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Q.No.4. State and prove weak law of large numbers

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Q.No.4. State and prove weak law of large numbers

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Solution

Sure, let's break down the Weak Law of Large Numbers (WLLN) and its proof.

The Weak Law of Large Numbers is a fundamental theorem in probability theory that describes the result of performing the same experiment a large number of times. It states that if you have a sequence of independent and identically distributed random variables, the average of these variables tends to the expected value as the number of variables goes to infinity.

Here's an analogy: Imagine you're flipping a fair coin. The expected value of a single flip (where Heads is 1 and Tails is 0) is 0.5. If you flip the coin only a few times, you might not get an average of 0.5. But if you flip the coin many, many times, the average of your flips will get closer and closer to 0.5. That's the essence of the Weak Law of Large Numbers.

Now, let's prove the Weak Law of Large Numbers. We'll use Chebyshev's inequality, which is a tool that gives us bounds on the probability that a random variable deviates from its expected value.

Let X1, X2, ..., Xn be independent and identically distributed random variables with expected value μ and variance σ^2. Let Sn = X1 + X2 + ... + Xn be the sum of these variables, and let An = Sn/n be the average.

We want to show that for any positive number ε, the probability that An deviates from μ by more than ε goes to zero as n goes to infinity. In mathematical terms, we want to show that P(|An - μ| > ε) -> 0 as n -> ∞.

By Chebyshev's inequality, we have P(|An - μ| > ε) ≤ Var(An) / ε^2. Since An is the average of the X's, its variance is Var(X1)/n = σ^2/n. So we have P(|An - μ| > ε) ≤ σ^2 / (nε^2).

As n goes to infinity, σ^2 / (nε^2) goes to zero (since the numerator is constant and the denominator is increasing). Therefore, P(|An - μ| > ε) -> 0 as n -> ∞, which is exactly what we wanted to show.

So, we've proven the Weak Law of Large Numbers: as the number of independent and identically distributed random variables increases, the average of these variables gets closer and closer to their expected value.

This problem has been solved

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