Comprehensive Exercises on the Law of Large Numbers and the Central Limit Theorem

Approach: Use Chebyshev’s inequality.

Steps:

1. Chebyshev’s inequality: $P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}$
2. Here $\mu = 0$, $\sigma^2 = 4$, and $\varepsilon = 3$
3. $P(|X| \geq 3) \leq \frac{4}{9} \approx 0.444$

Answer: $\frac{4}{9}$

Approach: Explain the intuitive meaning of Bernoulli’s law of large numbers.

Steps:

1. Bernoulli’s law of large numbers shows that over a large number of independent repeated trials, the frequency of occurrence of an event converges in probability to its probability
2. This means that when the number of trials is sufficiently large, the frequency of occurrence of the event is very close to its true probability
3. This provides the theoretical foundation for the statistical definition of probability

Answer: over a large number of independent repeated trials, the frequency of occurrence of an event tends to the probability

Approach: Write the central limit theorem for the sample mean.

Steps:

1. Sample mean: $\overline{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$
2. Standardization: $\frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}}$
3. Central limit theorem: $\frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}} \xrightarrow{d} N(0,1)$

Answer: $\frac{\overline{X}_n - \mu}{\sigma/\sqrt{n}} \xrightarrow{d} N(0,1)$

Approach: Use the De Moivre-Laplace theorem.

Steps:

1. The expectation of the binomial distribution $B(n,p)$: $E(X) = np$
2. The variance of the binomial distribution $B(n,p)$: $D(X) = np(1-p)$
3. By the De Moivre-Laplace theorem, $B(n,p) \approx N(np, np(1-p))$

Answer: $B(n,p) \approx N(np, np(1-p))$

Approach: Compare the differences between the law of large numbers and the central limit theorem.

Steps:

1. Law of large numbers: focuses on the sample mean converging to the expectation, and only states convergence
2. Central limit theorem: focuses on the distribution of the sample mean tending to a normal distribution, and describes the shape of the distribution
3. Law of large numbers: is the foundation of the central limit theorem
4. Central limit theorem: provides more precise information than the law of large numbers

Answer: the law of large numbers focuses on the sample mean converging to the expectation, whereas the central limit theorem focuses on the distribution of the sum tending to a normal distribution

Approach: First find an upper bound for $P(|X - 5| \geq 3)$, then use the relation of complementary events.

Steps:

1. $P(|X - 5| \geq 3) \leq \frac{9}{9} = 1$
2. $P(|X - 5| < 3) \geq 1 - 1 = 0$
3. Since $2 < X < 8$ is equivalent to $|X - 5| < 3$, we get $P(2 < X < 8) \geq 0$

Answer: $0$

Approach: Prove it using Chebyshev’s inequality.

Steps:

1. $E(\overline{X}_n) = \mu$
2. $D(\overline{X}_n) = \frac{\sigma^2}{n}$
3. By Chebyshev’s inequality: $P(|\overline{X}_n - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{n\varepsilon^2}$
4. As $n \to \infty$, $\frac{\sigma^2}{n\varepsilon^2} \to 0$
5. Hence $\overline{X}_n \xrightarrow{P} \mu$

Answer: the proof is complete

Approach: Use the De Moivre-Laplace theorem for the approximation.

Steps:

1. $E(X) = 100 \times 0.3 = 30$
2. $D(X) = 100 \times 0.3 \times 0.7 = 21$
3. $\sigma = \sqrt{21} \approx 4.58$
4. $P(25 \leq X \leq 35) \approx P(\frac{25-30}{4.58} \leq Z \leq \frac{35-30}{4.58})$
5. $= P(-1.09 \leq Z \leq 1.09) = 2\Phi(1.09) - 1 \approx 0.724$

Answer: approximately $0.724$

Approach: Use the properties of expectation and variance.

Steps:

1. $E(\overline{X}_n) = E(\frac{1}{n}\sum_{i=1}^n X_i) = \frac{1}{n}\sum_{i=1}^n E(X_i) = \mu$
2. $D(\overline{X}_n) = D(\frac{1}{n}\sum_{i=1}^n X_i) = \frac{1}{n^2}\sum_{i=1}^n D(X_i) = \frac{\sigma^2}{n}$

Answer: $E(\overline{X}_n) = \mu$, $D(\overline{X}_n) = \frac{\sigma^2}{n}$

Approach: Compute the actual probability of the normal distribution and compare it with the upper bound given by Chebyshev’s inequality.

Steps:

1. For the normal distribution, $P(|X - \mu| \geq 2\sigma) = 2(1 - \Phi(2)) \approx 0.0455$
2. Chebyshev’s inequality gives: $P(|X - \mu| \geq 2\sigma) \leq \frac{1}{4} = 0.25$
3. The actual probability $0.0455$ is far smaller than the upper bound $0.25$, showing that Chebyshev’s inequality is fairly conservative

Answer: the upper bound given by Chebyshev’s inequality is much larger than the actual probability, showing that it is fairly conservative

Approach: Use the central limit theorem for the sample mean.

Steps:

1. Sample mean: $\overline{X}_n = \frac{1}{n}\sum_{i=1}^n X_i$
2. By the central limit theorem, $\overline{X}_n \xrightarrow{d} N(\mu, \frac{\sigma^2}{n})$
3. That is, the asymptotic distribution of $\overline{X}_n$ is $N(\mu, \frac{\sigma^2}{n})$

Answer: $N(\mu, \frac{\sigma^2}{n})$

Approach: Use the Lindeberg-Lévy theorem.

Steps:

1. Let $S_n = \sum_{i=1}^n X_i$
2. $E(S_n) = n\mu$, $D(S_n) = n\sigma^2$
3. Standardization: $\frac{S_n - n\mu}{\sigma\sqrt{n}}$
4. By the Lindeberg-Lévy theorem, it tends to $N(0,1)$

Answer: the proof is complete

Approach: Apply Chebyshev’s inequality directly.

Steps:

1. $P(|X - 10| \geq 8) \leq \frac{16}{64} = \frac{1}{4}$
2. Substituting into the general form of Chebyshev’s inequality $P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}$, with $k\sigma = 8$ and $k^2 = 4$, the upper bound is $\frac{1}{4}$.

Answer: $P(|X - 10| \geq 8) \leq \frac{1}{4}$, that is, the probability of this event occurring is at most one quarter.

Approach: Use Chebyshev’s inequality with $\varepsilon = k\sigma$.

Steps:

1. Chebyshev’s inequality: $P(|X - \mu| \geq \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}$
2. Setting $\varepsilon = k\sigma$ gives: $P(|X - \mu| \geq k\sigma) \leq \frac{\sigma^2}{(k\sigma)^2} = \frac{1}{k^2}$

Answer: the proof is complete

Approach: Use the De Moivre-Laplace theorem for the approximation.

Steps:

1. $E(X) = 200 \times 0.4 = 80$
2. $D(X) = 200 \times 0.4 \times 0.6 = 48$
3. $\sigma = \sqrt{48} \approx 6.93$
4. $P(70 \leq X \leq 90) \approx P(\frac{70-80}{6.93} \leq Z \leq \frac{90-80}{6.93})$
5. $= P(-1.44 \leq Z \leq 1.44) = 2\Phi(1.44) - 1 \approx 0.850$

Answer: approximately $0.850$