Let X1, X2, and X3 be independent random variables such that E(X1) ‚E(X2) and E(X3) Further, suppose that Var(X1) = 5, Var(X2) = 10, and Var(X3) = 15. 1.) Compute E(TX1 + 4X2 + V3X3+ 15) 2.) Compute Var(/2X1 + V3X2 + 7X3+
Let X1, X2, and X3 be independent random variables such that E(X1) ‚E(X2) and E(X3) Further, suppose that Var(X1) = 5, Var(X2) = 10, and Var(X3) = 15. 1.) Compute E(TX1 + 4X2 + V3X3+ 15) 2.) Compute Var(/2X1 + V3X2 + 7X3+
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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![Let's consider three independent random variables \(X_1\), \(X_2\), and \(X_3\). The expected values and variances of these random variables are given as follows:
- \(E(X_1) = \frac{7}{8}\)
- \(E(X_2) = \frac{5}{9}\)
- \(E(X_3) = \frac{4}{7}\)
Further, the variances are given by:
- \(Var(X_1) = 5\)
- \(Var(X_2) = 10\)
- \(Var(X_3) = 15\)
We are tasked with computing two quantities:
1. \(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15)\)
2. \(Var(\sqrt{2}X_1 + \sqrt{3}X_2 + 7X_3 + \frac{4}{5})\)
### Solutions:
1. **Expected Value Calculation:**
To compute \(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15)\), we use the linearity property of expectation:
\(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15) = \pi E(X_1) + 4E(X_2) + \sqrt{3}E(X_3) + E(15)\)
Since the expected value of a constant is the constant itself, \(E(15) = 15\):
So,
\[
E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15) = \pi \cdot \frac{7}{8} + 4 \cdot \frac{5}{9} + \sqrt{3} \cdot \frac{4}{7} + 15
\]
2. **Variance Calculation:**
To compute \(Var(\sqrt{2}X_1 + \sqrt{3}X_2 + 7X_3 + \frac{4}{5})\), we use the fact that the variance of a constant is zero and that for independent random variables \(X\) and \(Y\), \(Var(aX + bY](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45fdd75f-924d-4ac4-9d15-03fc1f26bc02%2Fdfda5e3e-8b01-4739-b8ee-01ce51cdc776%2Fo6hz39ge_processed.png&w=3840&q=75)
Transcribed Image Text:Let's consider three independent random variables \(X_1\), \(X_2\), and \(X_3\). The expected values and variances of these random variables are given as follows:
- \(E(X_1) = \frac{7}{8}\)
- \(E(X_2) = \frac{5}{9}\)
- \(E(X_3) = \frac{4}{7}\)
Further, the variances are given by:
- \(Var(X_1) = 5\)
- \(Var(X_2) = 10\)
- \(Var(X_3) = 15\)
We are tasked with computing two quantities:
1. \(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15)\)
2. \(Var(\sqrt{2}X_1 + \sqrt{3}X_2 + 7X_3 + \frac{4}{5})\)
### Solutions:
1. **Expected Value Calculation:**
To compute \(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15)\), we use the linearity property of expectation:
\(E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15) = \pi E(X_1) + 4E(X_2) + \sqrt{3}E(X_3) + E(15)\)
Since the expected value of a constant is the constant itself, \(E(15) = 15\):
So,
\[
E(\pi X_1 + 4X_2 + \sqrt{3}X_3 + 15) = \pi \cdot \frac{7}{8} + 4 \cdot \frac{5}{9} + \sqrt{3} \cdot \frac{4}{7} + 15
\]
2. **Variance Calculation:**
To compute \(Var(\sqrt{2}X_1 + \sqrt{3}X_2 + 7X_3 + \frac{4}{5})\), we use the fact that the variance of a constant is zero and that for independent random variables \(X\) and \(Y\), \(Var(aX + bY
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