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Chapter 4: Problem 13

Suppose a random sample of size 2 is obtained from a distribution that has pdf \(f(x)=2(1-x), 0

Short Answer

Expert verified
The required probability is \(1/3\).

Step by step solution

01

Recognize the task and integrate the function within the right bounds

Since we sample 2 observations \(X\) and \(Y\) from the considered distribution, we have:- For \(X < Y/2\): \(\int_0^{2x} \int_x^{y/2} 2(1-x) \cdot 2(1-y) dx dy\)- For \(Y < X/2\): \(\int_0^{2y} \int_y^{x/2} 2(1-x) \cdot 2(1-y) dx dy\)Notice that for \(X < Y/2\) we integrate \(x\) first from 0 to \(y/2\), then \(y\) from \(2x\) to 1, and the same type of bounds for \(Y < X/2\) changing appropriately the variables.
02

Calculate the integrals

Proceeding with the double integral computation (for both cases since they are symmetrical, the result will be multiplied by 2):\[(1/4)\int_0^{1} (1-y)^2 dy = (1/4) \left[ y - y^2 + (1/3)y^3 \right]_0^{1}\] Since the integrals are symmetrical, the result of this integral is multiplied by two, yielding the total probability.
03

Solve the equation

Substitute the upper and lower bounds into the antiderivative. Evaluate the expression to get the final result: \((1/4)[ 1 - 1 + (1/3) ]*2 = (1/6)*2\)

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Most popular questions from this chapter

Prove the converse of Theorem MCT. That is, let \(X\) be a random variable with a continuous cdf \(F(x)\). Assume that \(F(x)\) is strictly increasing on the space of \(X .\) Consider the random variable \(Z=F(X)\). Show that \(Z\) has a uniform distribution on the interval \((0,1)\)

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