Derive The Most Powerful Test

Introduction

In the realm of hypothesis testing, the most powerful test is a crucial concept that helps us determine the best possible test for a given hypothesis. In this article, we will delve into the world of hypothesis testing and derive the most powerful test for a specific scenario. We will consider two sample problems where $X_1, \ldots, X_n$ are i.i.d with CDF F and $Y_1, \ldots, Y_m$ are i.i.d from CDF $F^\delta$ for some $\delta > 0$. Assuming F to be a CDF on the real line with density f, we will derive the most powerful test for the following hypothesis:

$$H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$$

where $F_0$ and $F_1$ are two distinct CDFs.

Preliminaries

Before we dive into the derivation of the most powerful test, let's establish some notation and assumptions.

• $X_1, \ldots, X_n$ are i.i.d with CDF F and density f.
• $Y_1, \ldots, Y_m$ are i.i.d from CDF $F^\delta$ for some $\delta > 0$.
• F is a CDF on the real line with density f.
• $F_0$ and $F_1$ are two distinct CDFs.

We will assume that the observations $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_m$ are independent.

The Neyman-Pearson Lemma

The Neyman-Pearson Lemma is a fundamental result in hypothesis testing that provides a way to derive the most powerful test for a given hypothesis. The lemma states that the most powerful test for the hypothesis $H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$ is given by:

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{f_1(x)}{f_0(x)} > k \\ 0 & \text{otherwise} \end{cases}$$

where $f_0$ and $f_1$ are the densities corresponding to the CDFs $F_0$ and $F_1$, respectively, and $k$ is a constant that depends on the desired significance level.

Derivation of the Most Powerful Test

To derive the most powerful test, we need to find the density $f_1$ corresponding to the CDF $F_1$. Since $F_1 = F^\delta$, we have:

$$f_1(x) = \delta f(x^\delta)$$

Now, we can apply the Neyman-Pearson Lemma to derive the most powerful test.

Let $\phi(x)$ be the most powerful test for the hypothesis $H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$. Then, we have:

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{\delta f(x^\delta)}{f_0(x)} > k \\ 0 & \text{otherwise} \end{cases}$$

To simplify the expression, we can rewrite it as:

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{\delta f(x^\delta)}{f_0(x)} > k \\ 0 & \text{otherwise} \end{cases}$$

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{f(x^\delta)}{f_0(x)} > \frac{k}{\delta} \\ 0 & \text{otherwise} \end{cases}$$

Now, we can see that the most powerful test depends on the ratio of the densities $f(x^\delta)$ and $f_0(x)$.

The Ratio of Densities

To find the ratio of the densities $f(x^\delta)$ and $f_0(x)$, we can use the following result:

$$\frac{f(x^\delta)}{f_0(x)} = \frac{\int_{-\infty}^{x^\delta} f(t) dt}{\int_{-\infty}^{x} f_0(t) dt}$$

Now, we can substitute this expression into the most powerful test.

The Most Powerful Test

Substituting the expression for the ratio of densities into the most powerful test, we get:

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{\int_{-\infty}^{x^\delta} f(t) dt}{\int_{-\infty}^{x} f_0(t) dt} > \frac{k}{\delta} \\ 0 & \text{otherwise} \end{cases}$$

This is the most powerful test for the hypothesis $H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$.

Conclusion

In this article, we derived the most powerful test for a specific scenario. We considered two sample problems where $X_1, \ldots, X_n$ are i.i.d with CDF F and $Y_1, \ldots, Y_m$ are i.i.d from CDF $F^\delta$ for some $\delta > 0$. Assuming F to be a CDF on the real line with density f, we derived the most powerful test for the hypothesis $H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$. The most powerful test depends on the ratio of the densities $f(x^\delta)$ and $f_0(x)$.

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.

Further Reading

• Hogg, R. V., & Craig, A. T. (1995). Introduction to mathematical statistics. Prentice Hall.
• Casella, G., & Berger, R. L. (2002). Statistical inference. Duxbury Press.
  Derive the Most Powerful Test: Q&A =====================================

Q: What is the most powerful test in hypothesis testing?

A: The most powerful test is a statistical test that has the highest probability of detecting a true alternative hypothesis when the null hypothesis is false. It is the best possible test for a given hypothesis.

Q: What is the Neyman-Pearson Lemma?

A: The Neyman-Pearson Lemma is a fundamental result in hypothesis testing that provides a way to derive the most powerful test for a given hypothesis. It states that the most powerful test for the hypothesis $H_0: F = F_0 \quad \text{vs} \quad H_1: F = F_1$ is given by:

$$\phi(x) = \begin{cases} 1 & \text{if } \frac{f_1(x)}{f_0(x)} > k \\ 0 & \text{otherwise} \end{cases}$$

where $f_0$ and $f_1$ are the densities corresponding to the CDFs $F_0$ and $F_1$, respectively, and $k$ is a constant that depends on the desired significance level.

Q: How do I derive the most powerful test?

A: To derive the most powerful test, you need to find the density $f_1$ corresponding to the CDF $F_1$. Since $F_1 = F^\delta$, you have:

$$f_1(x) = \delta f(x^\delta)$$

Then, you can apply the Neyman-Pearson Lemma to derive the most powerful test.

Q: What is the ratio of densities in the most powerful test?

A: The ratio of densities in the most powerful test is given by:

$$\frac{f(x^\delta)}{f_0(x)} = \frac{\int_{-\infty}^{x^\delta} f(t) dt}{\int_{-\infty}^{x} f_0(t) dt}$$

This ratio depends on the CDFs $F$ and $F_0$.

Q: How do I determine the constant $k$ in the most powerful test?

A: The constant $k$ in the most powerful test depends on the desired significance level. You can determine $k$ using the following formula:

$$k = \frac{\alpha}{1 - \alpha}$$

where $\alpha$ is the desired significance level.

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

A: The most powerful test assumes that the observations $X_1, \ldots, X_n$ and $Y_1, \ldots, Y_m$ are independent and identically distributed with CDFs $F$ and $F^\delta$, respectively.

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

A: The most powerful test has several limitations. It assumes that the CDFs $F$ and $F^\delta$ are known, which is often not the case in practice. Additionally, the test may not be robust to non-normality or non-identical distributions.

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

A: To apply the most powerful test in, you need to:

1. Determine the CDFs $F$ and $F^\delta$.
2. Find the density $f_1$ corresponding to the CDF $F_1$.
3. Apply the Neyman-Pearson Lemma to derive the most powerful test.
4. Determine the constant $k$ using the desired significance level.
5. Test the null hypothesis using the most powerful test.

Conclusion

In this article, we provided a Q&A guide to the most powerful test in hypothesis testing. We discussed the Neyman-Pearson Lemma, the ratio of densities, and the assumptions and limitations of the most powerful test. We also provided a step-by-step guide on how to apply the most powerful test in practice.

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.

Further Reading

• Hogg, R. V., & Craig, A. T. (1995). Introduction to mathematical statistics. Prentice Hall.
• Casella, G., & Berger, R. L. (2002). Statistical inference. Duxbury Press.