Math Problem Statement

If the variables Xi assume the values 2k-2log(k) with probabilities 2-k , k=1, 2, 3…, i=1, 2, …, examine if the law of large numbers holds for the sequence {xi} of independent and identically distributed random variables.

Solution

To determine whether the Law of Large Numbers (LLN) holds for the sequence {Xi}\{X_i\} of independent and identically distributed (i.i.d.) random variables, we need to evaluate whether the expected value E[Xi]\mathbb{E}[X_i] exists and is finite. The Law of Large Numbers states that the sample average converges to the expected value as the number of observations increases, provided that the expected value exists and is finite.

Given the random variables XiX_i with the distribution defined as:

Xi=2k2log(k)with probability2k,k=1,2,3,X_i = 2k - 2\log(k) \quad \text{with probability} \quad 2^{-k}, \quad k = 1, 2, 3, \dots

Step 1: Calculate the Expected Value E[Xi]\mathbb{E}[X_i]

The expected value E[Xi]\mathbb{E}[X_i] is calculated as:

E[Xi]=k=1(2k2log(k))2k\mathbb{E}[X_i] = \sum_{k=1}^{\infty} \left(2k - 2\log(k)\right) \cdot 2^{-k}

This expression can be split into two separate sums:

E[Xi]=2k=1k2k2k=1log(k)2k\mathbb{E}[X_i] = 2\sum_{k=1}^{\infty} k \cdot 2^{-k} - 2\sum_{k=1}^{\infty} \log(k) \cdot 2^{-k}

Step 2: Evaluate the First Sum

The first sum is:

k=1k2k\sum_{k=1}^{\infty} k \cdot 2^{-k}

This is a known sum, and its result is:

k=1k2k=2\sum_{k=1}^{\infty} k \cdot 2^{-k} = 2

Thus, the first part of the expected value becomes:

22=42 \cdot 2 = 4

Step 3: Evaluate the Second Sum

The second sum is more complex:

k=1log(k)2k\sum_{k=1}^{\infty} \log(k) \cdot 2^{-k}

To evaluate this sum, we recognize that it's related to a series that involves the logarithm weighted by an exponentially decreasing probability. This sum converges to a finite value, but its exact evaluation typically requires advanced methods, such as generating functions or numerical approximation. However, the important point is whether this sum converges, not necessarily its exact value.

Given that the series:

k=1log(k)2k\sum_{k=1}^{\infty} \log(k) \cdot 2^{-k}

does converge (as it behaves similarly to the sum of a convergent series), the second part contributes a finite value to the expected value.

Step 4: Conclusion on the Expected Value

Since both parts of the expected value sum are finite, we conclude that:

E[Xi]=42(finite value) is finite.\mathbb{E}[X_i] = 4 - 2 \cdot \text{(finite value)} \text{ is finite.}

Step 5: Applying the Law of Large Numbers

Since E[Xi]\mathbb{E}[X_i] is finite, the Law of Large Numbers does indeed apply to the sequence {Xi}\{X_i\}. Therefore, the sample average of the XiX_i will converge to the expected value E[Xi]\mathbb{E}[X_i] as the number of observations increases.

Summary

The Law of Large Numbers holds for the sequence {Xi}\{X_i\} because the expected value E[Xi]\mathbb{E}[X_i] is finite.

Would you like more details on the sum convergence or have any other questions?

Follow-up Questions

  1. How is the convergence of a series evaluated in probability theory?
  2. What is the significance of the Law of Large Numbers in statistics?
  3. Can you provide an example where the Law of Large Numbers fails?
  4. How does the Central Limit Theorem relate to the Law of Large Numbers?
  5. What are some practical applications of the Law of Large Numbers?

Tip

When evaluating the expected value of a random variable, always ensure that the sum converges, as the existence of a finite expected value is key to applying many important statistical theorems.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Expected Value
Convergence of Series

Formulas

-

Theorems

Law of Large Numbers

Suitable Grade Level

Advanced Undergraduate