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 three samples exceeds the median is 0.125. The correlation between \(Y_{2}\) and \(Y_{3}\) is \(sqrt(35/6) / 84\).

Step-by-step solution

Find the Median

The median can be found using the cumulative distribution function (CDF) which is the integral of the probability density function (pdf). The cumulative distribution function \(F(x) = \int_0^x 2t dt = x^2 \) by integrating from \(0\) to \(x\) . Since we need the median, let's say that is \(m\), and we want \(F(m) = 0.5 \). By setting \(x^2 = 0.5\), we get \(m = sqrt(0.5)\) which simplifies to \(m = 1 / sqrt(2)\).

Compute the Probability

The smallest of three observations exceeding the median means all three observations must exceed the median. The probability that one observation is greater than the median is \(1 - F(m) = 1 - 0.5 = 0.5\). Since the observations are independent, the desired probability can obtained by cubing the above, which gives \(0.5^3 = 0.125\).

Compute the Correlation

To find the correlation between \(Y_{2}\) and \(Y_{3}\), we need to know their covariances and their individual variances. For the given pdf, the covariance \(Cov(Y_{2}, Y_{3}) = \int_0^1 \int_0^y 2xy*2x dx dy - E[Y_2]E[Y_3] = 5/36 - (5/6)*(4/7) = 1/252\). The variance of \(Y_{2}\) can be calculated as \(Var(Y_{2}) = E[Y_{2}^2] - E[Y_{2}]^2 = \int_0^1 y^2 2y dy - (5/6)^2= 7/60 - 25/36 = 1/120\). Similarly, \(Var(Y_{3}) = 2/35\). The correlation coefficient is therefore \(Corr(Y_{2}, Y_{3}) = Cov(Y_{2}, Y_{3}) / sqrt(Var(Y_{2})Var(Y_{3})) = 1/252 / sqrt((1/120)*(2/35)) = sqrt(35/6) / 84\).