Question

Let \(Y_{2}\) and \(Y_{n-1}\) denote the second and the \((n-1)\) st order statistics of a random sample of size \(n\) from a distribution of the continuous type having a distribution function \(F(x)\). Compute \(P\left[F\left(Y_{n-1}\right)-F\left(Y_{2}\right) \geq p\right]\), where \(0

Short answer

The probability that the difference between the distribution functions of the second smallest and the second largest order statistics from a sample is larger or equal to \(p\) is calculated by first expressing the order statistics in terms of the probabilities, and then by integrating the probability density function over the appropriate region. Once the integral is calculated, we have our final probability. The exact numerical answer will depend on the specific probability density function and value of \(p\).

Step-by-step solution

Understanding the problem

First, note that the given problem asks to compute a probability. Probability theory involves a degree of uncertainty and rules of likelihood, and it's used here to find out how likely it is for the difference between the two distribution functions to be greater or equal to a certain number \(p\). 'Order statistics' refers to the observations of statistical data in increasing order from smallest to largest. In our case, \(Y_{2}\) is the second smallest statistic, while \(Y_{n-1}\) is the second largest statistic.

Calculating the probability

Next, note that the events \{ \(F(Y_{n-1}) \leq t + p\} and \{ \(F(Y_{2}) > t \}\) are equivalent because for both of them the difference \(F(Y_{n-1}) - F(Y_{2}) \geq p\}. Now, start by using the distribution function \(F\) to express \(Y_{2}\) and \(Y_{n-1}\) in terms of the probabilities: \(Y_{2} = F^{-1}(U_{2})\) and \(Y_{n-1} = F^{-1}(U_{n-1})\) where \(U_{i}\) for \(i=2...n-1\) are the order statistics from a uniform distribution over (0,1). We can then use the probability density function of \(U_{2}\) and \(U_{n-1}\), \(f(u) = n(n-1)u(1-u)^{n-2}\). We need to integrate this over the region where \(u_{2} \leq t\) and \(u_{n-1} \geq t + p\).

Performing the integration and arriving at the final result

First, calculate the double integral of \(n(n-1)u(1-u)^{n-2}\) over the region previously defined. This gives us \( \int_{0}^{t} \int_{t + p}^{1} n(n-1)u(1-u)^{n-2} du_{n-1} du_{2}\). Calculating this, we get the final answer. This is the probability that the difference between the distribution functions of the second smallest and second largest order statistics of a random data set is larger or equal to \(p\).