Question

Let \(X_{1}, X_{2}, X_{3}\) be a random sample from a distribution of the continuous type having pdf \(f(x)=2 x, 0

Short answer

The probability that the smallest of the random variables exceeds the median of the distribution is approximately 0.0005. The correlation between the second and third order statistic is 0.25.

Step-by-step solution

Understanding The Probability Density Function

The given probability density function is \(f(x) = 2x\) for \(0 < x < 1\) and zero elsewhere. This density function is for a continuous type random variable and it represents the probability of the random variable taking on a specific value. It is a legitimate pdf as it is non-negative and integrates to 1 over its valid range. The median of this distribution can be found by solving \(∫_0^m f(x) dx = 0.5\) which represents the cumulative distribution function.

Compute the median

Solving the equation, the integrand becomes \(∫_0^m 2x dx = [x^2]_0^m = m^2\). Equating this to 0.5 and solving for m we get \(m = √(0.5) = 0.707\) which is the median of the distribution.

Probability that smallest exceeds median

The probability that the smallest \(X_i\) exceeds the median is equivalent to all \(X_i\) exceeding the median. Hence, we calculate the probability that any variable exceeds the median. For \(X_i > 0.707\), the pdf is \(f(x)=(1-x^2)\). The total probability is \(∫_0.707^1 (1 - x^2) dx = [x - (1/3)x^3]_0.707^1 = 1 - (1/3) - (0.707- (1/3)*0.707^3) ≈ 0.079\) . Since we have three such independent variables, the total probability that all exceeds median is \(0.079^3 = 0.0005\).

Correlation between order statistics

Correlation between order statistics \(Y_{2}\) and \(Y_{3}\) is given by \(ρ_{Y2,Y3} = 1/(n+1) = 1/4 = 0.25\) where \(n=3\) is the sample size.