91Ó°ÊÓ

The joint pdf of two order statistics \(Y_{j}\) and \(Y_{k}\) from a continuous pdf \(f(x)\) is given by \(f_{Y_{j}, Y_{k}}(y_{j}, y_{k}) = n!/[j!(k-j)!(n-k)!] \cdot f(y_{j})f(y_{k})(F(y_{k}) - F(y_{j}))^{k-j-1}\cdot[F(y_{j})]^{j-1}\cdot[1 - F(y_{k})]^{n-k}\). Here \(f(x) = 2x\) which is the given pdf and \(F(x)\) is the corresponding cumulative distribution function (CDF) which can be computed by integrating \(f(x)\) from 0 to x, resulting in \(F(x) = x^2\). Substituting \(j = 3\), \(k = 4\), and corresponding \(f(x)\) and \(F(x)\) values in the formula, we obtain the joint pdf \(f_{Y_{3}, Y_{4}}(y_{3}, y_{4})\). It is also needed to ensure \(y_{j} \leq y_{k}\) and both order statistics fall in the given domain.

Once we find the joint pdf, we can compute the conditional pdf. The conditional pdf of \(Y_{j}\) given that \(Y_{k} = y_{k}\) is given by \(f_{Y_{j}|Y_{k}}(y_{j}|y_{k}) = f_{Y_{j}, Y_{k}}(y_{j}, y_{k}) / f_{Y_{k}}(y_{k})\). To quickly compute \(f_{Y_{k}}(y_{k})\), we use the formula for the pdf of the \(k\)th order statistic, that is \(f_{Y_{k}}(y_{k}) = n!/[k!(n-k)!] \cdot f(y_{k}) \cdot [F(y_{k})]^{k-1} \cdot [1 - F(y_{k})]^{n-k}\). Using the found joint pdf in the numerator and the computed \(f_{Y_{k}}(y_{k})\) in the denominator, we get the conditional pdf \(f_{Y_{3}|Y_{4}}(y_{3}|y_{4})\), ensuring \(0 < y_{3} < y_{4} < 1\).

Finally, we compute the conditional expectation \(E\left(Y_{3} | y_{4}\right)\) using the conditional pdf obtained in step 2. This is computed by integrating over the domain \(0 \leq y_{3} \leq y_{4}\), the product \(y_{3} \cdot f_{Y_{3}|Y_{4}}(y_{3}|y_{4})\). Note that this integral must be evaluated from 0 to \(y_{4}\), since the variable \(Y_{3}\) should be less than given \(y_{4}\).