Attempt the questions before opening the solutions. Each exercise connects at least two steps in the probability-to-inference chain. These are a curated learning set, not a verbatim replacement for every numbered textbook exercise.
A. Summation and moment algebra
A1. Show \(\sum_i(x_i-\bar x)=0\) and \(\sum_i(x_i-\bar x)^2=\sum_i x_i^2-n\bar x^2\). Which facts are deterministic?
Solution A1
\(\sum_i x_i-\sum_i\bar x=n\bar x-n\bar x=0\). Expanding the square gives \(\sum_i x_i^2-2\bar x\sum_i x_i+n\bar x^2=\sum_i x_i^2-n\bar x^2\). Both hold for any numerical sequence and require no probability model. The elementary rules are \(\sum_i a=na\), \(\sum_i ax_i=a\sum_i x_i\), and \(\sum_i(x_i+y_i)=\sum_i x_i+\sum_i y_i\).
A2. For \(Y\sim\)Bernoulli(.3), derive its first four raw moments, variance, skewness, and kurtosis.
Solution A2
\(Y^{r}=Y\) for every positive integer r, so all four raw moments equal \(p=.3\). Variance\(=p(1-p)=.21\). The third central moment is \(p(1-p)(1-2p)\); skewness\(=(1-2p)/\sqrt{p(1-p)}\approx .87287\). The fourth central moment is \(p(1-p)[(1-p)^{3}+p^{3}]\); kurtosis\(=[(1-p)^{3}+p^{3}]/[p(1-p)]=1/[p(1-p)]-3\approx 1.76190\). Bernoulli need not be leptokurtic.
A3. With probability .9 draw S from \(\mathcal{N}(0,4)\), and otherwise from \(\mathcal{N}(0,16)\). Find \(\mathbb{E}[S], \operatorname{Var}(S)\), skewness, and kurtosis. Why is this relevant to the stock-return box?
Solution A3
Condition on the component\(. \mathbb{E}[S]=0\) and \(\mathbb{E}[S^{2}]=.9\cdot 4+.1\cdot 16=5.2\). Odd third moments are zero by symmetry\(. \mathbb{E}[S^{4}]=.9\cdot 3\cdot 4^{2}+.1\cdot 3\cdot 16^{2}=120\). Kurtosis\(=120/5.2^{2}\approx 4.43787>3\). Mixing normal observations with different variances can generate a heavy-tailed marginal distribution. A normal conditional model and a normal marginal model are different claims.
B. Conditional information and dependence
B1. Use the commute table in Module 3 to compute \(\mathbb{E}[X], \mathbb{E}[Y], \operatorname{Cov}(X,Y)\), and Corr(X,Y). Then verify the law of total variance for \(Y\).
Solution B1
\(\mathbb{E}[X]=.7, \mathbb{E}[Y]=.78, \mathbb{E}[\)XY\(]=.63. \operatorname{Cov}=.63-.7\cdot .78=.084. \operatorname{Var}(X)=.21\) and \(\operatorname{Var}(Y)=.1716\), so \(\operatorname{Corr}=.084/\sqrt{.21\cdot .1716}\approx .44250. \mathbb{E}[\operatorname{Var}(Y\mid X)]=.3\cdot .25+.7\cdot .09=.138\); \(\operatorname{Var}(\mathbb{E}[Y\mid X])=.0336\). Their sum is .1716.
B2. Let \(m(X)=\mathbb{E}[Y\mid X], u=Y-m(X)\), and \(\mathbb{E}[Y^{2}]<\infty\). Prove \(\mathbb{E}[u]=0\) and \(\mathbb{E}[\)uX]=0 when the product is integrable. Explain why \(\mathbb{E}[u]=0\) alone is insufficient for the second conclusion.
Solution B2
\(\mathbb{E}[u\mid X]=0\). Iterated expectations gives \(\mathbb{E}[u]=0\), and \(\mathbb{E}[\)uX\(]=\mathbb{E}[\)XE\([u|X]]=0. A\) counterexample to using only \(\mathbb{E}[u]=0\) is \(u=X\) with \(\mathbb{E}[X]=0\) and \(\operatorname{Var}(X)>0\); then \(\mathbb{E}[\)uX\(]=\mathbb{E}[X^{2}]>0\).
B3. Independent \(\varepsilon\) and \(X\) satisfy \(\mathbb{E}[\varepsilon ]=0, \operatorname{Var}(\varepsilon )=1\); \(X\) takes 1 and 2 equally often. With \(Y=X\varepsilon\), determine whether mean independence implies independence here.
Solution B3
\(\mathbb{E}[Y\mid X]=0=\mathbb{E}[Y]\), while \(\operatorname{Var}(Y|X)=X^{2}\) takes 1 and 4. Thus the distribution changes with \(X\), so independence fails. Mean independence places no constant-variance restriction.
C. Sampling, consistency, and inference
C1. For \(i.i.d. Y_{i}\sim \mathcal{N}(10,4)\), compute \(\Pr(9.6\le \bar{Y} \le 10.4)\) at \(n=20\) and \(n=100\). Explain whether a CLT is needed.
Solution C1
\(\operatorname{SD}(\bar{Y} )=2/\sqrt{n}\), so the probability is \(2\Phi (.2\sqrt{n})-1\). It is approximately .6289 at \(n=20\) and .9545 at \(n=100\). The result is exact because the observations are normal; no asymptotic approximation is needed.
C2. Suppose observations have common variance 9 and pairwise correlation .1. Find \(\operatorname{Var}(\bar{Y} )\) at \(n=100\) and its limit under this common-correlation model.
Solution C2
\(\operatorname{Var}(\bar{Y} )=9/100+(99/100)(.1)(9)=.981\); the limit is .9. The i.i.d. formula would incorrectly give .09. More observations do not eliminate the common component.
C3. For \(n=100, \bar{Y} =52\), and \(s=10\), test \(H_{0}\):\(\mu =50\) against \((i) \mu \ne 50\) and (ii\() \mu >50\). Construct a 95% interval using the large-sample method.
Solution C3
\(\operatorname{SE}=1\) and \(t=2\). Two-sided \(p\approx .04550\); upper-sided \(p\approx .02275\). Both reject at 5%. The interval is [50.04,53.96]. The one-sided direction must have been selected in advance; it cannot be chosen to improve significance after seeing \(t\).
C4. Thirty independent 90% confidence intervals are constructed for the same fixed parameter. What is the expected number covering it? What is the probability all cover, if each has exactly 90% coverage?
Solution C4
The expected count is 30(.9)=27 by linearity; independence is not needed for that expectation. Independence gives \(\Pr(\)all cover\()=.9^{30}\approx .04239\). Marginal confidence levels do not automatically provide simultaneous coverage.
C5. Construct an unbiased estimator of \(\mu ^{2}\) from \(\bar{Y}\) and \(s^{2}\) under i.i.d. finite-variance sampling. Is \(\bar{Y} ^{2}\) consistent?
Solution C5
\(\mathbb{E}[\bar{Y} ^{2}]=\mu ^{2}+\sigma ^{2}/n\) and \(\mathbb{E}[s^{2}]=\sigma ^{2}\), so \(\bar{Y} ^{2}-s^{2}/n\) is unbiased for \(\mu ^{2}\). Also \(\bar{Y} ^{2}\xrightarrow{p} \mu ^{2}\) by continuous mapping. The unbiased estimator can be negative in a realized sample even though \(\mu ^{2}\ge 0\).
C6. Suppose \(Y_{i}\) are i.i.d. normal with known \(\sigma =10\) and \(n=100. A\) one-sided 5% test compares \(H_{0}\):\(\mu =50\) with \(\mu >50\). Find its cutoff for \(\bar{Y}\) and its power at \(\mu =52\). What happens to power if the significance level is reduced to 1%?
Solution C6
\(\operatorname{SD}(\bar{Y} )=1\), so reject above 50+1.64485=51.64485. At \(\mu =52\), power\(=1-\Phi (51.64485-52)=\Phi (.35515)\approx .6388\). At 1%, the cutoff is 52.32635 and power\(=1-\Phi (.32635)\approx .3721\). Reducing Type I error raises the hurdle and lowers power at this fixed alternative.
D. Group comparisons and identification
D1. Two independent groups have (\(n\),mean\(,\operatorname{SD})=(100,12,4\)) and (80,10,5). Estimate their difference and construct a large-sample 95% interval. State which assumption allows variances to add.
Solution D1
Difference=2; \(\operatorname{SE}=\sqrt{16/100+25/80}=\sqrt{.4725}\approx .68739\). The interval is [ .6527,3.3473 ]. Independence between group samples sets covariance between their means to zero. Equal variances are unnecessary.
D2. For 64 independent pairs, the mean before–after change is 3 and the SD of the changes is 8. Compute a 95% interval. Does it identify a treatment effect?
Solution D2
\(\operatorname{SE}=8/8=1\); interval=[1.04,4.96]. This measures uncertainty about the mean change. Without a design that rules out other causes, practice and time effects can explain the change.
D3. Show why equal two-group sample sizes equate pooled and unpooled SEs. Does this prove an exact Student \(t\) null distribution?
Solution D3
At common size \(m, s_{p}^{2}=(s_{A}^{2}+s_{B}^{2})/2\). Multiply by \(2/m\) to get \(SE_{p}^{2}=(s_{A}^{2}+s_{B}^{2})/m\), equal to the unpooled variance estimate. Exact \(t\) still generally requires normality, independence, and common population variance.
E. Theorem and paper connections
E1. In \(Y=a+\)bX\(+U\), let \(\operatorname{Var}(X)=4, \operatorname{Cov}(X,U)=2\), and \(b=3\). What does \(\operatorname{Cov}(X,Y)/\operatorname{Var}(X)\) equal? What restrictions on \(Z\) would recover b through another covariance ratio?
Solution E1
The observed ratio is 3+2/4=3.5. If \(\operatorname{Cov}(Z,U)=0\) and \(\operatorname{Cov}(Z,X)\ne 0\), then \(\operatorname{Cov}(Z,Y)/\operatorname{Cov}(Z,X)=3. A\) numerical correlation with \(X\) alone does not establish the exclusion restriction.
E2. In the sports-collectible design, let the probability of receiving A be r rather than .5, independently of preferences. A fraction q prefers A. Derive the no-endowment-effect trading rate.
Solution E2
The trading rate is \(q(1-r)+(1-q)r\). At \(r=.5\) it equals .5 for every q. Otherwise it depends on preferences, so testing against .5 without checking assignment probabilities can be wrong.
E3. Why does resampling the sample maximum produce a mass at zero in the centered bootstrap statistic? What kind of error remains as \(B\to \infty\)?
Solution E3
The original maximum is included with probability \(1-(1-1/n)^{n}\); whenever included, the bootstrap maximum equals it. Increasing B estimates this bootstrap mass more accurately but does not remove the mismatch with the target continuous distribution.
E4. Use LLN, continuous mapping, and Slutsky to explain the sequence “estimate variance \(\to\) studentize \(\to\) use normal quantiles.” Identify where a zero population variance would cause trouble.
Solution E4
LLNs give \(s^{2}\xrightarrow{p} \sigma ^{2}\). For \(\sigma ^{2}>0\), continuous mapping yields \(\sigma /s\xrightarrow{p} 1\). The CLT gives \(\sqrt{n}(\bar{Y} -\mu )/\sigma \xrightarrow{d} \mathcal{N}(0,1)\); Slutsky gives the same limit after replacing \(\sigma\) by \(s\). At \(\sigma ^{2}=0\), standardization divides by zero and this argument is unavailable.
Exercise design draws on SW Exercises 2.4, 2.13, 2.15, 2.19–2.27; 3.5–3.6, 3.15–3.21; Schellenberg Section 2; and Princeton bootstrap p. 12. Questions are adapted or newly constructed. All solutions are supplied locally.