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Chapter 6: Problem 44

If \(X_{1}, X_{2}, X_{3}\) are independent random variables that are uniformly distributed over \((0,1),\) compute the probability that the largest of the three is greater than the sum of the other two.

Short Answer

Expert verified
The probability that the largest of the three independent random variables is greater than the sum of the other two is \(\frac{3}{4}\).

Step by step solution

01

Set up the inequality

We want to find the probability that the largest of \(X_1, X_2, X_3\) is greater than the sum of the other two. Let's denote the largest value as \(X_{max}\). We can write the inequality as: \(X_{max} > X_{1} + X_{2} + X_{3} - X_{max}\) This simplifies to: \(X_{max} > \frac{1}{2}(X_{1} + X_{2} + X_{3})\)
02

Solve the inequality for different cases

There are three potential cases depending on which of the variables is the largest. We will solve the inequality for each of these cases. Case 1: \(X_1\) is the largest. In this case, the inequality becomes: \(X_{1} > \frac{1}{2}(X_{2} + X_{3})\) Case 2: \(X_2\) is the largest. In this case, the inequality becomes: \(X_{2} > \frac{1}{2}(X_{1} + X_{3})\) Case 3: \(X_3\) is the largest. In this case, the inequality becomes: \(X_{3} > \frac{1}{2}(X_{1} + X_{2})\)
03

Define the region in the 3-dimensional space

For each case above, we can define a region in the 3-dimensional space \((X_1, X_2, X_3)\) that satisfies the inequality. Case 1: The region is defined by: \(X_1 > X_2, X_1 > X_3, X_1 > \frac{1}{2}(X_2 + X_3)\) Case 2: The region is defined by: \(X_2 > X_1, X_2 > X_3, X_2 > \frac{1}{2}(X_1 + X_3)\) Case 3: The region is defined by: \(X_3 > X_1, X_3 > X_2, X_3 > \frac{1}{2}(X_1 + X_2)\)
04

Integrate over the defined region

To find the probability, we need to integrate the probability density function over the given region. Since the random variables are uniformly distributed in \((0,1)\), their joint probability density function is: \(f(X_{1}, X_{2}, X_{3}) = 1\) So, the probability for each case is the volume of the region defined above. We will integrate the function over the defined region in each case. The probability we are looking for is the sum of probabilities from all three cases. Case 1: \(\int_0^1 \int_0^{x_1} \int_0^{2x_1 - x_2} 1 dx_3 dx_2 dx_1\) Case 2: \(\int_0^1 \int_0^{x_2} \int_0^{2x_2 - x_1} 1 dx_3 dx_1 dx_2\) Case 3: \(\int_0^1 \int_0^{x_3} \int_0^{2x_3 - x_1} 1 dx_2 dx_1 dx_3\)
05

Calculate the probability

Compute the integrals to get the probability for each case. Add them up to get the required probability. \(P = \int_0^1 \int_0^{x_1} \int_0^{2x_1 - x_2} 1 dx_3 dx_2 dx_1 + \int_0^1 \int_0^{x_2} \int_0^{2x_2 - x_1} 1 dx_3 dx_1 dx_2 + \int_0^1 \int_0^{x_3} \int_0^{2x_3 - x_1} 1 dx_2 dx_1 dx_3\) \(P = \frac{1}{4} + \frac{1}{4} + \frac{1}{4}\) \(P = \frac{3}{4}\) The probability that the largest of the three independent random variables is greater than the sum of the other two is \(\frac{3}{4}\).

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