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Finite IID sample-mean variance and Chebyshev bound

Statement

For a finite repeated-observation model of Repeated-observation sampling models with IID and finite-variance hypotheses with $n \geq 1$ , write its variables as $Y_{1}, \dots, Y_{n}$ , where $Y_{i}$ here denotes the supplier's $Y_{i - 1}$ for $1 \leq i \leq n$ . They are IID square-integrable real random variables on one probability space, with common mean $μ$ and variance $σ^{2}$ . Put $\overline{Y} = n^{- 1} \sum_{k = 1}^{n} Y_{k}$ . Then $E [\overline{Y}] = μ, E ∣ \overline{Y} - μ ∣^{2} = \frac{σ^{2}}{n}, P (∣ \overline{Y} - μ ∣ \geq a) \leq \frac{σ^{2}}{n a^{2}} (a > 0) .$ No extension to an infinite IID sequence is required; no convergence assertion is made for a single finite family.

Facts & Assumptions

Given: The finite IID family and hypotheses in the Statement.

[L1]

The finite sampling model adopts IID and finite second moment, and its variables have common mean and variance. (Repeated-observation sampling models with IID and finite-variance hypotheses)

[L2]

For independent integrable variables, expectations of products factor; for the identity functions, $E [Y_{i} Y_{j}] = E [Y_{i}] E [Y_{j}]$ at distinct indices. (Expectations factor over finite products of independent random variables)

[L3]

Covariance is bilinear on finite square-integrable combinations and $Cov (V, W) = E [V W] - E [V] E [W]$ . (Variance and covariance identities for random variables)

[L4]

Expectations of finite integrable linear combinations are linear. (Linearity, monotonicity, and the modulus bound for expectation)

[L5]

If $Z$ is square-integrable and $a > 0$ , then $P (∣ Z - E [Z] ∣ \geq a) \leq Var (Z) / a^{2}$ . (Chebyshev's inequality for random variables)

Proof

1.1givenL1L4algebra

Square-integrability makes each $Y_{k}$ integrable by $2 ∣ Y_{k} ∣ \leq 1 + Y_{k}^{2}$ , and the finite mean is square-integrable by $(\sum_{k} y_{k})^{2} \leq n \sum_{k} y_{k}^{2}$ . Therefore [L4] gives $E [\overline{Y}] = n^{- 1} \sum_{k} μ = μ$ .

2.1givenstep 1.1L2L3algebra

For $i \neq j$ , mutual independence and [L2] give $E [Y_{i} Y_{j}] = μ^{2}$ , hence $Cov (Y_{i}, Y_{j}) = 0$ by [L3]. Bilinearity gives $Var (\overline{Y}) = n^{- 2} \sum_{i, j} Cov (Y_{i}, Y_{j}) = n^{- 2} \sum_{i = 1}^{n} σ^{2} = σ^{2} / n$ . Together with step 1.1 this is $E ∣ \overline{Y} - μ ∣^{2} = σ^{2} / n$ .

3.1step 1.1step 2.1L5algebra∎

Apply [L5] to $\overline{Y}$ using steps 1.1 and 2.1 to obtain the stated probability bound for every $a > 0$ . The finite calculation includes $n = 1$ and $σ^{2} = 0$ without dividing by a variance.

Depends on

Dependency tree · 2 levels

Sources

Rick Durrett, Probability: Theory and Examples, 5th ed., Section 2.2.1: finite variance sum and Chebyshev argument