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+

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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
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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