Problem:
Let X be a random variable of continuous type with probability density function f(x). Then, based on single observation X, the most powerful test of size α = 0.1 for testing H0 : f(X) = 2x, 0 ≤ x ≤ 1, against H1 : f(X) = 4x^2, 0 ≤ x ≤ 1, has power

Solution:
To find the most powerful test for testing the given null and alternative hypotheses, we need to follow the Neyman-Pearson lemma. According to the lemma, the most powerful test is obtained by comparing the likelihood ratio with a threshold. Let's go through the steps to find the power of the test.

Step 1: Formulating the likelihood ratio:
The likelihood ratio is defined as the ratio of the likelihood of the alternative hypothesis to the likelihood of the null hypothesis. In this case, the likelihood ratio is given by:

LR(x) = L(H1) / L(H0)

Since we have a continuous random variable, we can rewrite the likelihood ratio as:

LR(x) = f1(x) / f0(x)

where f1(x) is the probability density function (pdf) of H1 and f0(x) is the pdf of H0.

Step 2: Finding the threshold:
To determine the threshold for the most powerful test, we need to specify the size of the test, which is α = 0.1 in this case. The size of the test is the maximum probability of rejecting the null hypothesis when it is true. Mathematically, it is given by:

α = max[Pr(reject H0 | H0 is true)]

Step 3: Calculating the power:
The power of the test is the probability of rejecting the null hypothesis when the alternative hypothesis is true. Mathematically, it is given by:

Power = Pr(reject H0 | H1 is true)

Step 4: Applying the likelihood ratio test:
To find the power, we need to compare the likelihood ratio with the threshold. If the likelihood ratio is greater than the threshold, we reject the null hypothesis; otherwise, we fail to reject the null hypothesis.

In this case, the likelihood ratio is:

LR(x) = (4x^2) / (2x) = 2x

To determine the threshold, we need to find the value of x such that the probability of observing a random variable greater than or equal to x under the null hypothesis is α. Since the null hypothesis is H0: f(X) = 2x, we can integrate the pdf f0(x) from x to 1 to find this probability.

∫[x to 1] 2x dx = [x^2] [x to 1] = 1 - x^2

Setting 1 - x^2 = α = 0.1, we have:

x^2 = 0.9

x = √0.9

Therefore, the threshold value is √0.9.

Now, we can calculate the power by finding the probability of the likelihood ratio being greater than the threshold.

Power = Pr(LR(x) > √0.9 | H1 is true)

= Pr(2x > √0.9