Let X1, X2, ⋯, Xn be a random sample from the 𝒩(μ, σ^2) distribution, where μ∈ℝ is unknown, but

For exercise 120A, we have Z equals Y sub 1 bar minus Y sub 2 bar minus mu sub 2. This will be d

Question

Let \( X_{1}, X_{2}, \cdots, X_{n} \) be a random sample from the \( \mathcal{N}\left(\mu, \sigma^{2}\right) \) distribution, where \( \mu \in \mathbb{R} \) is unknown, but \( \sigma>0 \) is known. As shown in Slide 4 of Chapter 7 , the probability that the Find the first derivative of \( g(u)=h(u)-u \) random interval \[ \left(\bar{X}-1.96 \frac{\sigma}{\sqrt{n}}, \bar{X}+1.96 \frac{\sigma}{\sqrt{n}}\right) \] using the chain rule of differentiation to contans \( \mu \) is \( 0.95 \). If we replace \( 0.95 \) by any number \( 1-\alpha \), where \( \alpha \in(0,1) \), and \( 1.96 \) calculate \( h^{\prime}(u) \). To keep track of algebraic probability that the random interval \[ \left(\bar{X}-z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}, \bar{X}+z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\right) \] contains \( \mu \) is \( 1-\alpha ; \) see slide 7 . Note that \( \Phi\left(z_{\beta}\right)=1-\beta \). It is important to recognize that for any pair of numbers \( 0<\gamma<\delta<1 \) such that \( \delta-\gamma=1-\alpha \), the probability that the random interval \[ \left(\bar{X}-z_{\gamma} \frac{\sigma}{\sqrt{n}}, \bar{X}-z_{\delta} \frac{\sigma}{\sqrt{n}}\right) \] contains \( \mu \) is \( 1-\alpha \). Since \[ \frac{X-\mu}{\sigma / \sqrt{n}} \sim \mathcal{N}(0,1), \] we obtain \[ \mathbb{P}\left(z_{\delta}<\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}<z_{\gamma}\right)=\Phi\left(z_{\gamma}\right)-\Phi\left(z_{\delta}\right)=1-\gamma-(1-\delta)=\delta-\gamma=1-\alpha, \] thereby obtaining (2). If we put \( \delta=1-\alpha / 2 \) and \( \gamma=\alpha / 2 \), then (2) reduces to (1), since \[ \Phi\left(z_{1-\alpha / 2}\right)=\alpha / 2=\Phi\left(-z_{\alpha / 2}\right) \] by the symmetry of the \( \mathcal{N}(0,1) \) distribution around 0 . The question is why our textbook, from among this whole class of random intervals described in (2), picks the random interval described in (1). The answer is the random interval described in (1) has the shortest width in the class of random intervals described in (2). The objective of this assignment is to establish the assertion made in bold in the above sentence. But first, let us formulate the proposition using simpler notation (the \( z_{\beta} \) notation is cumbersome, to say the least). Proposition Let \( f: \mathbb{R}^{2} \mapsto \mathbb{R} \) be defined as \( f(a, b)=b-a \). Subject to the constraint \[ \Phi(b)-\Phi(a)=\beta \in(0,1) \] the function \( f \) is minimized at \( \left(a_{0}, b_{0}\right) \) such that \( \Phi\left(a_{0}\right)=(1-\beta) / 2 \).

          Let \( X_{1}, X_{2}, \cdots, X_{n} \) be a random sample from the \( \mathcal{N}\left(\mu, \sigma^{2}\right) \) distribution, where \( \mu \in \mathbb{R} \) is unknown, but \( \sigma>0 \) is known. As shown in Slide 4 of Chapter 7 , the probability that the Find the first derivative of \( g(u)=h(u)-u \) random interval
\[
\left(\bar{X}-1.96 \frac{\sigma}{\sqrt{n}}, \bar{X}+1.96 \frac{\sigma}{\sqrt{n}}\right)
\]
using the chain rule of differentiation to
contans \( \mu \) is \( 0.95 \). If we replace \( 0.95 \) by any number \( 1-\alpha \), where \( \alpha \in(0,1) \), and \( 1.96 \) calculate \( h^{\prime}(u) \). To keep track of algebraic probability that the random interval
\[
\left(\bar{X}-z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}, \bar{X}+z_{\alpha / 2} \frac{\sigma}{\sqrt{n}}\right)
\]
contains \( \mu \) is \( 1-\alpha ; \) see slide 7 . Note that \( \Phi\left(z_{\beta}\right)=1-\beta \).
It is important to recognize that for any pair of numbers \( 0<\gamma<\delta<1 \) such that \( \delta-\gamma=1-\alpha \), the probability that the random interval
\[
\left(\bar{X}-z_{\gamma} \frac{\sigma}{\sqrt{n}}, \bar{X}-z_{\delta} \frac{\sigma}{\sqrt{n}}\right)
\]
contains \( \mu \) is \( 1-\alpha \). Since
\[
\frac{X-\mu}{\sigma / \sqrt{n}} \sim \mathcal{N}(0,1),
\]
we obtain
\[
\mathbb{P}\left(z_{\delta}<\frac{\bar{X}-\mu}{\sigma / \sqrt{n}}<z_{\gamma}\right)=\Phi\left(z_{\gamma}\right)-\Phi\left(z_{\delta}\right)=1-\gamma-(1-\delta)=\delta-\gamma=1-\alpha,
\]
thereby obtaining (2). If we put \( \delta=1-\alpha / 2 \) and \( \gamma=\alpha / 2 \), then (2) reduces to (1), since
\[
\Phi\left(z_{1-\alpha / 2}\right)=\alpha / 2=\Phi\left(-z_{\alpha / 2}\right)
\]
by the symmetry of the \( \mathcal{N}(0,1) \) distribution around 0 .
The question is why our textbook, from among this whole class of random intervals described in (2), picks the random interval described in (1).

The answer is the random interval described in (1) has the shortest width in the class of random intervals described in (2).

The objective of this assignment is to establish the assertion made in bold in the above sentence. But first, let us formulate the proposition using simpler notation (the \( z_{\beta} \) notation is cumbersome, to say the least).
Proposition
Let \( f: \mathbb{R}^{2} \mapsto \mathbb{R} \) be defined as \( f(a, b)=b-a \). Subject to the constraint
\[
\Phi(b)-\Phi(a)=\beta \in(0,1)
\]
the function \( f \) is minimized at \( \left(a_{0}, b_{0}\right) \) such that \( \Phi\left(a_{0}\right)=(1-\beta) / 2 \).

Let X1, X2, ⋯, Xn be a random sample from the 𝒩(μ, σ^2) distribution, where μ∈ℝ is unknown, but σ>0 is known. As shown in Slide 4 of Chapter 7 , the probability that the Find the first derivative of g(u)=h(u)-u random interval

(X̅-1.96 (σ)/(√(n)), X̅+1.96 (σ)/(√(n)))

using the chain rule of differentiation to
contans μ is 0.95. If we replace 0.95 by any number 1-α, where α∈(0,1), and 1.96 calculate h^'(u). To keep track of algebraic probability that the random interval

(X̅-zα / 2(σ)/(√(n)), X̅+zα / 2(σ)/(√(n)))

contains μ is 1-α ; see slide 7 . Note that Φ(zβ)=1-β.
It is important to recognize that for any pair of numbers 0<γ<δ<1 such that δ-γ=1-α, the probability that the random interval

(X̅-zγ(σ)/(√(n)), X̅-zδ(σ)/(√(n)))

contains μ is 1-α. Since

(X-μ)/(σ / √(n))∼𝒩(0,1),

we obtain

ℙ(zδ<(X̅-μ)/(σ / √(n))<zγ)=Φ(zγ)-Φ(zδ)=1-γ-(1-δ)=δ-γ=1-α,

thereby obtaining (2). If we put δ=1-α / 2 and γ=α / 2, then (2) reduces to (1), since

Φ(z1-α / 2)=α / 2=Φ(-zα / 2)

by the symmetry of the 𝒩(0,1) distribution around 0 .
The question is why our textbook, from among this whole class of random intervals described in (2), picks the random interval described in (1).

The answer is the random interval described in (1) has the shortest width in the class of random intervals described in (2).

The objective of this assignment is to establish the assertion made in bold in the above sentence. But first, let us formulate the proposition using simpler notation (the zβ notation is cumbersome, to say the least).
Proposition
Let f: ℝ^2↦ℝ be defined as f(a, b)=b-a. Subject to the constraint

Φ(b)-Φ(a)=β∈(0,1)

the function f is minimized at (a0, b0) such that Φ(a0)=(1-β) / 2.

Added by Charles M.

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