Derivation and Proof of Chebyshev's Inequality

Chebyshev's inequality states that for a random variable X with finite expected value μ and finite non-zero variance σ²:

P(|X - μ| ≥ kσ) ≤ 1/k² for any real number k > 0

Derivation and Proof:

Step 1: Define an indicator function

Let's define an indicator function I:

I = 1 if |X - μ| ≥ kσ

I = 0 if |X - μ| < kσ

Step 2: Express the probability using the indicator function

P(|X - μ| ≥ kσ) = E[I]

Step 3: Use the properties of expectation

E[(X - μ)²] ≥ E[(X - μ)² · I]

This is true because (X - μ)² is always non-negative, and we're only considering the cases where I = 1.

Step 4: Apply the definition of I

E[(X - μ)² · I] ≥ E[(kσ)² · I] = k²σ² · E[I]

This is because when I = 1, we know that |X - μ| ≥ kσ, so (X - μ)² ≥ (kσ)².

Step 5: Use the definition of variance

E[(X - μ)²] = σ²

Step 6: Combine the inequalities

σ² ≥ k²σ² · E[I]

Step 7: Solve for E[I]

E[I] ≤ 1/k²

Step 8: Conclusion

Since E[I] = P(|X - μ| ≥ kσ), we have proven Chebyshev's inequality:

P(|X - μ| ≥ kσ) ≤ 1/k²

This completes the derivation and proof of Chebyshev's inequality.