ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pass

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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Finite check of the IID sample-mean variance and Chebyshev bound

Example

Let $Ω = {(σ_{1}, σ_{2}) : σ_{1}, σ_{2} \in {- 1, 1}}$ be the uniform finite probability space with four outcomes of weight $1 / 4$ , and let $X_{1} (σ_{1}, σ_{2}) : = σ_{1}, X_{2} (σ_{1}, σ_{2}) : = σ_{2} .$ Then $(X_{1}, X_{2})$ is an IID family with $μ = E [X_{1}] = 0$ and $σ^{2} = Var (X_{1}) = 1$ ; the sample mean $\overline{X} = (X_{1} + X_{2}) / 2$ takes the values $- 1, 0, 1$ with probabilities $1 / 4, 1 / 2, 1 / 4$ respectively. The exact second moment and the Chebyshev bound computed directly below read $E ∣ \overline{X} ∣^{2} = \frac{1}{2} = \frac{σ^{2}}{2}, P (∣ \overline{X} ∣ \geq 1) = \frac{1}{2} = \frac{σ^{2}}{2 \cdot 1^{2}},$ so the bound is attained at $a = 1$ , while $P (∣ \overline{X} ∣ \geq \frac{1}{2}) = \frac{1}{2} \leq 2 = \frac{σ^{2}}{2 \cdot (1 / 2)^{2}}$ is strict. This is a finite check inside the stated model; it evaluates no measurement and estimates no physical quantity.

Facts & Assumptions

Given: The uniform finite probability space on ${- 1, 1}^{2}$ and the coordinate maps $X_{1}, X_{2}$ of the Example.

[L3]

On a finite probability space, expectations are the finite weighted sums of Expectation of a real random variable on a finite probability space, and the finite model agrees with the probability measure $P (A) = \sum_{ω \in A} w (ω)$ , whose expectations coincide with those integral expectations. (Finite probability spaces are exactly finite full-power-set probability spaces, Finite probability spaces, outcome weights, events, and event probabilities)

[L4]

A finite family of finite-valued random variables is mutually independent exactly when $P (X_{j} = x_{j} for all j \in J) = \prod_{j \in J} P (X_{j} = x_{j})$ for every nonempty $J$ and every choice of values; it is identically distributed when the individual distributions agree. (Pairwise and mutual independence of finite-valued random variables, Identical distribution and IID families)

Verification

technique · direct enumeration of four outcomes

1.1givenL3L4algebra

The four outcomes have weight $1 / 4$ each, so $P (X_{1} = - 1) = P (X_{1} = 1) = 1 / 2$ and the same for $X_{2}$ ; for every pair $(a, b) \in {- 1, 1}^{2}$ one has $P (X_{1} = a, X_{2} = b) = 1 / 4 = P (X_{1} = a) P (X_{2} = b)$ . By [L4] the family is independent, the two distributions agree, and $E [X_{1}] = \frac{1}{2} (- 1) + \frac{1}{2} (1) = 0$ , $E [X_{1}^{2}] = \frac{1}{2} (1) + \frac{1}{2} (1) = 1$ by [L3].

1.2givenL3algebra

Enumerating $\overline{X} = (X_{1} + X_{2}) / 2$ on the four outcomes gives the values $1, 0, 0, - 1$ , each with weight $1 / 4$ ; hence $P (∣ \overline{X} ∣ \geq 1) = \frac{2}{4} = \frac{1}{2}, P (∣ \overline{X} ∣ \geq \frac{1}{2}) = \frac{2}{4} = \frac{1}{2} .$

2.1step 1.1step 1.2L3algebra

By the enumeration in step 1.2 and the expectation formula [L3], $E [\overline{X}] = 0$ and $E ∣ \overline{X} ∣^{2} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} = σ^{2} / 2$ . For any $a > 0$ , the pointwise inequality $a^{2} 1_{{∣ \overline{X} ∣ \geq a}} \leq ∣ \overline{X} ∣^{2}$ and summation with nonnegative weights give $P (∣ \overline{X} ∣ \geq a) \leq E ∣ \overline{X} ∣^{2} / a^{2} = σ^{2} / (2 a^{2})$ .

3.1step 1.2step 2.1algebra∎

At $a = 1$ the bound of step 2.1 is $1 / 2$ , equal to the exact probability from step 1.2; at $a = 1 / 2$ it is $2$ , strictly larger than the exact probability $1 / 2$ . These establish the displayed finite-sample identity and bounds without requiring an infinite IID extension.

Depends on

Dependency tree · 2 levels

Sources

Rick Durrett, Probability: Theory and Examples, 5th ed., Section 2.2.1 (the variance of a sum of independent variables and the Chebyshev bound for the i.i.d. sample mean)