1000} \approx 31.6. So the wobble around -0.15 collapses by a factor of about 31.6. Punchline: the larger sample buys her a tighter confidence interval around the wrong center — more precision, zero reduction in bias.

(c) [4 pts] As N grows, the standard error shrinks while the (biased) estimate stays at -0.15, so t = \hat{\tilde\beta}_1/\widehat{\operatorname{se}} explodes into the hundreds. This is exactly Maya’s “t = 4.1” from the chapter opening, only louder: a giant t-statistic certifies that the estimate is far from zero and measured sharply, but says nothing about whether the estimate is the right quantity. A precise estimate of a biased number is still a wrong answer. This problem lives entirely in the Bias column of the §2.5.11 ledger — the column no standard-error machinery can touch.

(d) [3 pts] If \delta_1 = 0 (the omitted variable is relevant but uncorrelated with D), then the bias term \beta_2\delta_1 = 0 and \hat{\tilde\beta}_1 is consistent — it converges to the true \beta_1. The cost of omitting it is only that it sits in the error and inflates \operatorname{Var}(u), widening the standard error. There, more data does help: the standard error shrinks like 1/\sqrt{N} and the estimate homes in on the truth. That scenario belongs to the SE column of the ledger, not the bias column.

Problem 4 — Classical measurement error: derive \lambda, predict the attenuated slope (18 pts)

(a) [8 pts] Substitute x^\* = x - m into the true model: y = \beta_0 + \beta_1(x - m) + \varepsilon = \beta_0 + \beta_1 x + (\varepsilon - \beta_1 m). We OLS-regress y on the observed x, so the probability limit of the slope is \operatorname{Cov}(x,y)/\operatorname{Var}(x).

Numerator. Using x = x^\* + m and y = \beta_0 + \beta_1 x^\* + \varepsilon, \operatorname{Cov}(x,y) = \operatorname{Cov}(x^\* + m,\ \beta_0 + \beta_1 x^\* + \varepsilon) = \beta_1 \operatorname{Var}(x^\*) + \beta_1\operatorname{Cov}(m,x^\*) + \operatorname{Cov}(x^\*,\varepsilon) + \operatorname{Cov}(m,\varepsilon). Under CEV the last three covariances are zero: \operatorname{Cov}(m,x^\*)=0 (error uncorrelated with the truth), \operatorname{Cov}(x^\*,\varepsilon)=0 (long model well-specified), and \operatorname{Cov}(m,\varepsilon)=0 (error uncorrelated with the equation error). So \operatorname{Cov}(x,y) = \beta_1\sigma_{x^\*}^2.

Denominator. Because x^\* and m are uncorrelated, the variance splits: \operatorname{Var}(x) = \operatorname{Var}(x^\* + m) = \sigma_{x^\*}^2 + \sigma_m^2.

Ratio. \hat\beta_1 \xrightarrow{p} \frac{\beta_1 \sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \sigma_m^2} = \beta_1 \cdot \lambda, \qquad \lambda = \frac{\sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \sigma_m^2}. Since variances are non-negative, the denominator is at least the numerator, so 0 < \lambda \le 1 (strictly less than 1 whenever \sigma_m^2 > 0). Multiplying by \lambda \in (0,1) shrinks the magnitude toward zero regardless of the sign of \beta_1 — this is attenuation bias, and its direction (toward zero) is always known under CEV.

(b) [4 pts] With \beta_1 = 1.0 and \lambda = 0.6: \hat\beta_1 \xrightarrow{p} 1.0 \times 0.6 = 0.6. Now set \sigma_m^2 = \tfrac{1}{4}\sigma_{x^\*}^2. Then \lambda = \frac{\sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \tfrac14\sigma_{x^\*}^2} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.8, so the new probability limit is 1.0 \times 0.8 = 0.8 — halving the noise variance moved the estimate from 0.6 up toward the truth 1.0, but it is still attenuated.

(c) [3 pts] Attenuation is a bias in the probability limit — the center of the sampling distribution sits at \beta_1\lambda, not at \beta_1. Sampling noise is the spread around that center, and it averages out: more data shrinks the spread like 1/\sqrt{N}. But more data only makes \hat\beta_1 converge more tightly onto the biased center \beta_1\lambda. Attenuation is the bias, not the noise; N does not touch it.

(d) [3 pts] Threat: classical measurement error in the activity regressor, biasing its coefficient toward zero by the reliability factor \lambda. Fixes: (1) measurement — get a less-noisy measure of true activity (a more reliable scraper, validated on-chain ground truth); (2) design — instrument for x^\* (Week 4 IV). A valid instrument would have to be correlated with true activity x^\* but uncorrelated with both the equation error \varepsilon and the measurement error m — for example, a second, independent measurement of activity, which under CEV provides exactly the leverage to undo \lambda.

Problem 5 — Mismeasured y versus mismeasured x: bias versus standard error (16 pts)

(a) [6 pts] With clean regressor x = x^\* and observed outcome y = y^\* + v, substitute the true relation y^\* = \beta_0 + \beta_1 x^\* + \varepsilon: y = y^\* + v = \beta_0 + \beta_1 x^\* + \varepsilon + v = \beta_0 + \beta_1 x + (\varepsilon + v). The composite error is \varepsilon + v. Under the classical assumption v is uncorrelated with x, and \varepsilon already is, so \operatorname{Cov}(x,\ \varepsilon + v) = 0 and \mathbb{E}[\varepsilon + v\mid x] = 0: the zero-conditional-mean assumption survives. Hence \hat\beta_1 remains consistent — it still converges to \beta_1. The only cost is a larger error variance, \operatorname{Var}(\varepsilon + v) = \sigma_\varepsilon^2 + \sigma_v^2, which inflates the standard error of \hat\beta_1 (wider confidence intervals), a precision loss, not bias.

(b) [4 pts]

Mismeasured quantity (classical)	Biases \hat\beta_1?	Direction	Bias or SE problem?
regressor x	Yes	toward zero (attenuation, factor \lambda)	Bias
outcome y	No	—	SE (wider intervals)

Rule: classical noise in x biases the slope toward zero; classical noise in y does not bias it at all — it only widens the standard error. The asymmetry comes from where the noise lands: noise in x enters both the regressor and the error and correlates them; noise in y enters only the error and stays uncorrelated with x.

(c) [3 pts] A noisily-measured control is worse than a noisy y because the control’s job is to absorb confounding from the key regressor. A noisy proxy only partially removes that confounding — the regression controls for the noisy proxy, not the true confounder — so residual confounding leaks back onto the coefficient of the key regressor, biasing it (re-opening OVB in the OVB direction). The name from Section 2.5.7: a noisily-measured control is a partially-omitted control. This is why “I controlled for it” and “I controlled for it well” are different claims.

(d) [3 pts] The “classical” assumption requires v uncorrelated with x. A non-classical case in Devon’s setting: suppose return is mismeasured because high-activity wallets (high x) are reported on exchanges that systematically over-state realized returns — then the outcome error v is correlated with x, \operatorname{Cov}(x,v)\neq 0, the zero-conditional-mean assumption fails, and \hat\beta_1 is biased (in an unknown direction). The protection in part (a) was the classical assumption, not the mere fact that the error was in y.

Problem 6 — Functional form, RESET, and a Monte Carlo design (16 pts)

(a) [5 pts] (i) If the truth is y = \beta_0 + \beta_1 x + \beta_2 x^2 + \varepsilon but Priya fits y = \tilde\beta_0 + \tilde\beta_1 x + u, then the fitted error must absorb the dropped term: u = \beta_2 x^2 + \varepsilon. Since x^2 is a deterministic function of x, it is correlated with x (for non-symmetric or non-zero-mean x, \operatorname{Cov}(x,x^2)\neq 0), so \operatorname{Cov}(x,u)\neq 0 and \mathbb{E}[u\mid x]\neq 0. This is “the same disease as OVB”: a relevant, correlated variable — here x^2, a transformation of a variable she already has — has been banished into the error term, breaking \mathbb{E}[\varepsilon\mid x]=0. (ii) Because the truth is convex (\beta_2 > 0), the best straight line understates the premium’s sensitivity at high emissions (where the curve is steep) and overstates it at low emissions (where the curve is flat). The single slope \hat{\tilde\beta}_1 is a misleading average that matches the true marginal effect \partial y/\partial x = \beta_1 + 2\beta_2 x at neither low nor high x.

(b) [4 pts] (i) p = 0.001 on the linear fit: reject the null of correct functional form — there is curvature the straight line is not capturing; “the shape is wrong.” p = 0.42 on the quadratic fit: fail to reject — the x^2 term has soaked up the detectable curvature, so “the shape now looks right.” RESET flagged the problem and confirmed the fix. (ii) A clean RESET does not rule out omitted variable bias (or measurement error). For example, a confounder like firm industry or physical climate exposure could still bias the emissions coefficient. RESET is structurally blind to it because RESET only checks for neglected curvature among the variables already in the model — it builds its test regressors from powers of the fitted values \hat y, which are functions of the included regressors only. It can say nothing about a variable that is not in the data at all. “Passing RESET means the shape looks right, not that the coefficient is causal.”

(c) [7 pts] A correct design earns full marks if it specifies all five required pieces unambiguously. Two model answers follow; either suffices.

Model answer — OVB simulation. - DGP. Choose true parameters \beta_0 = 1, \beta_1 = 0 (the demographic effect, set to zero so any nonzero short coefficient is pure bias), \beta_2 = 0.30. Draw x \sim \mathcal{N}(0,1) (the included regressor, e.g. group indicator standardized). Generate the omitted z correlated with x via z = \delta_1 x + \nu with \delta_1 = -0.50 and \nu \sim \mathcal{N}(0,1) independent — this \delta_1 is the auxiliary slope by construction. Draw \varepsilon \sim \mathcal{N}(0,1) independent of x,z. Form y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon. Use N = 50{,}000. - Two regressions. Long: y on x and z (should recover \hat\beta_1 \approx 0). Short: y on x alone (should recover \hat{\tilde\beta}_1 \approx \beta_1 + \beta_2\delta_1). - Theoretical prediction. \hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\delta_1 = 0 + (0.30)(-0.50) = -0.15; the long-model coefficient \to 0. - Sweep. Flip the sign of \delta_1 to +0.50. By the two-sign rule the bias flips sign: predict the short coefficient moves from -0.15 to +0.15 before running. Optionally sweep \delta_1 over a grid and plot \hat{\tilde\beta}_1 against \beta_2\delta_1 — points should lie on the 45° line. - Single run vs. many replications. A single fixed-N run gives one \hat{\tilde\beta}_1 near -0.15 with sampling wobble; averaging over many (say 1000) Monte Carlo replications shows the mean of \hat{\tilde\beta}_1 sits at -0.15 — i.e. the bias is in the center of the sampling distribution, not its spread. (More replications sharpen the estimate of the center; they do not move it.)

Model answer — attenuation simulation. - DGP. True \beta_1 = 2, \beta_0 = 1. Draw x^\* \sim \mathcal{N}(0,1) so \sigma_{x^\*}^2 = 1, \varepsilon \sim \mathcal{N}(0,1), y = \beta_0 + \beta_1 x^\* + \varepsilon. Corrupt the regressor: x = x^\* + m, m \sim \mathcal{N}(0,\sigma_m^2) independent of x^\* and \varepsilon. N = 50{,}000. - Two regressions. Clean: y on x^\* (recovers \hat\beta_1 \approx 2). Mismeasured: y on x (recovers \hat\beta_1 \approx 2\lambda). - Theoretical prediction. \hat\beta_1 \xrightarrow{p} \beta_1\lambda with \lambda = 1/(1 + \sigma_m^2) (since \sigma_{x^\*}^2 = 1). - Sweep. Run \sigma_m \in \{0, 0.5, 1.0, 2.0\}. Predict, before running, that \hat\beta_1 slides 2.00 \to 1.60 \to 1.00 \to 0.40 (i.e. \beta_1\lambda for \lambda = 1, 0.8, 0.5, 0.2). Plot \hat\beta_1 against \lambda — a straight line through the origin with slope \beta_1. - Single run vs. many replications. One run at fixed N gives a noisy \hat\beta_1 near \beta_1\lambda; averaging over many replications shows the mean estimate equals \beta_1\lambda — confirming attenuation is a shift of the distribution’s center, while N (or replication count) only tightens the spread.

A full-credit answer states the expected pattern before running (the prediction is the point of the exercise) and articulates the center-versus-spread distinction in the last bullet.

End of solutions for Problem Set 2.5. Every “Bias”-column threat rehearsed here — OVB, attenuation, mismeasured controls, functional form — is answered not by a better standard error but by a research design: potential outcomes and randomization (Week 3), then instrumental variables, difference-in-differences, and regression discontinuity (Week 4). Maya’s fair-lending question returns in Week 4 with the tools to hold creditworthiness genuinely fixed.

# Solutions — Problem Set 2.5 (OVB Sign Predictions and the Measurement-Error Simulation) Full worked solutions to `book/weeks/week-02/ps2.5.md`, covering Chapter 2.5. Notation follows the Conventions: the true (long) model is $y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon$ with $\mathbb{E}[\varepsilon \mid x,z] = 0$; the short model is $y = \tilde\beta_0 + \tilde\beta_1 x + u$; the auxiliary slope is $\delta_1 = \operatorname{Cov}(x,z)/\operatorname{Var}(x)$; for measurement error $x = x^\* + m$ and $\lambda = \sigma_{x^\*}^2/(\sigma_{x^\*}^2 + \sigma_m^2)$. The recurring theme of the key: **bias lives in the probability limit and moves the center of the sampling distribution; precision lives in the standard error and controls only the spread. No amount of data, and no standard-error flavor, fixes a biased center — only a research design does.** --- ## Problem 1 — Derive the OVB formula, then read the bias term (14 pts) **(a)** [6 pts] By definition the short-regression slope is $$\hat{\tilde\beta}_1 = \frac{\widehat{\operatorname{Cov}}(x,y)}{\widehat{\operatorname{Var}}(x)}.$$ Substitute the true model $y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon$ into the covariance. Covariance is linear in each argument and ignores additive constants ($\operatorname{Cov}(x,\beta_0)=0$), so $$\operatorname{Cov}(x,y) = \operatorname{Cov}(x,\ \beta_0 + \beta_1 x + \beta_2 z + \varepsilon) = \beta_1 \operatorname{Var}(x) + \beta_2 \operatorname{Cov}(x,z) + \operatorname{Cov}(x,\varepsilon).$$ The last term is zero because the long model is well-specified: $\mathbb{E}[\varepsilon\mid x,z]=0$ implies $\operatorname{Cov}(x,\varepsilon)=0$. Divide through by $\operatorname{Var}(x)$: $$\frac{\operatorname{Cov}(x,y)}{\operatorname{Var}(x)} = \beta_1 + \beta_2 \cdot \frac{\operatorname{Cov}(x,z)}{\operatorname{Var}(x)} = \beta_1 + \beta_2\,\delta_1,$$ recognizing the fraction as the slope $\delta_1$ from the auxiliary regression of the omitted $z$ on the included $x$. The **Law of Large Numbers** (Chapter 1.4) guarantees the sample covariance and variance converge in probability to their population counterparts, so the *sample* slope inherits the limit: $$\hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\,\delta_1.$$ **(b)** [4 pts] The bias is the *product* $\beta_2\,\delta_1$; a product is zero whenever *either* factor is zero. So OVB requires the omitted variable to **both** genuinely affect the outcome ($\beta_2 \neq 0$) **and** be correlated with the included regressor ($\delta_1 \neq 0$). Example of harmless omission: a variable that drives $y$ but is *uncorrelated* with $x$ — say, omitting a firm's headquarters time zone from a return regression on momentum, when the time zone affects returns through some channel but is unrelated to the momentum signal. Here $\delta_1 = 0$, so the bias term vanishes and $\hat{\tilde\beta}_1$ stays consistent. The cost is *not* bias but **precision**: the omitted relevant variable sits in the error, inflating $\operatorname{Var}(u)$ and therefore the standard error of $\hat{\tilde\beta}_1$. **(c)** [4 pts] The short-model error is $u = \beta_2 z + \varepsilon$. Then $$\operatorname{Cov}(x,u) = \beta_2 \operatorname{Cov}(x,z) + \operatorname{Cov}(x,\varepsilon) = \beta_2 \operatorname{Cov}(x,z),$$ which is nonzero whenever $\beta_2 \neq 0$ and $\operatorname{Cov}(x,z)\neq 0$ — i.e. whenever $\delta_1 \neq 0$ (and $z$ matters). A nonzero $\operatorname{Cov}(x,u)$ means $\mathbb{E}[u\mid x]\neq 0$: $x$ carries information about the error. This is exactly the failure of the Chapter 2.2 zero-conditional-mean assumption — **OVB is endogeneity** caused by banishing a relevant, correlated variable into the error term, and it is the canonical example of why $\mathbb{E}[\varepsilon\mid x]=0$ breaks. --- ## Problem 2 — The two-sign rule in three finance scenarios (20 pts) The rule: sign of bias $= \operatorname{sign}(\beta_2)\times\operatorname{sign}(\delta_1)$. "Same signs push up, opposite signs push down." A *downward* bias makes the coefficient too negative; an *upward* bias makes it too positive — relative to the included regressor's own true effect. **(a)** [7 pts] **Maya.** (i) Omitted variable: creditworthiness $C$. (ii) $\beta_2 > 0$ — more creditworthy applicants are approved more often. (iii) $\delta_1 < 0$ — regressing $C$ on $D$ gives a negative slope because the $D=1$ group has lower average measured creditworthiness in this sample. (iv) Bias $= (+)(-) = -$: biased **downward**, too negative. As an estimate of $D$'s *own* effect on approval, the short coefficient is **too negative** — it *overstates* the approval penalty against the $D=1$ group, because part of the lower approval is really the lender responding to lower measured $C$, which the regression mistakenly loads onto $D$. (v) Honest conclusion: *"The raw $D$ coefficient is an upper bound (in magnitude) on the disadvantage attributable to $D$ itself; it is contaminated by omitted creditworthiness, and I cannot separate discrimination from creditworthiness without controlling for $C$ — or, better, a design (Week 4) that compares applicants identical on $C$."* Note we have **not** shown there is no discrimination; we have shown the number is biased. **(b)** [7 pts] **Sam.** (i) Omitted variable: firm size, entered as "smallness." (ii) $\beta_2 > 0$ — smaller firms earn higher returns (smallness raises return), by stipulation. (iii) $\delta_1 > 0$ — momentum scores are higher for smaller firms, so regressing smallness on momentum gives a positive slope. (iv) Bias $= (+)(+) = +$: biased **upward**, too positive. Sam's momentum coefficient *overstates* momentum's own profitability — a structural cause of the kind of mirage the multiple-testing discussion warned about. (v) Honest conclusion: *"The momentum coefficient is biased upward by omitted size; some of the apparent momentum premium is the small-firm premium traveling with the momentum signal. I should control for size (a bet) or use a design that isolates momentum variation orthogonal to size."* Caveat: the empirical sign of the size–momentum relationship must be verified in the *actual* sample before asserting it in print — the mechanics of the two-sign rule hold regardless of which sign is true, but the *direction of the conclusion* depends on the real $\delta_1$. **(c)** [6 pts] **Priya.** Omitted variable: physical climate exposure. $\beta_2 > 0$ (higher exposure raises premiums). $\delta_1 > 0$ (high-ESG firms tend to have higher physical exposure in her sample). Bias $= (+)(+) = +$: the ESG coefficient is biased **upward**. If the true ESG effect on premiums is negative or zero, the upward bias pulls the estimate toward positive, which **undercuts** a naive "ESG lowers premiums" claim — the omitted exposure is masking (or even reversing) the apparent ESG effect. Priya cannot claim ESG reduces premiums until physical exposure is held fixed (control, or a design that does so). --- ## Problem 3 — OVB does not shrink with N (the precisely-wrong number) (16 pts) **(a)** [4 pts] By the OVB formula, $$\hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\,\delta_1 = 0 + (0.30)(-0.50) = -0.15.$$ Maya reads a $-0.15$ coefficient on $D$ — apparently a $15$-percentage-point approval penalty against the $D=1$ group — **in a world with literally zero discrimination** ($\beta_1 = 0$). The entire "effect" is borrowed from creditworthiness through the correlation between $C$ and $D$. **(b)** [5 pts] (i) The point estimate $\hat{\tilde\beta}_1$ stays centered on $-0.15$ at *both* sample sizes. Bias is a property of the probability limit; it does not depend on $N$. At $N = 5{,}000{,}000$ the estimate is, if anything, *more reliably* $-0.15$ — the wrong number, measured precisely. (ii) The standard error shrinks like $1/\sqrt{N}$. Multiplying $N$ by $1000$ divides the standard error by $\sqrt{1000} \approx 31.6$. So the wobble around $-0.15$ collapses by a factor of about $31.6$. Punchline: the larger sample buys her a tighter confidence interval **around the wrong center** — more precision, zero reduction in bias. **(c)** [4 pts] As $N$ grows, the standard error shrinks while the (biased) estimate stays at $-0.15$, so $t = \hat{\tilde\beta}_1/\widehat{\operatorname{se}}$ explodes into the hundreds. This is exactly Maya's "$t = 4.1$" from the chapter opening, only louder: a giant t-statistic certifies that the estimate is far from zero *and measured sharply*, but says nothing about whether the estimate is the *right* quantity. A precise estimate of a biased number is still a wrong answer. This problem lives entirely in the **Bias** column of the §2.5.11 ledger — the column no standard-error machinery can touch. **(d)** [3 pts] If $\delta_1 = 0$ (the omitted variable is relevant but uncorrelated with $D$), then the bias term $\beta_2\delta_1 = 0$ and $\hat{\tilde\beta}_1$ is **consistent** — it converges to the true $\beta_1$. The cost of omitting it is only that it sits in the error and inflates $\operatorname{Var}(u)$, widening the standard error. There, more data **does** help: the standard error shrinks like $1/\sqrt{N}$ and the estimate homes in on the truth. That scenario belongs to the **SE** column of the ledger, not the bias column. --- ## Problem 4 — Classical measurement error: derive $\lambda$, predict the attenuated slope (18 pts) **(a)** [8 pts] Substitute $x^\* = x - m$ into the true model: $$y = \beta_0 + \beta_1(x - m) + \varepsilon = \beta_0 + \beta_1 x + (\varepsilon - \beta_1 m).$$ We OLS-regress $y$ on the observed $x$, so the probability limit of the slope is $\operatorname{Cov}(x,y)/\operatorname{Var}(x)$. *Numerator.* Using $x = x^\* + m$ and $y = \beta_0 + \beta_1 x^\* + \varepsilon$, $$\operatorname{Cov}(x,y) = \operatorname{Cov}(x^\* + m,\ \beta_0 + \beta_1 x^\* + \varepsilon) = \beta_1 \operatorname{Var}(x^\*) + \beta_1\operatorname{Cov}(m,x^\*) + \operatorname{Cov}(x^\*,\varepsilon) + \operatorname{Cov}(m,\varepsilon).$$ Under CEV the last three covariances are zero: $\operatorname{Cov}(m,x^\*)=0$ (error uncorrelated with the truth), $\operatorname{Cov}(x^\*,\varepsilon)=0$ (long model well-specified), and $\operatorname{Cov}(m,\varepsilon)=0$ (error uncorrelated with the equation error). So $\operatorname{Cov}(x,y) = \beta_1\sigma_{x^\*}^2$. *Denominator.* Because $x^\*$ and $m$ are uncorrelated, the variance splits: $$\operatorname{Var}(x) = \operatorname{Var}(x^\* + m) = \sigma_{x^\*}^2 + \sigma_m^2.$$ *Ratio.* $$\hat\beta_1 \xrightarrow{p} \frac{\beta_1 \sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \sigma_m^2} = \beta_1 \cdot \lambda, \qquad \lambda = \frac{\sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \sigma_m^2}.$$ Since variances are non-negative, the denominator is at least the numerator, so $0 < \lambda \le 1$ (strictly less than 1 whenever $\sigma_m^2 > 0$). Multiplying by $\lambda \in (0,1)$ shrinks the magnitude toward zero regardless of the sign of $\beta_1$ — this is **attenuation bias**, and its direction (toward zero) is always known under CEV. **(b)** [4 pts] With $\beta_1 = 1.0$ and $\lambda = 0.6$: $\hat\beta_1 \xrightarrow{p} 1.0 \times 0.6 = 0.6$. Now set $\sigma_m^2 = \tfrac{1}{4}\sigma_{x^\*}^2$. Then $$\lambda = \frac{\sigma_{x^\*}^2}{\sigma_{x^\*}^2 + \tfrac14\sigma_{x^\*}^2} = \frac{1}{1 + 0.25} = \frac{1}{1.25} = 0.8,$$ so the new probability limit is $1.0 \times 0.8 = 0.8$ — halving the noise variance moved the estimate from $0.6$ up toward the truth $1.0$, but it is still attenuated. **(c)** [3 pts] Attenuation is a bias in the **probability limit** — the *center* of the sampling distribution sits at $\beta_1\lambda$, not at $\beta_1$. Sampling noise is the *spread* around that center, and it averages out: more data shrinks the spread like $1/\sqrt{N}$. But more data only makes $\hat\beta_1$ converge more tightly onto the *biased* center $\beta_1\lambda$. Attenuation is the bias, not the noise; $N$ does not touch it. **(d)** [3 pts] Threat: *classical measurement error in the activity regressor, biasing its coefficient toward zero by the reliability factor $\lambda$.* Fixes: (1) **measurement** — get a less-noisy measure of true activity (a more reliable scraper, validated on-chain ground truth); (2) **design** — instrument for $x^\*$ (Week 4 IV). A valid instrument would have to be correlated with true activity $x^\*$ but uncorrelated with both the equation error $\varepsilon$ and the measurement error $m$ — for example, a *second, independent* measurement of activity, which under CEV provides exactly the leverage to undo $\lambda$. --- ## Problem 5 — Mismeasured $y$ versus mismeasured $x$: bias versus standard error (16 pts) **(a)** [6 pts] With clean regressor $x = x^\*$ and observed outcome $y = y^\* + v$, substitute the true relation $y^\* = \beta_0 + \beta_1 x^\* + \varepsilon$: $$y = y^\* + v = \beta_0 + \beta_1 x^\* + \varepsilon + v = \beta_0 + \beta_1 x + (\varepsilon + v).$$ The composite error is $\varepsilon + v$. Under the classical assumption $v$ is uncorrelated with $x$, and $\varepsilon$ already is, so $\operatorname{Cov}(x,\ \varepsilon + v) = 0$ and $\mathbb{E}[\varepsilon + v\mid x] = 0$: the zero-conditional-mean assumption **survives**. Hence $\hat\beta_1$ remains **consistent** — it still converges to $\beta_1$. The only cost is a larger error variance, $\operatorname{Var}(\varepsilon + v) = \sigma_\varepsilon^2 + \sigma_v^2$, which inflates the **standard error** of $\hat\beta_1$ (wider confidence intervals), a *precision* loss, not bias. **(b)** [4 pts] | Mismeasured quantity (classical) | Biases $\hat\beta_1$? | Direction | Bias or SE problem? | |---|---|---|---| | regressor $x$ | Yes | toward zero (attenuation, factor $\lambda$) | **Bias** | | outcome $y$ | No | — | **SE** (wider intervals) | Rule: **classical noise in $x$ biases the slope toward zero; classical noise in $y$ does not bias it at all — it only widens the standard error.** The asymmetry comes from *where the noise lands*: noise in $x$ enters both the regressor and the error and correlates them; noise in $y$ enters only the error and stays uncorrelated with $x$. **(c)** [3 pts] A noisily-measured *control* is worse than a noisy $y$ because the control's job is to *absorb confounding* from the key regressor. A noisy proxy only *partially* removes that confounding — the regression controls for the noisy proxy, not the true confounder — so residual confounding leaks back onto the coefficient of the key regressor, **biasing** it (re-opening OVB in the OVB direction). The name from Section 2.5.7: *a noisily-measured control is a **partially-omitted control**.* This is why "I controlled for it" and "I controlled for it *well*" are different claims. **(d)** [3 pts] The "classical" assumption requires $v$ uncorrelated with $x$. A **non-classical** case in Devon's setting: suppose return is mismeasured because high-activity wallets (high $x$) are reported on exchanges that systematically over-state realized returns — then the outcome error $v$ is *correlated* with $x$, $\operatorname{Cov}(x,v)\neq 0$, the zero-conditional-mean assumption fails, and $\hat\beta_1$ is **biased** (in an unknown direction). The protection in part (a) was the classical assumption, not the mere fact that the error was in $y$. --- ## Problem 6 — Functional form, RESET, and a Monte Carlo design (16 pts) **(a)** [5 pts] (i) If the truth is $y = \beta_0 + \beta_1 x + \beta_2 x^2 + \varepsilon$ but Priya fits $y = \tilde\beta_0 + \tilde\beta_1 x + u$, then the fitted error must absorb the dropped term: $u = \beta_2 x^2 + \varepsilon$. Since $x^2$ is a deterministic function of $x$, it is correlated with $x$ (for non-symmetric or non-zero-mean $x$, $\operatorname{Cov}(x,x^2)\neq 0$), so $\operatorname{Cov}(x,u)\neq 0$ and $\mathbb{E}[u\mid x]\neq 0$. This is "the same disease as OVB": a relevant, correlated variable — here $x^2$, a transformation of a variable she already has — has been banished into the error term, breaking $\mathbb{E}[\varepsilon\mid x]=0$. (ii) Because the truth is convex ($\beta_2 > 0$), the best straight line **understates** the premium's sensitivity at high emissions (where the curve is steep) and **overstates** it at low emissions (where the curve is flat). The single slope $\hat{\tilde\beta}_1$ is a misleading average that matches the true marginal effect $\partial y/\partial x = \beta_1 + 2\beta_2 x$ at *neither* low nor high $x$. **(b)** [4 pts] (i) $p = 0.001$ on the linear fit: **reject** the null of correct functional form — there is curvature the straight line is not capturing; "the shape is wrong." $p = 0.42$ on the quadratic fit: **fail to reject** — the $x^2$ term has soaked up the detectable curvature, so "the shape now looks right." RESET flagged the problem and confirmed the fix. (ii) A clean RESET does **not** rule out **omitted variable bias** (or measurement error). For example, a confounder like firm industry or physical climate exposure could still bias the emissions coefficient. RESET is structurally blind to it because RESET only checks for *neglected curvature among the variables already in the model* — it builds its test regressors from powers of the fitted values $\hat y$, which are functions of the included regressors only. It can say nothing about a variable that is not in the data at all. "Passing RESET means the shape looks right, not that the coefficient is causal." **(c)** [7 pts] A correct design earns full marks if it specifies all five required pieces unambiguously. Two model answers follow; either suffices. **Model answer — OVB simulation.** - *DGP.* Choose true parameters $\beta_0 = 1$, $\beta_1 = 0$ (the demographic effect, set to zero so any nonzero short coefficient is *pure bias*), $\beta_2 = 0.30$. Draw $x \sim \mathcal{N}(0,1)$ (the included regressor, e.g. group indicator standardized). Generate the omitted $z$ *correlated with $x$* via $z = \delta_1 x + \nu$ with $\delta_1 = -0.50$ and $\nu \sim \mathcal{N}(0,1)$ independent — this $\delta_1$ is the auxiliary slope by construction. Draw $\varepsilon \sim \mathcal{N}(0,1)$ independent of $x,z$. Form $y = \beta_0 + \beta_1 x + \beta_2 z + \varepsilon$. Use $N = 50{,}000$. - *Two regressions.* **Long:** $y$ on $x$ and $z$ (should recover $\hat\beta_1 \approx 0$). **Short:** $y$ on $x$ alone (should recover $\hat{\tilde\beta}_1 \approx \beta_1 + \beta_2\delta_1$). - *Theoretical prediction.* $\hat{\tilde\beta}_1 \xrightarrow{p} \beta_1 + \beta_2\delta_1 = 0 + (0.30)(-0.50) = -0.15$; the long-model coefficient $\to 0$. - *Sweep.* Flip the sign of $\delta_1$ to $+0.50$. By the two-sign rule the bias flips sign: predict the short coefficient moves from $-0.15$ to $+0.15$ *before* running. Optionally sweep $\delta_1$ over a grid and plot $\hat{\tilde\beta}_1$ against $\beta_2\delta_1$ — points should lie on the 45° line. - *Single run vs. many replications.* A single fixed-$N$ run gives one $\hat{\tilde\beta}_1$ near $-0.15$ with sampling wobble; averaging over many (say 1000) Monte Carlo replications shows the *mean* of $\hat{\tilde\beta}_1$ sits at $-0.15$ — i.e. the bias is in the **center** of the sampling distribution, not its spread. (More replications sharpen the estimate of the center; they do not move it.) **Model answer — attenuation simulation.** - *DGP.* True $\beta_1 = 2$, $\beta_0 = 1$. Draw $x^\* \sim \mathcal{N}(0,1)$ so $\sigma_{x^\*}^2 = 1$, $\varepsilon \sim \mathcal{N}(0,1)$, $y = \beta_0 + \beta_1 x^\* + \varepsilon$. Corrupt the regressor: $x = x^\* + m$, $m \sim \mathcal{N}(0,\sigma_m^2)$ independent of $x^\*$ and $\varepsilon$. $N = 50{,}000$. - *Two regressions.* **Clean:** $y$ on $x^\*$ (recovers $\hat\beta_1 \approx 2$). **Mismeasured:** $y$ on $x$ (recovers $\hat\beta_1 \approx 2\lambda$). - *Theoretical prediction.* $\hat\beta_1 \xrightarrow{p} \beta_1\lambda$ with $\lambda = 1/(1 + \sigma_m^2)$ (since $\sigma_{x^\*}^2 = 1$). - *Sweep.* Run $\sigma_m \in \{0, 0.5, 1.0, 2.0\}$. Predict, before running, that $\hat\beta_1$ slides $2.00 \to 1.60 \to 1.00 \to 0.40$ (i.e. $\beta_1\lambda$ for $\lambda = 1, 0.8, 0.5, 0.2$). Plot $\hat\beta_1$ against $\lambda$ — a straight line through the origin with slope $\beta_1$. - *Single run vs. many replications.* One run at fixed $N$ gives a noisy $\hat\beta_1$ near $\beta_1\lambda$; averaging over many replications shows the mean estimate equals $\beta_1\lambda$ — confirming attenuation is a shift of the distribution's center, while $N$ (or replication count) only tightens the spread. A full-credit answer states the *expected pattern before running* (the prediction is the point of the exercise) and articulates the center-versus-spread distinction in the last bullet. --- *End of solutions for Problem Set 2.5. Every "Bias"-column threat rehearsed here — OVB, attenuation, mismeasured controls, functional form — is answered not by a better standard error but by a research **design**: potential outcomes and randomization (Week 3), then instrumental variables, difference-in-differences, and regression discontinuity (Week 4). Maya's fair-lending question returns in Week 4 with the tools to hold creditworthiness genuinely fixed.*