Question

If \(U \sim U(0,1)\), show that \(\min (U, 1-U) \sim U\left(0, \frac{1}{2}\right)\). Hence justify the computation of a two-sided significance level as \(2 \min \left(P^{-}, P^{+}\right)\).

Short answer

\(\min(U, 1-U)\) follows a uniform distribution \(U(0, \frac{1}{2})\). The two-sided significance level is computed as \(2 \min(P^{-}, P^{+})\).

Step-by-step solution

Define the Random Variables

We start by considering two random variables: \(U\) which is uniformly distributed over \((0, 1)\) and \(1-U\). Our task is to find the distribution of \(\min(U, 1-U)\). This minimum will take values between 0 and 0.5, since the highest it can be is when \(U = 0.5\) and \(1-U = 0.5\).

Determine the Cumulative Distribution Function (CDF)

To find the probability distribution, we compute the cumulative distribution function (CDF) of \(\min(U, 1-U)\). For any \(x\) in \((0, 0.5)\), the CDF \(F(x)\) is the probability that \(\min(U, 1-U) \leq x\). This occurs if either \(U \leq x\) or \(1-U \leq x\). Thus, \(F(x) = P(U \leq x) + P(1-U \leq x) - P(U \leq x \text{ and } 1-U \leq x)\).

Calculate Probabilities in CDF Expression

- \(P(U \leq x) = x\) since \(U\) is uniformly distributed over \((0,1)\).- \(P(1-U \leq x) = P(U \geq 1-x) = 1-(1-x) = x\) since \(U\) is uniform.- \(P(U \leq x \text{ and } 1-U \leq x) = P(U \leq x \text{ and } U \geq 1-x) = 0\) because both conditions cannot hold simultaneously.Therefore, the CDF is \(F(x) = x + x = 2x\) for \(x < 0.5\).

Verify Uniform Distribution

Given the CDF \(F(x) = 2x\), we can confirm the probability density function (PDF) by differentiating the CDF, which leads to \(f(x) = \frac{d}{dx}F(x) = 2\), showing a constant PDF which is characteristic of a uniform distribution. Therefore, \(\min(U, 1-U)\) is uniformly distributed over \((0, 0.5)\).

Justify the Two-Sided Significance Level Calculation

In hypothesis testing, calculating a two-sided significance level involves computing the extreme probabilities of \(\min\) between the observed and expected values. Thus, using symmetry, the significance level can be calculated as \(2 \min(P^{-}, P^{+})\), where \(P^{-}\) and \(P^{+}\) represent the left and right tail probabilities of the distribution, ensuring all extreme outcomes are considered.

Key concepts

Cumulative Distribution Function

The cumulative distribution function (CDF) is a fundamental concept in probability theory used to describe the probability that a random variable is less than or equal to a certain value. It's an essential tool because it encapsulates the entire probability distribution of a random variable, thereby allowing us to easily understand all possible outcomes and their likelihoods. For a uniform distribution, the CDF is particularly straightforward.

In this exercise, we are examining the minimum of two random variables, \(U\) and \(1-U\), both stemming from a uniform distribution over \((0, 1)\). To find the CDF of \(\min(U, 1-U)\), we calculate the probability that \(\min(U, 1-U) \leq x\). This condition implies that either \(U\) or \(1-U\) must both be less than or equal to \(x\).

To derive this, we add the probabilities of each individual condition and subtract the overlap because they cannot both hold true at the same time. Thus, we compute: \(F(x) = P(U \leq x) + P(1-U \leq x) - P(U \leq x \text{ and } 1-U \leq x)\). Since \(U\) is uniformly distributed, these probabilities simplify to \(x\), leading to the CDF, \(F(x) = 2x\) for \(x \in (0, 0.5)\).

This derived function, \(F(x) = 2x\), implies that as \(x\) increases from 0 to 0.5, the likelihood of \(\min(U, 1-U)\) being less than or equal to \(x\) increases linearly, confirming the uniform nature of the distribution.

Probability Density Function

Once we have the cumulative distribution function (CDF) for a random variable, we can derive its probability density function (PDF). The PDF is an essential function in probability theory that describes the likelihood of a random variable to take on a particular value. It is especially crucial for continuous random variables, like those uniformly distributed.

In our scenario with the variable \(\min(U, 1-U)\), we already determined that the CDF is \(F(x) = 2x\) for \(x \in (0, 0.5)\). To find the PDF, we take the derivative of the CDF with respect to \(x\).

Calculating the derivative, we find that the PDF, \(f(x) = \frac{d}{dx}F(x) = 2\) for \(x < 0.5\). This constant value of the PDF illustrates a uniform distribution over \((0, 0.5)\).

In a uniform distribution, each value within the given range is equally probable. Therefore, for \(\min(U, 1-U)\), any value between 0 and 0.5 is equally likely to occur, underscoring the characteristic feature of a uniform distribution. This uniformity is crucial for ensuring consistency in random experiments, leading to more refined predictions and analyses.

Hypothesis Testing

Hypothesis testing is a statistical method used to decide the plausibility of a hypothesis based on sample data. It is a core component of statistical inference, aiming to determine whether observed data falls within a pre-determined range of expected outcomes, under a specific hypothesis.

In the context of the exercise, hypothesis testing requires computing a two-sided significance level. This concept involves calculating the probabilities associated with extreme deviations of the test statistic—either higher or lower than expected.

For our uniformly distributed variable \(\min(U, 1-U)\), we compute the significance level by evaluating the tail probabilities. We denote these as \(P^{-}\) and \(P^{+}\), representing the left and right tails, respectively. The significance level for a two-sided hypothesis test is determined as \(2 \min(P^{-}, P^{+})\).

This calculation ensures that we comprehensively account for extreme outcomes that could reject the null hypothesis. Using this symmetry-inspired method, we are able to consider all potential deviations of the statistic, thus improving the robustness of the hypothesis test.