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Opposed most powerful tests for two normal alternatives

Example

Assume AC and let $X \sim N (θ, 1)$ . For the simple null $H_{0} : θ = 0$ and alternatives $θ \in {- 1, 1}$ , consider the level- $1 / 2$ randomized tests with rejection functions $ϕ_{+} (x) = 1_{{x > 0}}, ϕ_{-} (x) = 1_{{x < 0}} .$ The upper-tail test favors $θ = 1$ , while the lower-tail test favors $θ = - 1$ ; their powers at the opposite alternative are strictly below $1 / 2$ .

Verification

Given: AC and the normal experiment in the statement.

[F1] For the two-sided normal-location problem, no UMP level- $α$ test exists for any $0 < α < 1$ . (A two-sided normal alternative need not have a UMP test)

[F2] $N (0, 1)$ has the even positive density $φ (u) = e^{- u^{2} / 2} / \sqrt{2 π}$ , and $N (θ, 1)$ is the law of $θ + Z$ for $Z \sim N (0, 1)$ . (Standard normal and normal laws)

[F3] Assuming countable choice, $L (R)$ is a sigma-algebra and $λ$ is a measure, so $(R, L (R), λ)$ is a measure space. (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume, Measure spaces)

[F4] Countable choice implies that a singleton is Lebesgue null, so a law with a Lebesgue density has no atom at zero. (The Axiom of Countable Choice ( ${A C}_{ω}$ ), A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in $R^{n}$ )

[A1] AC states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F5] A randomized test is a binary-action probability kernel; its test function is a measurable rejection probability in $[0, 1]$ . (Randomized tests)

[F6] Under countable choice, a nondegenerate interval has positive Lebesgue measure equal to its length. (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

[F7] Power is $E_{θ} ψ$ and level $α$ means the null rejection probability is at most $α$ . (Test size and power)

[F8] A law with density $f$ integrates a nonnegative function $g$ as $\int g f d λ$ . (The measure with density $f$ relative to $μ$ , Integrating against a density agrees with integrating the product)

[F9] The nonnegative integral is monotone and homogeneous for nonnegative scalars. (Monotonicity and nonnegative homogeneity of the nonnegative integral)

[F10] A measurable indicator has nonnegative integral equal to the measure of its set. (Nonnegative simple measurable functions, The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions)

[F11] The Borel sigma-algebra on $R$ contains open rays, so both threshold indicators are measurable. (The Borel sigma-algebra of a topological space)

[F12] A probability measure has total mass one and is countably additive on disjoint measurable sets. (Probability measures and probability spaces, Measures on sigma-algebras)

[F13] For a measurable self-map, measurability means inverse images of measurable sets are measurable; it preserves measure when every measurable set and its inverse image have equal measure. (A measurable function between measurable spaces, Measure-preserving transformations and systems)

[F14] Under countable choice, every Borel subset of $R$ is Lebesgue measurable. (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable)

[F15] A measure-preserving measurable map preserves the integral of every nonnegative measurable function. (Integral invariance under measure-preserving maps)

[F16] Products, sums, differences, and scalar multiples of measurable functions are measurable whenever the operations are defined. (Arithmetic and lattice operations preserve measurability whenever they are defined)

[F17] Integrable real functions form a vector space and the integral is linear on it. (Integrable real and complex functions, and their integrals, The Lebesgue integral is linear on $L^{1} (μ)$ )

[F18] Translation $τ_{θ} (u) = u + θ$ is a measurable measure-preserving map of Lebesgue measure, by translation invariance. (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Measure-preserving transformations and systems)

[F19] Reflection $ρ (u) = - u$ is a measurable measure-preserving map of Lebesgue measure. (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it, Measure-preserving transformations and systems)

[F20] A probability kernel assigns a probability measure to each observation and has measurable event-evaluation functions. (Measure kernel and probability kernel)

1.1A1F2F3F8F13F14F15F16F18algebra

Apply AC [A1] to the values in any sequence of nonempty sets and compose the choice function with that sequence; this supplies the countable-choice hypothesis in [F3], [F4], [F14], and [F18]–[F19]. For every Borel set $B$ and $θ \in R$ , the pushforward definition [F2] gives $P_{θ} (B) = \int 1_{B} (u + θ) φ (u) d λ (u)$ . Apply [F15] to $τ_{θ}$ and the nonnegative measurable function $h (x) = 1_{B} (x) φ (x - θ)$ ; [F14], continuity of $φ$ , and [F16] give its measurability, and $h (u + θ) = 1_{B} (u + θ) φ (u)$ . Hence $P_{θ} (B) = \int_{B} φ (x - θ) d λ (x)$ , so $p_{θ} (x) = φ (x - θ)$ is a density and [F8] gives $E_{θ} g = \int g p_{θ} d λ$ for every nonnegative measurable test function $g$ .

2.1F4F5F7F8F11F12F13F14F15F16F19F20step 1.1algebraconstruct

Let $ρ (u) = - u$ and $h (u) = φ (u) 1_{(0, \infty)} (u)$ . By [F14] and [F16], $h$ is nonnegative and Lebesgue measurable; by evenness, $h \circ ρ = φ 1_{(- \infty, 0)}$ . The map $ρ$ is measurable and measure-preserving by [F19], so [F15] gives equal integrals of these two functions. By [F8], $P_{0} (X > 0) = P_{0} (X < 0)$ . The singleton at zero is null by [F4] and step 1.1, while [F12] gives total mass one; therefore both probabilities equal $1 / 2$ . The indicators are measurable by [F11]. For $B \subseteq {0, 1}$ , define $K_{\pm} (x, B) = ϕ_{\pm} (x) 1_{{1 \in B}} + (1 - ϕ_{\pm} (x)) 1_{{0 \in B}}$ . The two weights are nonnegative and sum to one, and the formula is countably additive because each point belongs to at most one event in a disjoint sequence; each event evaluation is measurable by [F16]. Thus [F20] gives probability kernels with test functions $ϕ_{\pm}$ , hence randomized tests by [F5] and level- $1 / 2$ tests by [F7].

3.1F7F8F9F16F17step 1.1step 2.1algebra

Let $ψ$ be any level- $1 / 2$ test for the simple null $θ = 0$ , put $D = ϕ_{+} - ψ$ , and set $p_{0} = φ$ and $p_{1} (x) = φ (x - 1)$ as in step 1.1. The ratio $p_{1} (x) / p_{0} (x) = e^{x - 1 / 2}$ is strictly increasing, and with $k = e^{- 1 / 2}$ the product $D (x) (p_{1} (x) - k p_{0} (x))$ is nonnegative: above zero both factors are nonnegative, below zero both are nonpositive, and at zero the second factor vanishes. By [F16] the product is measurable. Since $∣ D ∣ \leq 1$ , $p_{0}, p_{1}$ are integrable with integral one, and $∣ D p_{i} ∣ \leq p_{i}$ , we have $D p_{0}, D p_{1} \in L^{1} (λ)$ ; then [F17] gives $D (p_{1} - k p_{0}) = D p_{1} - k D p_{0} \in L^{1} (λ)$ and yields $E_{1} ϕ_{+} - E_{1} ψ - k (E_{0} ϕ_{+} - E_{0} ψ) = \int D (p_{1} - k p_{0}) d λ \geq 0.$ Step 2.1 gives $E_{0} ϕ_{+} = 1 / 2$ , and level gives $E_{0} ψ \leq 1 / 2$ , so $k > 0$ implies $E_{1} ϕ_{+} \geq E_{1} ψ$ . Thus $ϕ_{+}$ is most powerful against $θ = 1$ .

3.2F7F8F9F16F17step 1.1step 2.1algebra

For the same arbitrary level- $1 / 2$ test $ψ$ , put $D = ϕ_{-} - ψ$ and $p_{- 1} (x) = φ (x + 1)$ . Then $p_{- 1} (x) / p_{0} (x) = e^{- x - 1 / 2}$ is strictly decreasing. With $k = e^{- 1 / 2}$ , $D (x) (p_{- 1} (x) - k p_{0} (x)) \geq 0$ : below zero both factors are nonnegative, above zero both are nonpositive, and at zero the second factor vanishes. The factors are measurable by [F16]. The bounds $∣ D ∣ \leq 1$ and $\int p_{i} d λ = 1$ give $D p_{- 1}, D p_{0} \in L^{1} (λ)$ by [F8]–[F9], so [F17] gives $D (p_{- 1} - k p_{0}) = D p_{- 1} - k D p_{0} \in L^{1} (λ)$ . Linearity yields $E_{- 1} ϕ_{-} - E_{- 1} ψ - k (E_{0} ϕ_{-} - E_{0} ψ) = \int D (p_{- 1} - k p_{0}) d λ \geq 0$ . Since [F7] and step 2.1 give $E_{0} ϕ_{-} = 1 / 2 \geq E_{0} ψ$ , this proves $E_{- 1} ϕ_{-} \geq E_{- 1} ψ$ . Thus $ϕ_{-}$ is most powerful against $θ = - 1$ .

3.3F2F4F6F7F8F9F10F12step 2.1algebra

Let $I = \int_{[0, 1]} φ d λ$ . On $[0, 1]$ , $φ (u) \geq e^{- 1 / 2} / \sqrt{2 π} > 0$ , so [F9]–[F10] give $I \geq e^{- 1 / 2} λ ([0, 1]) / \sqrt{2 π} > 0$ using [F6]. By [F8], this is $P (0 < Z < 1)$ since the endpoints are null by [F4]. By [F12] and step 2.1, $P (Z < 1) = P (Z < 0) + P (0 < Z < 1) = \frac{1}{2} + I$ , and symmetry with the null endpoints [F4] gives $P (Z > - 1) = P (Z < 1)$ and $P (Z < - 1) = P (Z > 1) = 1 - P (Z < 1) = \frac{1}{2} - I$ . Thus for $X = 1 + Z$ , $E_{1} ϕ_{+} = \frac{1}{2} + I$ and $E_{1} ϕ_{-} = \frac{1}{2} - I$ ; reflection gives $E_{- 1} ϕ_{-} = \frac{1}{2} + I$ and $E_{- 1} ϕ_{+} = \frac{1}{2} - I$ . These expectations are powers by [F7], so the two most powerful tests just proved have strict power reversal at the opposite alternatives.

4.1F1step 3.1step 3.2step 3.3∎

By [F1] at $α = 1 / 2$ , no test maximizes power at both alternatives; step 3.3 exhibits the strict reversal for the two most powerful tests established in steps 3.1–3.2.

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Sources

George Deligiannidis, Foundations of Statistical Inference, Neyman–Pearson Theorem 13.2 and UMP discussion, PDF pp. 77–78 (standard reference, not scraped)