Appendix z — Solutions — Problem Set 1.4 (The LLN and the CLT)

Full worked solutions to book/weeks/week-01/ps1.4.md. Notation follows CONVENTIONS §3: the standardized sample mean is z_N=(\bar{x}-\mu)/(\sigma/\sqrt{N}), \xrightarrow{p} is convergence in probability, \xrightarrow{d} convergence in distribution. Normal-table values used: \Pr(Z>1.645)\approx 0.05, \Pr(|Z|>1.96)\approx 0.05, \Pr(Z>2.33)\approx 0.01, \Pr(Z<-2.33)\approx 0.01, plus \Pr(Z>0.5)\approx 0.309, \Pr(Z>2)\approx 0.0228, \Pr(Z>2.5)\approx 0.0062.

Problem 1 — Chebyshev forces the Weak Law (14 pts)

(a) (4 pts) The sample mean is unbiased and its variance falls like 1/N: \mathbb{E}[\bar{x}] = \mu = 42, \qquad \operatorname{Var}(\bar{x}) = \frac{\sigma^2}{N} = \frac{30^2}{N} = \frac{900}{N}. The fact from Ch 1.3 is that for N independent draws, averaging cancels noise so that \operatorname{Var}(\bar{x})=\sigma^2/N exactly (and \mathbb{E}[\bar{x}]=\mu).

(b) (5 pts) Chebyshev applied to the random variable \bar{x} (which has mean \mu and variance \sigma^2/N): \Pr\!\big(|\bar{x}-\mu|\ge \epsilon\big) \le \frac{\operatorname{Var}(\bar{x})}{\epsilon^2} = \frac{\sigma^2}{N\,\epsilon^2} = \frac{900}{N\cdot 2^2} = \frac{225}{N}. Set the bound \le 0.05: \frac{225}{N}\le 0.05 \quad\Longrightarrow\quad N \ge \frac{225}{0.05} = 4500. So Chebyshev guarantees the miss-probability is at most 0.05 once \boxed{N \ge 4500}.

(Remark for graders: 4500 is large precisely because Chebyshev is loose; the CLT-based answer would be far smaller. Full marks for N\ge 4500 with the correct bound.)

(c) (3 pts) The bound \sigma^2/(N\epsilon^2) has \epsilon and \sigma^2 fixed and N in the denominator, so it \to 0 as N\to\infty for any fixed \epsilon>0. A probability squeezed below a quantity that vanishes must itself vanish; that is exactly the statement \bar{x}\xrightarrow{p}\mu. Looseness is harmless here because we only need an upper bound that goes to zero, not a tight one — a crude bound that vanishes settles the limit just as decisively as a sharp one.

(d) (2 pts) The bound is 225/N; replacing N by 4N gives 225/(4N), so the bound shrinks by a factor of 4. To cut the miss-probability bound to a quarter you must quadruple the sample — the standard error scales like \sigma/\sqrt{N}, so halving uncertainty costs a fourfold increase in N.

Problem 2 — Standardize a sample mean and read the normal table (16 pts)

(a) (3 pts) \mathbb{E}[\bar{x}] = \mu = 120, \qquad \text{SE}(\bar{x}) = \frac{\sigma}{\sqrt{N}} = \frac{200}{\sqrt{400}} = \frac{200}{20} = 10.

(b) (4 pts) z_N = \frac{\bar{x}-\mu}{\sigma/\sqrt{N}} = \frac{\bar{x}-120}{10} \;\xrightarrow{d}\; N(0,1). By the CLT, z_N is approximately standard normal for large N. The skew of the raw policy costs does not block this: the CLT promises that the standardized average becomes normal regardless of the population’s shape, as long as the population has a finite mean and finite variance — both of which hold here. Skewness only slows convergence; it does not prevent it.

(c) (6 pts) Standardize the threshold \bar{x}=130: \Pr(\bar{x}>130) = \Pr\!\left(\frac{\bar{x}-120}{10} > \frac{130-120}{10}\right) = \Pr(Z > 1) \approx 0.159. So there is about a 16% chance the portfolio’s average monthly cost exceeds $130. (Using \Pr(Z>1)\approx 0.1587.)

(d) (3 pts) The colleague is wrong. The Central Limit Theorem concerns the average of the 400 policies, not a single policy. As long as the single-policy cost has a finite variance (it does — \sigma=200 is large but finite), the standardized average \bar{x} is approximately normal even though each individual cost is wildly skewed. Skew slows the bell’s arrival; with N=400 it has had ample opportunity to form.

Problem 3 — How large must N be for a skewed population? (18 pts)

(a) (4 pts) \text{SE}(\bar{x}) = \sigma/\sqrt{N} = 40/\sqrt{N}.

For SE =1: 40/\sqrt{N}=1 \Rightarrow \sqrt{N}=40 \Rightarrow N = 1600.
For SE =0.5: 40/\sqrt{N}=0.5 \Rightarrow \sqrt{N}=80 \Rightarrow N = 6400.

Note the fourfold jump (1600\to 6400) to halve the SE — the \sqrt{N} law again.

(b) (5 pts) We want \Pr(\bar{x}<0)\le 0.01. Standardize: \Pr(\bar{x}<0) = \Pr\!\left(\frac{\bar{x}-\mu}{\sigma/\sqrt{N}} < \frac{0-\mu}{\sigma/\sqrt{N}}\right) = \Pr\!\left(Z < \frac{-5}{40/\sqrt{N}}\right) = \Pr\!\left(Z < -\frac{5\sqrt{N}}{40}\right). We need \Pr(Z<-2.33)\approx 0.01, so require -\frac{5\sqrt{N}}{40} \le -2.33 \quad\Longrightarrow\quad \frac{5\sqrt{N}}{40}\ge 2.33 \quad\Longrightarrow\quad \sqrt{N} \ge \frac{2.33\cdot 40}{5} = 18.64. Thus N \ge 18.64^2 \approx 347.5, i.e. the smallest integer is \boxed{N = 348}.

(c) (5 pts) The formula in (a) uses only \sigma, and it remains arithmetically true even for the jackpot population (same \mu and \sigma would give the same SE). But the answer to (b) relied on the CLT having already kicked in — i.e., on z_N actually being approximately normal at N\approx 348. With one-in-a-thousand +\$5{,}000 jackpots, the population is savagely right-skewed: a single jackpot draw can dominate the whole sum, so the standardized mean is still far from normal at N=348 (too peaked, too heavy in the right tail), and the normal probability 0.01 is not trustworthy. The unstated work is being done by the “N\ge 30” folk rule, which quietly assumes the population is not too skewed or heavy-tailed. Here that assumption is false, so Sam likely needs N in the thousands or more.

(d) (4 pts) The standard error \sigma/\sqrt{N} controls how tight the cloud is, but the shape of the population controls how fast the standardized mean becomes normal. The more skewed or heavy-tailed the population, the larger N must be before z_N looks like N(0,1) and a normal-table probability can be trusted. Two populations with identical \mu and \sigma can demand wildly different N for the CLT to bite.

Problem 4 — Devon’s heavy-tails hierarchy (18 pts)

(a) (6 pts) Using “mean exists iff \nu>1, variance finite iff \nu>2”:

Model	\nu	Mean exists?	Variance finite?
A	5	Yes (5>1)	Yes (5>2)
B	2	Yes (2>1)	No (\nu=2 not >2; variance infinite)
C	1	No (\nu=1 not >1; Cauchy)	No (undefined)

(b) (6 pts)

Model A (\nu=5): Both LLN and CLT hold (and are guaranteed). Mean exists \Rightarrow LLN applies; variance finite \Rightarrow CLT applies. Convergence is slow (heavy tails) but it does occur.
Model B (\nu=2): LLN holds (the mean exists, so \bar{x}\xrightarrow{p}\mu=0), but the CLT fails: the variance is infinite, so there is no \sigma to standardize by and the classical CLT’s hypothesis is violated. z_N never settles onto N(0,1).
Model C (\nu=1, Cauchy): Both fail. The mean does not exist, so the LLN has nothing to converge to; with no mean there is certainly no finite variance, so the CLT fails too.

(c) (3 pts) For the Cauchy, the average of N i.i.d. draws has the same distribution as a single draw — averaging a million of them is no more informative than looking at one. This is why the chapter calls it the “canonical horror story”: the LLN itself fails, so “more data” is not a cure.

(d) (3 pts) Yes, his sample mean is eventually consistent and asymptotically normal: his returns have a finite (if very large) variance, which satisfies both the LLN and the classical CLT, so in the limit \bar{x}\xrightarrow{p}\mu and z_N\xrightarrow{d}N(0,1). But N=300 is not likely enough: tails heavier than \nu=5 make convergence very slow (the chapter notes you may need hundreds-to-thousands even for \nu=5), so any normal-table claim at N=300 is built on sand.

Problem 5 — Design a Monte Carlo to show CLT convergence (16 pts)

This is a design problem; full marks for a clear, implementable procedure. A reference design follows; the runnable version lives in nb1.4.

(a) (8 pts) A complete loop:

Population. Draw from the exponential with scale 1: \mu = 1 and \sigma = 1. It is strongly right-skewed (piled at zero, long right tail), so it is a good stress test for the CLT.
One replication. Draw a sample of size N from the exponential. Compute its sample mean \bar{x}. Standardize it: z_N = \frac{\bar{x}-\mu}{\sigma/\sqrt{N}} = \frac{\bar{x}-1}{1/\sqrt{N}} = (\bar{x}-1)\sqrt{N}. Store this single number z_N.
Many replications. Repeat step 2 a large number of times — e.g. \text{reps}=50{,}000 — to map out the sampling distribution of z_N. (More reps = smoother histogram.)
Plot. Histogram the 50{,}000 stored z_N values with density=True and overlay the standard-normal pdf N(0,1) as a reference curve on the same axes.

Pseudocode:

fix seed
for N in N_list:
    z_values = []
    repeat reps times:
        sample = draw N exponential(scale=1) values
        xbar   = mean(sample)
        z      = (xbar - 1) / (1/sqrt(N))
        append z to z_values
    histogram(z_values, density=True)
    overlay standard normal pdf

(b) (4 pts) Sweep N \in \{1, 2, 5, 30, 100\}. Expected appearance vs. the overlaid N(0,1):

N=1: looks like the raw exponential — sharp left wall, long right tail; not bell-like.
N=2: wall softens, peak pulled toward center, still lopsided.
N=5: skew fading, a bell trying to form.
N=30: histogram and normal curve nearly coincide.
N=100: indistinguishable from the bell.

Smallest N where the bell is indistinguishable by eye: about N=30, matching the “N\ge 30” folk rule, which holds for a mildly skewed population like the exponential.

(c) (4 pts) A quantitative diagnostic (any one): - Sample skewness and excess kurtosis of the stored z_N values: both should head toward 0 as N grows (standard normal has skew 0, excess kurtosis 0). Reporting them as numbers turns “looks bell-shaped” into a measurement. - Alternatively: a Q–Q plot of z_N against the standard normal — points hugging the 45^\circ line indicate normality, and systematic departure in the tails flags slow convergence.

Either is acceptable with a one-sentence justification.

Problem 6 — Aggregating many small returns (18 pts)

(a) (4 pts) For N=100: \mathbb{E}[\bar{r}] = \mu = 0.0004, \qquad \text{SD}(\bar{r}) = \frac{\sigma}{\sqrt{N}} = \frac{0.02}{\sqrt{100}} = \frac{0.02}{10} = 0.002. Diversification (averaging N independent assets) shrinks volatility from 0.02 to 0.002 — a factor of \sqrt{N}=10 — because averaging cancels noise (the LLN force: \operatorname{Var}(\bar{r})=\sigma^2/N\to 0), but the expected return is the average of identical means and stays at \mu. The LLN drives \bar{r} toward \mu; it does not change \mu.

(b) (6 pts) Standardize the threshold \bar{r}=0: \Pr(\bar{r}<0) = \Pr\!\left(\frac{\bar{r}-\mu}{\sigma/\sqrt{N}} < \frac{0-0.0004}{0.002}\right) = \Pr(Z < -0.2). From the table \Pr(Z<-0.2)\approx 0.421. So there is about a 42% chance of a negative-return day. (The edge \mu is tiny relative to the daily SE, so a losing day is nearly a coin flip — only slightly better than even.)

(c) (4 pts) - Dependence (assets move together): correlation breaks the independence assumption behind \operatorname{Var}(\bar{r})=\sigma^2/N. When returns reinforce rather than cancel, the variance of the average falls slower than 1/N (in the extreme, not at all), so the true SE is larger than 0.002 and the “\sigma/\sqrt{N}” rate is wrong. (This is why time-series/clustered standard errors exist — Week 2.) - Fat tails: even if independent, heavy tails mean a single monster day dominates the average, so the normal approximation in (b) fails — \bar{r}’s standardized distribution is too peaked and too heavy in the tails at N=100, and the \Pr(Z<-0.2) reading is untrustworthy. The mechanism is the CLT’s “no single contribution dominates” clause being violated.

(d) (4 pts) The LLN promises that as N\to\infty, \bar{r}\xrightarrow{p}\mu — with enough independent assets the portfolio’s average return collapses onto the true mean 0.0004. The CLT adds the shape: rescaled by the SE \sigma/\sqrt{N}, the misses of \bar{r} around \mu follow the standard normal bell. It is the CLT that licenses the calibrated probability statement in (b): without knowing the shape of the cloud around \mu, Sam could say where \bar{r} lands but never attach a number like “42%” to how often it dips below zero.

End of solutions for PS 1.4. Cross-references: Ch 1.3 (\operatorname{Var}(\bar{x})=\sigma^2/N), Ch 1.4 (LLN, Chebyshev, CLT, fat-tail hierarchy), nb1.4 (simulations for Problem 5). Looks ahead to Ch 1.5, where \sigma is replaced by \hat\sigma and the normal becomes Student’s t.

# Solutions — Problem Set 1.4 (The LLN and the CLT) *Full worked solutions to `book/weeks/week-01/ps1.4.md`. Notation follows CONVENTIONS §3: the standardized sample mean is $z_N=(\bar{x}-\mu)/(\sigma/\sqrt{N})$, $\xrightarrow{p}$ is convergence in probability, $\xrightarrow{d}$ convergence in distribution. Normal-table values used: $\Pr(Z>1.645)\approx 0.05$, $\Pr(|Z|>1.96)\approx 0.05$, $\Pr(Z>2.33)\approx 0.01$, $\Pr(Z<-2.33)\approx 0.01$, plus $\Pr(Z>0.5)\approx 0.309$, $\Pr(Z>2)\approx 0.0228$, $\Pr(Z>2.5)\approx 0.0062$.* --- ## Problem 1 — Chebyshev forces the Weak Law (14 pts) **(a) (4 pts)** The sample mean is unbiased and its variance falls like $1/N$: $$ \mathbb{E}[\bar{x}] = \mu = 42, \qquad \operatorname{Var}(\bar{x}) = \frac{\sigma^2}{N} = \frac{30^2}{N} = \frac{900}{N}. $$ The fact from Ch 1.3 is that for $N$ **independent** draws, averaging cancels noise so that $\operatorname{Var}(\bar{x})=\sigma^2/N$ exactly (and $\mathbb{E}[\bar{x}]=\mu$). **(b) (5 pts)** Chebyshev applied to the random variable $\bar{x}$ (which has mean $\mu$ and variance $\sigma^2/N$): $$ \Pr\!\big(|\bar{x}-\mu|\ge \epsilon\big) \le \frac{\operatorname{Var}(\bar{x})}{\epsilon^2} = \frac{\sigma^2}{N\,\epsilon^2} = \frac{900}{N\cdot 2^2} = \frac{225}{N}. $$ Set the bound $\le 0.05$: $$ \frac{225}{N}\le 0.05 \quad\Longrightarrow\quad N \ge \frac{225}{0.05} = 4500. $$ So Chebyshev *guarantees* the miss-probability is at most $0.05$ once $\boxed{N \ge 4500}$. *(Remark for graders: $4500$ is large precisely because Chebyshev is loose; the CLT-based answer would be far smaller. Full marks for $N\ge 4500$ with the correct bound.)* **(c) (3 pts)** The bound $\sigma^2/(N\epsilon^2)$ has $\epsilon$ and $\sigma^2$ fixed and $N$ in the denominator, so it $\to 0$ as $N\to\infty$ for *any* fixed $\epsilon>0$. A probability squeezed below a quantity that vanishes must itself vanish; that is exactly the statement $\bar{x}\xrightarrow{p}\mu$. Looseness is harmless here because we only need an upper bound that goes to zero, not a tight one — a crude bound that vanishes settles the limit just as decisively as a sharp one. **(d) (2 pts)** The bound is $225/N$; replacing $N$ by $4N$ gives $225/(4N)$, so the bound shrinks by a **factor of 4**. To cut the miss-probability bound to a quarter you must quadruple the sample — the standard error scales like $\sigma/\sqrt{N}$, so halving uncertainty costs a fourfold increase in $N$. --- ## Problem 2 — Standardize a sample mean and read the normal table (16 pts) **(a) (3 pts)** $$ \mathbb{E}[\bar{x}] = \mu = 120, \qquad \text{SE}(\bar{x}) = \frac{\sigma}{\sqrt{N}} = \frac{200}{\sqrt{400}} = \frac{200}{20} = 10. $$ **(b) (4 pts)** $$ z_N = \frac{\bar{x}-\mu}{\sigma/\sqrt{N}} = \frac{\bar{x}-120}{10} \;\xrightarrow{d}\; N(0,1). $$ By the CLT, $z_N$ is approximately standard normal for large $N$. The skew of the raw policy costs does **not** block this: the CLT promises that the standardized *average* becomes normal regardless of the population's shape, as long as the population has a **finite mean and finite variance** — both of which hold here. Skewness only slows convergence; it does not prevent it. **(c) (6 pts)** Standardize the threshold $\bar{x}=130$: $$ \Pr(\bar{x}>130) = \Pr\!\left(\frac{\bar{x}-120}{10} > \frac{130-120}{10}\right) = \Pr(Z > 1) \approx 0.159. $$ So there is about a **16% chance** the portfolio's average monthly cost exceeds \$130. (Using $\Pr(Z>1)\approx 0.1587$.) **(d) (3 pts)** The colleague is wrong. The **Central Limit Theorem** concerns the *average* of the 400 policies, not a single policy. As long as the single-policy cost has a **finite variance** (it does — $\sigma=200$ is large but finite), the standardized average $\bar{x}$ is approximately normal even though each individual cost is wildly skewed. Skew slows the bell's arrival; with $N=400$ it has had ample opportunity to form. --- ## Problem 3 — How large must $N$ be for a skewed population? (18 pts) **(a) (4 pts)** $\text{SE}(\bar{x}) = \sigma/\sqrt{N} = 40/\sqrt{N}$. - For SE $=1$: $40/\sqrt{N}=1 \Rightarrow \sqrt{N}=40 \Rightarrow N = 1600$. - For SE $=0.5$: $40/\sqrt{N}=0.5 \Rightarrow \sqrt{N}=80 \Rightarrow N = 6400$. Note the fourfold jump ($1600\to 6400$) to halve the SE — the $\sqrt{N}$ law again. **(b) (5 pts)** We want $\Pr(\bar{x}<0)\le 0.01$. Standardize: $$ \Pr(\bar{x}<0) = \Pr\!\left(\frac{\bar{x}-\mu}{\sigma/\sqrt{N}} < \frac{0-\mu}{\sigma/\sqrt{N}}\right) = \Pr\!\left(Z < \frac{-5}{40/\sqrt{N}}\right) = \Pr\!\left(Z < -\frac{5\sqrt{N}}{40}\right). $$ We need $\Pr(Z<-2.33)\approx 0.01$, so require $$ -\frac{5\sqrt{N}}{40} \le -2.33 \quad\Longrightarrow\quad \frac{5\sqrt{N}}{40}\ge 2.33 \quad\Longrightarrow\quad \sqrt{N} \ge \frac{2.33\cdot 40}{5} = 18.64. $$ Thus $N \ge 18.64^2 \approx 347.5$, i.e. the smallest integer is $\boxed{N = 348}$. **(c) (5 pts)** The formula in (a) uses only $\sigma$, and it remains *arithmetically* true even for the jackpot population (same $\mu$ and $\sigma$ would give the same SE). But the answer to (b) relied on the **CLT having already kicked in** — i.e., on $z_N$ actually being approximately normal at $N\approx 348$. With one-in-a-thousand $+\$5{,}000$ jackpots, the population is *savagely* right-skewed: a single jackpot draw can dominate the whole sum, so the standardized mean is still far from normal at $N=348$ (too peaked, too heavy in the right tail), and the normal probability $0.01$ is not trustworthy. The unstated work is being done by the **"$N\ge 30$" folk rule**, which quietly assumes the population is not too skewed or heavy-tailed. Here that assumption is false, so Sam likely needs $N$ in the thousands or more. **(d) (4 pts)** The standard error $\sigma/\sqrt{N}$ controls how *tight* the cloud is, but the **shape** of the population controls how fast the standardized mean becomes *normal*. The more skewed or heavy-tailed the population, the larger $N$ must be before $z_N$ looks like $N(0,1)$ and a normal-table probability can be trusted. Two populations with identical $\mu$ and $\sigma$ can demand wildly different $N$ for the CLT to bite. --- ## Problem 4 — Devon's heavy-tails hierarchy (18 pts) **(a) (6 pts)** Using "mean exists iff $\nu>1$, variance finite iff $\nu>2$": | Model | $\nu$ | Mean exists? | Variance finite? | |-------|------|--------------|------------------| | A | 5 | Yes ($5>1$) | Yes ($5>2$) | | B | 2 | Yes ($2>1$) | **No** ($\nu=2$ not $>2$; variance infinite) | | C | 1 | **No** ($\nu=1$ not $>1$; Cauchy) | No (undefined) | **(b) (6 pts)** - **Model A ($\nu=5$):** Both **LLN and CLT hold** (and are guaranteed). Mean exists $\Rightarrow$ LLN applies; variance finite $\Rightarrow$ CLT applies. Convergence is *slow* (heavy tails) but it does occur. - **Model B ($\nu=2$):** **LLN holds** (the mean exists, so $\bar{x}\xrightarrow{p}\mu=0$), but the **CLT fails**: the variance is infinite, so there is no $\sigma$ to standardize by and the classical CLT's hypothesis is violated. $z_N$ never settles onto $N(0,1)$. - **Model C ($\nu=1$, Cauchy):** **Both fail.** The mean does not exist, so the LLN has nothing to converge to; with no mean there is certainly no finite variance, so the CLT fails too. **(c) (3 pts)** For the Cauchy, the average of $N$ i.i.d. draws has the **same distribution as a single draw** — averaging a million of them is no more informative than looking at one. This is why the chapter calls it the **"canonical horror story"**: the LLN itself fails, so "more data" is not a cure. **(d) (3 pts)** **Yes**, his sample mean is *eventually* consistent and asymptotically normal: his returns have a finite (if very large) variance, which satisfies both the LLN and the classical CLT, so in the limit $\bar{x}\xrightarrow{p}\mu$ and $z_N\xrightarrow{d}N(0,1)$. But **$N=300$ is not likely enough**: tails heavier than $\nu=5$ make convergence very slow (the chapter notes you may need hundreds-to-thousands even for $\nu=5$), so any normal-table claim at $N=300$ is built on sand. --- ## Problem 5 — Design a Monte Carlo to show CLT convergence (16 pts) *This is a design problem; full marks for a clear, implementable procedure. A reference design follows; the runnable version lives in **nb1.4**.* **(a) (8 pts)** A complete loop: 1. **Population.** Draw from the exponential with scale $1$: $\mu = 1$ and $\sigma = 1$. It is strongly right-skewed (piled at zero, long right tail), so it is a good stress test for the CLT. 2. **One replication.** Draw a sample of size $N$ from the exponential. Compute its sample mean $\bar{x}$. Standardize it: $$ z_N = \frac{\bar{x}-\mu}{\sigma/\sqrt{N}} = \frac{\bar{x}-1}{1/\sqrt{N}} = (\bar{x}-1)\sqrt{N}. $$ Store this single number $z_N$. 3. **Many replications.** Repeat step 2 a large number of times — e.g. $\text{reps}=50{,}000$ — to map out the sampling distribution of $z_N$. (More reps = smoother histogram.) 4. **Plot.** Histogram the $50{,}000$ stored $z_N$ values with `density=True` and overlay the standard-normal pdf $N(0,1)$ as a reference curve on the same axes. Pseudocode: ``` fix seed for N in N_list: z_values = [] repeat reps times: sample = draw N exponential(scale=1) values xbar = mean(sample) z = (xbar - 1) / (1/sqrt(N)) append z to z_values histogram(z_values, density=True) overlay standard normal pdf ``` **(b) (4 pts)** Sweep $N \in \{1, 2, 5, 30, 100\}$. Expected appearance vs. the overlaid $N(0,1)$: - $N=1$: looks like the raw exponential — sharp left wall, long right tail; not bell-like. - $N=2$: wall softens, peak pulled toward center, still lopsided. - $N=5$: skew fading, a bell trying to form. - $N=30$: histogram and normal curve nearly coincide. - $N=100$: indistinguishable from the bell. Smallest $N$ where the bell is indistinguishable by eye: about **$N=30$**, matching the "$N\ge 30$" folk rule, which holds for a *mildly* skewed population like the exponential. **(c) (4 pts)** A quantitative diagnostic (any one): - **Sample skewness and excess kurtosis** of the stored $z_N$ values: both should head toward $0$ as $N$ grows (standard normal has skew $0$, excess kurtosis $0$). Reporting them as numbers turns "looks bell-shaped" into a measurement. - *Alternatively:* a **Q–Q plot** of $z_N$ against the standard normal — points hugging the $45^\circ$ line indicate normality, and systematic departure in the tails flags slow convergence. Either is acceptable with a one-sentence justification. --- ## Problem 6 — Aggregating many small returns (18 pts) **(a) (4 pts)** For $N=100$: $$ \mathbb{E}[\bar{r}] = \mu = 0.0004, \qquad \text{SD}(\bar{r}) = \frac{\sigma}{\sqrt{N}} = \frac{0.02}{\sqrt{100}} = \frac{0.02}{10} = 0.002. $$ Diversification (averaging $N$ independent assets) shrinks volatility from $0.02$ to $0.002$ — a factor of $\sqrt{N}=10$ — because **averaging cancels noise** (the LLN force: $\operatorname{Var}(\bar{r})=\sigma^2/N\to 0$), but the *expected* return is the average of identical means and stays at $\mu$. The LLN drives $\bar{r}$ toward $\mu$; it does not change $\mu$. **(b) (6 pts)** Standardize the threshold $\bar{r}=0$: $$ \Pr(\bar{r}<0) = \Pr\!\left(\frac{\bar{r}-\mu}{\sigma/\sqrt{N}} < \frac{0-0.0004}{0.002}\right) = \Pr(Z < -0.2). $$ From the table $\Pr(Z<-0.2)\approx 0.421$. So there is about a **42% chance** of a negative-return day. (The edge $\mu$ is tiny relative to the daily SE, so a losing day is nearly a coin flip — only slightly better than even.) **(c) (4 pts)** - **Dependence (assets move together):** correlation breaks the independence assumption behind $\operatorname{Var}(\bar{r})=\sigma^2/N$. When returns reinforce rather than cancel, the variance of the average falls *slower* than $1/N$ (in the extreme, not at all), so the true SE is **larger than $0.002$** and the "$\sigma/\sqrt{N}$" rate is wrong. *(This is why time-series/clustered standard errors exist — Week 2.)* - **Fat tails:** even if independent, heavy tails mean a single monster day dominates the average, so the **normal approximation** in (b) fails — $\bar{r}$'s standardized distribution is too peaked and too heavy in the tails at $N=100$, and the $\Pr(Z<-0.2)$ reading is untrustworthy. The mechanism is the CLT's "no single contribution dominates" clause being violated. **(d) (4 pts)** The **LLN** promises that as $N\to\infty$, $\bar{r}\xrightarrow{p}\mu$ — with enough independent assets the portfolio's average return collapses onto the true mean $0.0004$. The **CLT** adds the *shape*: rescaled by the SE $\sigma/\sqrt{N}$, the misses of $\bar{r}$ around $\mu$ follow the standard normal bell. It is the **CLT** that licenses the calibrated probability statement in (b): without knowing the shape of the cloud around $\mu$, Sam could say where $\bar{r}$ lands but never attach a number like "42%" to how often it dips below zero. --- *End of solutions for PS 1.4. Cross-references: Ch 1.3 ($\operatorname{Var}(\bar{x})=\sigma^2/N$), Ch 1.4 (LLN, Chebyshev, CLT, fat-tail hierarchy), nb1.4 (simulations for Problem 5). Looks ahead to Ch 1.5, where $\sigma$ is replaced by $\hat\sigma$ and the normal becomes Student's $t$.*