Math Problem Statement
solve l
Solution
Let's carefully solve this problem step by step. We'll tackle the sub-questions systematically.
(a) Find the values of and .
From the problem, the joint probability mass function is a table of probabilities. The total probability must sum to 1:
Substituting the values in the table:
Simplify:
\tag{1}$$ Additionally, the expected value of $$X$$, $$E(X)$$, is given as 1: $$E(X) = \sum_{X=0}^{2} X \cdot P(X)$$ Calculate $$P(X)$$, the marginal probabilities for $$X = 0, 1, 2$$: $$P(X=0) = 0.15 + 0.05 + 0 = 0.2$$ $$P(X=1) = 0.25 + 0.3 + \beta = 0.55 + \beta$$ $$P(X=2) = 0.1 + \alpha + 0.1 = 0.2 + \alpha$$ The expectation is: $$E(X) = 0 \cdot 0.2 + 1 \cdot (0.55 + \beta) + 2 \cdot (0.2 + \alpha) = 1$$ Simplify: $$0 + 0.55 + \beta + 0.4 + 2\alpha = 1$$ $$0.95 + \beta + 2\alpha = 1$$ $$\beta + 2\alpha = 0.05 \tag{2}$$ Solve the system of equations (1) and (2): From (1): $$\beta = -\alpha$$ Substitute into (2): $$-\alpha + 2\alpha = 0.05$$ $$\alpha = 0.05$$ Substitute $$\alpha = 0.05$$ into (1): $$\alpha + \beta = 0 \implies 0.05 + \beta = 0 \implies \beta = -0.05$$ Thus: $$\alpha = 0.05, \quad \beta = -0.05$$ --- ### (b) Find $$P(X < Y)$$. Here, we sum the probabilities where $$X < Y$$: $$P(X < Y) = P(X=0, Y=1) + P(X=0, Y=2) + P(X=1, Y=2)$$ From the table: $$P(X < Y) = 0.05 + 0 + (-0.05) = 0$$ --- ### (c) Find $$E(Y)$$. The expectation of $$Y$$ is: $$E(Y) = \sum_{Y=0}^{2} Y \cdot P(Y)$$ First, calculate the marginal probabilities for $$Y = 0, 1, 2$$: $$P(Y=0) = 0.15 + 0.25 + 0.1 = 0.5$$ $$P(Y=1) = 0.05 + 0.3 + 0.05 = 0.4$$ $$P(Y=2) = 0 + (-0.05) + 0.1 = 0.05$$ Now: $$E(Y) = 0 \cdot 0.5 + 1 \cdot 0.4 + 2 \cdot 0.05 = 0 + 0.4 + 0.1 = 0.5$$ --- ### (d) Find $$\text{Skew}(X)$$. The skewness formula is: $$\text{Skew}(X) = \frac{E[(X - \mu_X)^3]}{\sigma_X^3}$$ We need $$\mu_X = E(X) = 1$$ and $$\sigma_X^2 = \text{Var}(X)$$. To calculate $$\text{Var}(X)$$, we first find $$E(X^2)$$: $$E(X^2) = \sum_{X=0}^{2} X^2 \cdot P(X)$$ $$E(X^2) = 0^2 \cdot 0.2 + 1^2 \cdot (0.55 - 0.05) + 2^2 \cdot (0.2 + 0.05)$$ $$E(X^2) = 0 + 1 \cdot 0.5 + 4 \cdot 0.25 = 0 + 0.5 + 1 = 1.5$$ Now: $$\text{Var}(X) = E(X^2) - [E(X)]^2 = 1.5 - 1^2 = 0.5$$ Thus: $$\sigma_X = \sqrt{0.5}$$ Now calculate $$E[(X - \mu_X)^3]$$: $$E[(X - \mu_X)^3] = \sum_{X=0}^{2} (X - 1)^3 \cdot P(X)$$ $$= (-1)^3 \cdot 0.2 + 0^3 \cdot 0.5 + 1^3 \cdot 0.25$$ $$= -0.2 + 0 + 0.25 = 0.05$$ Finally: $$\text{Skew}(X) = \frac{0.05}{(0.5)^{3/2}} = \frac{0.05}{\sqrt{0.125}} = \frac{0.05}{0.3536} \approx 0.1413$$ --- Would you like to continue with the remaining parts or need clarification on these solutions? Here are **5 follow-up questions**: 1. What does the skewness value tell us about the distribution of $$X$$? 2. How can we verify the independence of $$X$$ and $$Y$$ from the given data? 3. What is the significance of $$\alpha$$ and $$\beta$$ in the joint PMF? 4. How does $$P(X < Y)$$ being 0 impact the relationship between $$X$$ and $$Y$$? 5. Can we interpret the marginal PMFs for $$X$$ and $$Y$$ as being symmetric or not? **Tip**: Always ensure your probabilities sum to 1 when working with a joint PMF!
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Conditional Probability
Cumulative Distribution Function (CDF)
Formulas
Conditional Probability: P(X | Y) = P(X, Y) / P(Y)
Cumulative Distribution Function: F_X(x) = P(X ≤ x)
Conditional CDF: F_{X | Y}(x) = P(X ≤ x | Y = y)
Theorems
Law of Total Probability
Definition of Joint and Marginal Distributions
Suitable Grade Level
Undergraduate (Statistics and Probability Course)
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