Math Problem Statement
Solution
Let's solve each part of the question:
Given Information
We have two independent random variables and with their joint probability distribution shown in the table. Specifically:
- Values for : 2 and 4
- Values for : 1, 3, and 5
The joint probabilities are given as follows:
f(x, y) & X=2 & X=4 \\ \hline Y=1 & 0.10 & 0.15 \\ Y=3 & 0.20 & 0.30 \\ Y=5 & 0.10 & 0.15 \\ \end{array}$$ #### Part (a): $$E(2X - 3Y)$$ The expected value $$E(2X - 3Y)$$ can be computed by breaking it down into: $$E(2X - 3Y) = 2E(X) - 3E(Y)$$ Since $$X$$ and $$Y$$ are independent, we can find $$E(X)$$ and $$E(Y)$$ separately using their marginal distributions. 1. **Calculate $$E(X)$$:** $$E(X) = \sum_{x} x \cdot P(X=x)$$ where - $$P(X=2) = 0.10 + 0.20 + 0.10 = 0.40$$ - $$P(X=4) = 0.15 + 0.30 + 0.15 = 0.60$$ Then, $$E(X) = 2 \cdot 0.40 + 4 \cdot 0.60 = 0.80 + 2.4 = 3.2$$ 2. **Calculate $$E(Y)$$:** $$E(Y) = \sum_{y} y \cdot P(Y=y)$$ where - $$P(Y=1) = 0.10 + 0.15 = 0.25$$ - $$P(Y=3) = 0.20 + 0.30 = 0.50$$ - $$P(Y=5) = 0.10 + 0.15 = 0.25$$ Then, $$E(Y) = 1 \cdot 0.25 + 3 \cdot 0.50 + 5 \cdot 0.25 = 0.25 + 1.5 + 1.25 = 3.0$$ 3. **Calculate $$E(2X - 3Y)$$:** $$E(2X - 3Y) = 2 \cdot 3.2 - 3 \cdot 3.0 = 6.4 - 9 = -2.6$$ #### Part (b): $$E(XY)$$ Since $$X$$ and $$Y$$ are independent, $$E(XY) = E(X) \cdot E(Y)$$. Using the values calculated above: $$E(XY) = E(X) \cdot E(Y) = 3.2 \cdot 3.0 = 9.6$$ ### Final Answers 1. $$E(2X - 3Y) = -2.6$$ 2. $$E(XY) = 9.6$$ Would you like further explanation or details on any step? --- Here are some related questions that may deepen your understanding: 1. How would you find the variance of $$X$$ and $$Y$$ using the marginal distributions? 2. What is the significance of $$X$$ and $$Y$$ being independent in this problem? 3. How would the calculation change if $$X$$ and $$Y$$ were not independent? 4. Can you calculate $$\text{Var}(2X - 3Y)$$ using the variances of $$X$$ and $$Y$$? 5. How would we approach this problem if it were given as a continuous joint distribution? **Tip:** When variables are independent, $$E(XY) = E(X) \cdot E(Y)$$ simplifies the calculation significantly, as we can treat their expectations separately.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Expected Value
Joint Probability Distribution
Independence of Random Variables
Formulas
E(aX + bY) = aE(X) + bE(Y)
E(XY) = E(X) * E(Y) (for independent X and Y)
Theorems
Linearity of Expectation
Expected Value of Product of Independent Variables
Suitable Grade Level
College/University
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