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uppose Company 1’s simple return X is a random variable and takes three possiblevalues {−0.1, 0.1, 0.2}. Company 2’s simple return Y is a random variable and takes twopossible values {−0.3, 0.4}. The joint probability distribution f (X, Y ) is as follows:f (X = −0.1, Y = −0.3) = 0.2, f (X = 0.1, Y = −0.3) = 0.2, f (X = 0.2, Y = −0.3) = 0.1,f (X = −0.1, Y = 0.4) = 0.1, f (X = 0.1, Y = 0.4) = 0.2, f (X = 0.2, Y = 0.4) = 0.2Please calculate the following:(a). Marginal distributions: fX (x) and fY (y);(b). Mean: E[X] and E[Y ];(c). Variance: var(X) and var(Y );(d). Covariance: cov(X, Y );(e). Correlation: corr(X, Y )

Question

uppose Company 1’s simple return X is a random variable and takes three possiblevalues {−0.1, 0.1, 0.2}. Company 2’s simple return Y is a random variable and takes twopossible values {−0.3, 0.4}. The joint probability distribution f (X, Y ) is as follows:f (X = −0.1, Y = −0.3) = 0.2, f (X = 0.1, Y = −0.3) = 0.2, f (X = 0.2, Y = −0.3) = 0.1,f (X = −0.1, Y = 0.4) = 0.1, f (X = 0.1, Y = 0.4) = 0.2, f (X = 0.2, Y = 0.4) = 0.2Please calculate the following:(a). Marginal distributions: fX (x) and fY (y);(b). Mean: E[X] and E[Y ];(c). Variance: var(X) and var(Y );(d). Covariance: cov(X, Y );(e). Correlation: corr(X, Y )

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Solution

Sure, let's calculate each of these step by step:

(a). Marginal distributions: fX(x) and fY(y);

The marginal distribution of X, fX(x), is the sum of the joint probabilities for each value of X, regardless of the value of Y. Similarly, the marginal distribution of Y, fY(y), is the sum of the joint probabilities for each value of Y, regardless of the value of X.

fX(-0.1) = f(X = -0.1, Y = -0.3) + f(X = -0.1, Y = 0.4) = 0.2 + 0.1 = 0.3 fX(0.1) = f(X = 0.1, Y = -0.3) + f(X = 0.1, Y = 0.4) = 0.2 + 0.2 = 0.4 fX(0.2) = f(X = 0.2, Y = -0.3) + f(X = 0.2, Y = 0.4) = 0.1 + 0.2 = 0.3

fY(-0.3) = f(X = -0.1, Y = -0.3) + f(X = 0.1, Y = -0.3) + f(X = 0.2, Y = -0.3) = 0.2 + 0.2 + 0.1 = 0.5 fY(0.4) = f(X = -0.1, Y = 0.4) + f(X = 0.1, Y = 0.4) + f(X = 0.2, Y = 0.4) = 0.1 + 0.2 + 0.2 = 0.5

(b). Mean: E[X] and E[Y ];

The expected value E[X] is the sum of each value of X multiplied by its probability, and similarly for E[Y].

E[X] = -0.1fX(-0.1) + 0.1fX(0.1) + 0.2fX(0.2) = -0.10.3 + 0.10.4 + 0.20.3 = 0.05 E[Y] = -0.3fY(-0.3) + 0.4fY(0.4) = -0.30.5 + 0.40.5 = 0.05

(c). Variance: var(X) and var(Y );

The variance var(X) is the sum of the squared difference between each value of X and the mean, multiplied by its probability, and similarly for var(Y).

var(X) = ((-0.1 - E[X])^2)*fX(-0.1) + ((0.1 - E[X])^2)*fX(0.1) + ((0.2 - E[X])^2)*fX(0.2) = ((-0.1 - 0.05)^2)*0.3 + ((0.1 - 0.05)^2)*0.4 + ((0.2 - 0.05)^2)*0.3 = 0.0075 var(Y) = ((-0.3 - E[Y])^2)*fY(-0.3) + ((0.4 - E[Y])^2)*fY(0.4) = ((-0.3 - 0.05)^2)*0.5 + ((0.4 - 0.05)^2)*0.5 = 0.2025

(d). Covariance: cov(X, Y );

The covariance cov(X, Y) is the sum of the product of the differences of X and Y from their means, multiplied by the joint probability of X and Y.

cov(X, Y) = sum((x - E[X])*(y - E[Y])f(x, y)) for all x, y = ((-0.1 - 0.05)(-0.3 - 0.05)0.2 + (-0.1 - 0.05)(0.4 - 0.05)0.1 + (0.1 - 0.05)(-0.3 - 0.05)0.2 + (0.1 - 0.05)(0.4 - 0.05)0.2 + (0.2 - 0.05)(-0.3 - 0.05)0.1 + (0.2 - 0.05)(0.4 - 0.05)*0.2) = -0.0075

(e). Correlation: corr(X, Y )

The correlation corr(X, Y) is the covariance of X and Y divided by the product of their standard deviations.

corr(X, Y) = cov(X, Y) / (sqrt(var(X)) * sqrt(var(Y))) = -0.0075 / (sqrt(0.0075) * sqrt(0.2025)) = -0.75

So, the marginal distributions are fX(x) = {0.3, 0.4, 0.3} and fY(y) = {0.5, 0.5}, the means are E[X] = 0.05 and E[Y] = 0.05, the variances are var(X) = 0.0075 and var(Y) = 0.2025, the covariance is cov(X, Y) = -0.0075, and the correlation is corr(X, Y) = -0.75.

This problem has been solved

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