Let X be a single (i.e. n=1 ) Gaussian random variable with unknown mean μ and variance 1 . Consider the following hypotheses:

H0:μ=0vsH1:μ≠0.

Define a test ψα:R→{0,1} with level α that is of the form

ψα=1{fα(X)>0},

for some function fα:R→R .

We want our test ψ above to satisfy the following:

symmetric in the value of X

its “acceptance region" is an interval. (The acceptance region of a test is the region in which the null hypothesis is not rejected, i.e. the complement of its rejection region.)

Specify the function fα(X) in terms of α below.

(Type alpha for α . If applicable, enter abs(x) for |x| , Phi(x) for Φ(x)=P(Z≤x) where Z∼N(0,1) , and q(alpha) for qα , the 1−α -quantile of a standard normal distribution, e.g. enter q(0.01) for q0.01 .)

f(X)=?

f(X) = |X| - q(α)

To specify the function fα(X) in terms of α for the given test, we can use the fact that X is a Gaussian random variable with unknown mean μ and variance 1.

Since our test is symmetric in the value of X, we can define the acceptance region as an interval symmetric around zero. Let's denote this interval as (-c, c), where c > 0.

To determine the value of c, we need to find the critical value that satisfies the level α requirement. The level α is the maximum probability of Type I error, which is the probability of rejecting the null hypothesis when it is actually true.

In this case, the null hypothesis is μ = 0. To find the critical value, we need to find the value of c that ensures the probability of X lying outside the interval (-c, c) when μ = 0 is equal to α.

Since X is a Gaussian random variable with mean 0 and variance 1, we can use the standard normal distribution. The probability of X lying outside (-c, c) can be calculated as the sum of the probabilities of X being less than -c and X being greater than c.

P(X < -c) + P(X > c) = 2(1 - Φ(c))

Where Φ(c) is the cumulative distribution function of the standard normal distribution.

To satisfy the level α requirement, we need:

2(1 - Φ(c)) = α

Solving for c:

1 - Φ(c) = α/2

Φ(c) = 1 - α/2

Using the properties of the cumulative distribution function, we can rewrite this as:

c = q(1 - α/2)

Therefore, the function fα(X) is defined as:

fα(X) = X - c,

where c = q(1 - α/2) is the critical value that defines the acceptance region (-c, c).

To specify the function fα(X) in terms of α for the test ψα, we need to define the acceptance region as an interval and make it symmetric in the value of X.

Let's assume that the acceptance region is an interval [-k, k], where k is a positive real number.

Since we want the test to be symmetric, we need to consider both tails of the distribution. Thus, we can split the interval [-k, k] into two equal parts: [-k, 0] and [0, k].

To determine the boundaries of the interval, we can use the properties of the standard normal distribution.

The area under the standard normal curve from -∞ to -k is α/2, and likewise, the area from k to +∞ is α/2.

So, we want to find the value of k such that the area under the standard normal curve from -k to k is equal to 1 - α.

Using the symmetry of the standard normal distribution, we can express this as:

P(-k ≤ Z ≤ k) = 1 - α

Here, Z is a standard normal random variable.

Using the symmetry property, we can rewrite the equation as:

P(Z ≤ k) - P(Z ≤ -k) = 1 - α

Since the standard normal distribution is symmetric, P(Z ≤ -k) = P(Z ≥ k).

So, we can rewrite the equation as:

2P(Z ≤ k) - 1 = 1 - α

Simplifying further, we get:

P(Z ≤ k) = 1 - α/2

Now, P(Z ≤ k) is the cumulative distribution function of the standard normal distribution, denoted as Φ(k).

Therefore, we have:

Φ(k) = 1 - α/2

To solve for k, we can use the inverse of the cumulative distribution function, denoted as qα/2, which gives us the value of k such that the area under the standard normal curve from -∞ to k is equal to 1 - α/2.

So, we have:

k = q(1 - α/2)

Therefore, the function fα(X) in terms of α is:

fα(X) = X - q(1 - α/2)

This function subtracts the value of q(1 - α/2) from X and compares the result to zero. If the result is greater than zero, the test rejects the null hypothesis H0, otherwise, it accepts H0.

Hi Ram J, these are very interesting questions. When you have the opportunity, can you please show your work for any questions you have attempted to answer, and let us know where you are getting stuck? Thanks.