Let X_1, \ldots , X_ n \stackrel{iid}{\sim } \mathsf{Ber}(p). We want to test
\begin{array}{rl} H_0: & p\le .5 \\ H_1: & p> .5\\ \end{array}
at asymptotic level 5%. Let \psi be the Wald test.
1) Assume that the true p is equal to 0.2. What is the limit of \mathbf{P}_{0.2}[\psi =1] as n \to \infty?
The Wald test statistic is given by:
\[\psi = \frac{\hat{p} - 0.5}{\sqrt{\frac{0.5(1-0.5)}{n}}}\]
where \(\hat{p}\) is the sample proportion.
1) Assuming the true probability is equal to 0.2, the estimate for \(p\) is simply the sample mean, which is given by:
\[\hat{p} = \frac{1}{n}\sum_{i=1}^{n}X_i \sim \mathcal{N}\left(p, \frac{p(1-p)}{n}\right)\]
Substituting \(p = 0.2\) into the formula, we get:
\[\hat{p} \sim \mathcal{N}\left(0.2, \frac{0.16}{n}\right)\]
So the test statistic becomes:
\[\psi = \frac{0.2 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{n}}} = \frac{-0.3}{\sqrt{\frac{0.25}{n}}} = \frac{-0.3}{\sqrt{\frac{1}{4n}}} = \frac{-0.3}{\frac{1}{2\sqrt{n}}} = -0.6\sqrt{n}\]
In the limit as \(n \to \infty\), the value of \(\psi\) approaches negative infinity. Thus, \(\mathbf{P}_{0.2}[\psi =1]\) goes to zero.
For every \lambda >0 we can show that
\lim _{n \to \infty }\mathbf{P}_{.5-\frac{\lambda }{\sqrt{n}}}[\psi =1]=\mathbf{P}[Z>1.645+ f(\lambda )]
where f(\lambda ) is a function that depends only on \lambda. What is this function?
To find the function f(λ), we need to find the probability \(\mathbf{P}[Z>1.645+ f(\lambda)]\) where Z is a standard normal random variable.
First, let's rewrite the limit expression:
\[\lim _{n \to \infty }\mathbf{P}_{.5-\frac{\lambda }{\sqrt{n}}}[\psi =1]=\mathbf{P}[Z>1.645+ f(\lambda)]\]
We know that the test statistic \(\psi\) follows a standard normal distribution under the null hypothesis. So we can rewrite the left-hand side as:
\[\mathbf{P}[Z>1.645+ f(\lambda)] = 1 - \mathbf{P}[Z \leq 1.645+ f(\lambda)]\]
Since the limit expression is true for every λ > 0, we can let λ tend to 0. This will give us:
\[\lim _{n \to \infty }\mathbf{P}_{.5}[\psi =1]=\mathbf{P}[Z>1.645+ f(0)]\]
We know that under the null hypothesis (p = 0.5), the test statistic \(\psi\) follows a standard normal distribution, so we can substitute \(f(0)\) with 0:
\[\mathbf{P}[Z>1.645+ f(0)] = \mathbf{P}[Z>1.645]\]
Therefore, for every λ > 0, the function f(λ) is equal to zero: f(λ) = 0.
what is the function
Apologies for the confusion. Let's derive the function f(λ) using the limit expression given earlier.
\[\lim _{n \to \infty }\mathbf{P}_{.5-\frac{\lambda }{\sqrt{n}}}[\psi =1]=\mathbf{P}[Z>1.645+ f(\lambda)]\]
We can rewrite the left-hand side as:
\[\lim _{n \to \infty }\mathbf{P}_{.5-\frac{\lambda }{\sqrt{n}}}[\psi =1] = 1 - \lim _{n \to \infty }\mathbf{P}_{.5-\frac{\lambda }{\sqrt{n}}}[\psi \leq 1]\]
To find the limit of the probability on the right-hand side, we can use the properties of the standard normal distribution.
Let \(F_{Z}(z)\) denote the cumulative distribution function (CDF) of the standard normal distribution.
Since \(\psi\) follows a standard normal distribution under the null hypothesis, we can rewrite the limit expression as:
\[1 - \lim _{n \to \infty } F_{Z}(.5-\frac{\lambda}{\sqrt{n}})[\psi \leq 1]\]
The CDF of the standard normal distribution is continuous, so we can rewrite the limit expression as:
\[1 - F_{Z}(\lim _{n \to \infty }(.5-\frac{\lambda}{\sqrt{n}}))[1-P[\psi = 1]]\]
Now, we know that \(\lim _{n \to \infty }(.5-\frac{\lambda}{\sqrt{n}}) = .5\), and \(P[\psi = 1] = P[\psi - .5 = .5] = P[Z = .5]\).
So the final limit expression becomes:
\[1 - F_{Z}(.5)[1-F_{Z}(.5)] = F_{Z}(.5)-F_{Z}(.5)^2\]
Therefore, the function f(λ) is given by:
\[f(\lambda) = F_{Z}(.5)-F_{Z}(.5)^2\]
in normal math notation
Apologies for the confusion. In normal mathematical notation, the function f(λ) is given by:
\[f(\lambda) = \Phi(.5) - \Phi(.5)^2\]
where Φ(·) represents the cumulative distribution function (CDF) of the standard normal distribution.
Phi is not permitted inthe answr
I apologize for the confusion. To avoid using Φ (Phi), we can rewrite the function using the standard normal quantile function, denoted as Q(·). The function f(λ) is then given by:
\[f(\lambda) = Q(1- \lambda) \]
Here, Q(·) represents the quantile function of the standard normal distribution.
For every 𝜆>0 we can show that
lim𝑛→∞𝐏.5−𝜆𝑛√[𝜓=1]=𝐏[𝑍>1.645+𝑓(𝜆)]
where 𝑓(𝜆) is a function that depends only on 𝜆. What is this function?