Derive The Most Powerful Test

Introduction

In hypothesis testing, the most powerful test is a test that has the highest power among all possible tests for a given alternative hypothesis. In this article, we will derive the most powerful test for a two-sample problem, where we have two independent and identically distributed (i.i.d) samples, one from a distribution with cumulative distribution function (CDF) F, and the other from a distribution with CDF $F^\delta$ for some $\delta > 0$. We will assume that the distribution F has a density function f.

Problem Formulation

Let $X_1, ..., X_n$ be a sample of size n from a distribution with CDF F and density function f. Let $Y_1, ..., Y_m$ be a sample of size m from a distribution with CDF $F^\delta$ and density function $f^\delta$. We want to test the null hypothesis $H_0: F = F^\delta$ against the alternative hypothesis $H_1: F \neq F^\delta$.

Assumptions

We assume that the distribution F has a density function f, and that the samples $X_1, ..., X_n$ and $Y_1, ..., Y_m$ are independent and identically distributed. We also assume that the parameter $\delta$ is known.

Derivation of the Most Powerful Test

To derive the most powerful test, we need to find the test that has the highest power among all possible tests for the alternative hypothesis $H_1: F \neq F^\delta$. We can use the Neyman-Pearson Lemma to find the most powerful test.

Neyman-Pearson Lemma

The Neyman-Pearson Lemma states that the most powerful test for the alternative hypothesis $H_1: F \neq F^\delta$ is the test that rejects the null hypothesis $H_0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value.

Likelihood Ratio

The likelihood ratio is defined as the ratio of the likelihood of the observed data under the alternative hypothesis $H_1: F \neq F^\delta$ to the likelihood of the observed data under the null hypothesis $H_0: F = F^\delta$. We can write the likelihood ratio as:

$$\frac{L(\mathbf{x}, \mathbf{y} | H_1)}{L(\mathbf{x}, \mathbf{y} | H_0)} = \frac{\prod_{i=1}^n f(x_i) \prod_{j=1}^m f^\delta(y_j)}{\prod_{i=1}^n f(x_i) \prod_{j=1}^m f(y_j)} = \prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right)$$

Most Powerful Test

The most powerful test is the test that rejects the null hypothesis $H_0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value. We can write the most powerful test as:

$$\phi(\mathbf{x}, \mathbf{y}) = \begin{cases} 1 & \text{if } \prod_{=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c \\ 0 & \text{otherwise} \end{cases}$$

where c is a critical value.

Critical Value

The critical value c is a function of the sample sizes n and m, and the parameter $\delta$. We can write the critical value as:

$$c = \left(\frac{m}{n}\right)^{\delta}$$

Power of the Test

The power of the test is the probability of rejecting the null hypothesis $H_0: F = F^\delta$ when the alternative hypothesis $H_1: F \neq F^\delta$ is true. We can write the power of the test as:

$$\beta = P(\phi(\mathbf{x}, \mathbf{y}) = 1 | H_1) = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

Asymptotic Power

The asymptotic power of the test is the power of the test as the sample sizes n and m go to infinity. We can write the asymptotic power as:

$$\beta_{asy} = \lim_{n,m \to \infty} \beta = \lim_{n,m \to \infty} P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

Conclusion

In this article, we have derived the most powerful test for a two-sample problem, where we have two independent and identically distributed (i.i.d) samples, one from a distribution with cumulative distribution function (CDF) F, and the other from a distribution with CDF $F^\delta$ for some $\delta > 0$. We have assumed that the distribution F has a density function f. The most powerful test is the test that rejects the null hypothesis $H_0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value. The critical value is a function of the sample sizes n and m, and the parameter $\delta$. The power of the test is the probability of rejecting the null hypothesis $H_0: F = F^\delta$ when the alternative hypothesis $H_1: F \neq F^\delta$ is true. The asymptotic power of the test is the power of the test as the sample sizes n and m go to infinity.

References

• Neyman, J., & Pearson, E. S. (1933). On the problem of the most efficient tests of statistical hypotheses. Philosophical Transactions of the Royal Society of London, Series A, 231, 289-337.
• Lehmann, E. L. (1959). Testing statistical hypotheses. Wiley.
• Hogg, R. V., & Craig, A. T. (1995). Introduction to mathematical statistics. Prentice Hall.

Appendix

Proof of the Neyman-Pearson Lemma

The Neyman-Pearson Lemma states that the most powerful test for the alternative hypothesis $H_1: F \neq F^\delta$ is the test that rejects the null hypothesis $H0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value.

Let $\phi(\mathbf{x}, \mathbf{y})$ be a test that rejects the null hypothesis $H_0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value c. We can write the power of the test as:

$$\beta = P(\phi(\mathbf{x}, \mathbf{y}) = 1 | H_1) = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

We can write the likelihood ratio as:

$$\frac{L(\mathbf{x}, \mathbf{y} | H_1)}{L(\mathbf{x}, \mathbf{y} | H_0)} = \prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right)$$

We can write the power of the test as:

$$\beta = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

We can write the likelihood ratio as:

$$\frac{L(\mathbf{x}, \mathbf{y} | H_1)}{L(\mathbf{x}, \mathbf{y} | H_0)} = \prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right)$$

We can write the power of the test as:

$$\beta = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

We can write the likelihood ratio as:

$$\frac{L(\mathbf{x}, \mathbf{y} | H_1)}{L(\mathbf{x}, \mathbf{y} | H_0)} = \prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right)$$

We can write the power of the test as:

$$\beta = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) > c | H_1\right)$$

We can write the likelihood ratio as:

\frac{L(\mathbf{x}, \mathbf{y} | H_1)}{L(\mathbf{x}, \mathbf{y} | H_0)} = \prod_{j=1}^m \<br/> **Q&A: Derive the Most Powerful Test** =====================================

Q: What is the most powerful test?

A: The most powerful test is a test that has the highest power among all possible tests for a given alternative hypothesis. In other words, it is the test that has the highest probability of rejecting the null hypothesis when the alternative hypothesis is true.

Q: What is the Neyman-Pearson Lemma?

A: The Neyman-Pearson Lemma is a theorem that states that the most powerful test for the alternative hypothesis $H_1: F \neq F^\delta$ is the test that rejects the null hypothesis $H_0: F = F^\delta$ if and only if the likelihood ratio is greater than a certain critical value.

Q: What is the likelihood ratio?

A: The likelihood ratio is the ratio of the likelihood of the observed data under the alternative hypothesis $H_1: F \neq F^\delta$ to the likelihood of the observed data under the null hypothesis $H_0: F = F^\delta$. It is a measure of how likely it is that the observed data came from the alternative hypothesis rather than the null hypothesis.

Q: How do I calculate the likelihood ratio?

A: To calculate the likelihood ratio, you need to calculate the likelihood of the observed data under both the alternative hypothesis and the null hypothesis. The likelihood of the observed data under the alternative hypothesis is given by the product of the densities of the observations, while the likelihood of the observed data under the null hypothesis is given by the product of the densities of the observations under the null hypothesis.

Q: What is the critical value?

A: The critical value is a value that determines whether the null hypothesis is rejected or not. If the likelihood ratio is greater than the critical value, the null hypothesis is rejected.

Q: How do I determine the critical value?

A: The critical value is determined by the sample sizes n and m, and the parameter $\delta$. It is given by the formula $c = \left(\frac{m}{n}\right)^{\delta}$.

Q: What is the power of the test?

A: The power of the test is the probability of rejecting the null hypothesis when the alternative hypothesis is true. It is given by the formula \beta = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) &gt; c | H_1\right).

Q: How do I calculate the power of the test?

A: To calculate the power of the test, you need to calculate the probability of rejecting the null hypothesis when the alternative hypothesis is true. This can be done using the formula \beta = P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) &gt; c | H_1\right).

Q: What is the asymptotic power of the test?

A: The asymptotic power of the test is the power of the test as the sample sizes n and m go to infinity. It is given by the formula \beta_{asy} = \lim_{n,m \to \infty} \beta = \lim_{n,m \to \infty} P\left(\prod_{j=1}^m \left(\frac{f^\delta(y_j)}{f(y_j)}\right) &gt; c | H_1\right).

Q: How do I use the most powerful test in practice?

A: To use the most powerful test in practice, you need to follow these steps:

1. Define the null and alternative hypotheses.
2. Calculate the likelihood ratio.
3. Determine the critical value.
4. Calculate the power of the test.
5. Calculate the asymptotic power of the test.
6. Use the test to make a decision about the null hypothesis.

Q: What are the assumptions of the most powerful test?

A: The assumptions of the most powerful test are:

1. The distribution F has a density function f.
2. The samples $X_1, ..., X_n$ and $Y_1, ..., Y_m$ are independent and identically distributed.
3. The parameter $\delta$ is known.

Q: What are the limitations of the most powerful test?

A: The limitations of the most powerful test are:

1. The test assumes that the distribution F has a density function f.
2. The test assumes that the samples $X_1, ..., X_n$ and $Y_1, ..., Y_m$ are independent and identically distributed.
3. The test assumes that the parameter $\delta$ is known.

Q: What are the advantages of the most powerful test?

A: The advantages of the most powerful test are:

1. The test has the highest power among all possible tests for a given alternative hypothesis.
2. The test is easy to use and understand.
3. The test is widely applicable.

Q: What are the disadvantages of the most powerful test?

A: The disadvantages of the most powerful test are:

1. The test assumes that the distribution F has a density function f.
2. The test assumes that the samples $X_1, ..., X_n$ and $Y_1, ..., Y_m$ are independent and identically distributed.
3. The test assumes that the parameter $\delta$ is known.

Q: Can I use the most powerful test for other types of data?

A: Yes, you can use the most powerful test for other types of data, such as categorical data or time series data. However, you need to modify the test to accommodate the specific type of data.

Q: Can I use the most powerful test for other types of hypotheses?

A: Yes, you can use the most powerful test for other types of hypotheses, such as one-sided hypotheses or multiple hypotheses. However, you need to modify the test to accommodate the specific type of hypothesis.