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In probability theory and statistics, covariance is a fundamental measure that quantifies the degree to which two random variables change together. When two discrete random variables X and Y are defined over finite sample spaces, their joint probability mass function (PMF) completely characterizes their probabilistic relationship.
Given the joint PMF represented as a 2D array where each entry joint_pmf[i][j] gives the probability P(X = x_values[i], Y = y_values[j]), your task is to compute the covariance between X and Y.
The covariance between X and Y is defined as:
$$\text{Cov}(X, Y) = E[XY] - E[X] \cdot E[Y]$$
Where:
To compute these expectations from the joint PMF, you must first derive the marginal distributions:
$$P(X = x_i) = \sum_{j} P(X = x_i, Y = y_j)$$ $$P(Y = y_j) = \sum_{i} P(X = x_i, Y = y_j)$$
Then: $$E[X] = \sum_{i} x_i \cdot P(X = x_i)$$ $$E[Y] = \sum_{j} y_j \cdot P(Y = y_j)$$ $$E[XY] = \sum_{i} \sum_{j} x_i \cdot y_j \cdot P(X = x_i, Y = y_j)$$
Your Task: Write a Python function that computes the covariance between two discrete random variables X and Y, given their possible values and joint PMF.
x_values = [0, 1]
y_values = [0, 1]
joint_pmf = [[0.4, 0.1], [0.1, 0.4]]0.15Step 1: Compute Marginal Probabilities • P(X=0) = 0.4 + 0.1 = 0.5 • P(X=1) = 0.1 + 0.4 = 0.5 • P(Y=0) = 0.4 + 0.1 = 0.5 • P(Y=1) = 0.1 + 0.4 = 0.5
Step 2: Compute Expected Values • E[X] = 0×0.5 + 1×0.5 = 0.5 • E[Y] = 0×0.5 + 1×0.5 = 0.5
Step 3: Compute E[XY] • E[XY] = (0×0×0.4) + (0×1×0.1) + (1×0×0.1) + (1×1×0.4) = 0.4
Step 4: Calculate Covariance • Cov(X,Y) = E[XY] - E[X]×E[Y] = 0.4 - 0.5×0.5 = 0.4 - 0.25 = 0.15
The positive covariance indicates that X and Y tend to increase together.
x_values = [1, 2, 3]
y_values = [1, 2, 3]
joint_pmf = [[0.111111, 0.111111, 0.111111], [0.111111, 0.111111, 0.111111], [0.111111, 0.111111, 0.111111]]0.0This represents a uniform joint distribution where each combination has equal probability (1/9 ≈ 0.111111).
Step 1: Marginal Probabilities Each value of X and Y has marginal probability 1/3.
Step 2: Expected Values • E[X] = 1×(1/3) + 2×(1/3) + 3×(1/3) = 2 • E[Y] = 1×(1/3) + 2×(1/3) + 3×(1/3) = 2
Step 3: Compute E[XY] E[XY] = (1/9) × (1×1 + 1×2 + 1×3 + 2×1 + 2×2 + 2×3 + 3×1 + 3×2 + 3×3) = (1/9) × (1 + 2 + 3 + 2 + 4 + 6 + 3 + 6 + 9) = 36/9 = 4
Step 4: Covariance Cov(X,Y) = 4 - 2×2 = 0
Zero covariance indicates that X and Y are statistically independent in this uniform distribution.
x_values = [0, 1]
y_values = [0, 1]
joint_pmf = [[0.1, 0.4], [0.4, 0.1]]-0.15Step 1: Compute Marginal Probabilities • P(X=0) = 0.1 + 0.4 = 0.5 • P(X=1) = 0.4 + 0.1 = 0.5 • P(Y=0) = 0.1 + 0.4 = 0.5 • P(Y=1) = 0.4 + 0.1 = 0.5
Step 2: Compute Expected Values • E[X] = 0.5, E[Y] = 0.5
Step 3: Compute E[XY] • E[XY] = (0×0×0.1) + (0×1×0.4) + (1×0×0.4) + (1×1×0.1) = 0.1
Step 4: Calculate Covariance • Cov(X,Y) = 0.1 - 0.25 = -0.15
The negative covariance indicates an inverse relationship: when X increases, Y tends to decrease, and vice versa. This is the opposite pattern from Example 1.
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