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Let \(Y_{1}

Short Answer

Expert verified
The joint pdf of \(Y_3\) and \(Y_4\) is \(f(y_3, y_4) = 360(y_3^2)(y_4 - y_3)(1 - y_4)y_3y_4(2y_4)(1-y_4)^2\), the conditional pdf of \(Y_3\) given \(Y_4 = y_4\) is \(\frac{30}{y_4}y_3(y_4 - y_3)\), and \(E(Y_3 | y_4) = \frac{1}{4}y_4\).

Step by step solution

01

Joint pdf of Y3 and Y4

To calculate the joint pdf of \(Y_3\) and \(Y_4\), one can use the formula for the joint pdf of two order statistics which turns out to be \(f(y_3, y_4) = n!(f(y_3)F(y_3))^{k-1}(f(y_4)(1 - F(y_4)))^{n-k}(1 - F(y_3))^{k-j}(F(y_4) - F(y_3))^{j-k-1}\), where \(F(x)\) is the CDF (Cumulative Distribution Function) of the distribution, and \(k\) and \(j\) are the order statistics. Here \(k=3\) and \(j=4\), and from \(f(x) = 2x\), we get \(F(x) = x^2\). Substituting these values, the joint pdf turns out to be \(f(y_3, y_4) = 360(y_3^2)(y_4 - y_3)(1 - y_4)y_3y_4(2y_4)(1-y_4)^2\).
02

Conditional pdf of Y3 given Y4=y4

The conditional pdf of \(Y_3\) given \(Y_4 = y_4\), can be found by treating \(Y_4 = y_4\) as a constant in the joint pdf. This implies that the integral of the joint pdf over all possible values of \(Y_3\) should be one. This gives us the conditional pdf \(f(y_3|y_4) = f(y_3,y_4) / \int_0^{y_4} f(y_3,y_4) \,dy_3\). Simplifying, we get \(f(y_3|y_4) = 30y_3*(y_4 - y_3) * (1 - y_4)/\int_0^{y_4} 30y_3*(y_4 - y_3) * (1 - y_4) \,dy_3\). On further simplification, this can be written as \(\frac{30}{y_4}y_3(y_4 - y_3)\).
03

Conditional expectation E(Y3 | Y4)

The conditional expectation \(E(Y_3 | y_4)\), is given by the integral of \(y_3\) multiplied by the conditional pdf, integrated over all possible values of \(y_3\). That is, \(E(Y_3 | y_4) = \int_0^{y_4} y_3 * f(y_3 | y_4) \,dy_3\). Substituting the expression obtained in step 2 and calculating the integral, we get \(E(Y_3 | y_4) = \frac{1}{4}y_4\).

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Most popular questions from this chapter

Let \(X_{1}, X_{2}\) be a random sample of size \(n=2\) from the distribution having pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

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