Please answer all parts of the question. Thank you
a) First, we compute the value of c here by using the property that the sum of all probabilities in the joint (a, b) space should be 1.
Therefore, we have here:
p(-2, 1) + p(-2, 2) + p(2, 1) + p(2, 2) = 1
4c + 8c + 4c + 8c = 1
24c = 1
c = 1/24
Therefore, the covariance of the two variables here is computed
as:
E(ab) = -2p(-2, 1) - 4p(-2, 2) + 2p(2, 1) + 4p(2, 2)
E(ab) = -8c - 32c + 8c + 32c = 0
E(a) = -2*12c + 2*12c = 0
Therefore, Cov(a, b) = E(ab) - E(a)E(b) = 0
As the covariance here is 0, therefore yes A and B are uncorrelated here. (because the numerator for correlation is covariance )
b) From the given obtained probabilities, we have here:
P(a = -2) = 12c = 0.5
P(a = -2 | b = 1) = 4c / 8c = 0.5 which is same as P(a = -2)
Therefore a and b are independent here.
Please answer all parts of the question. Thank you [1] The joint probability mass function of...
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[1] The joint probability mass function of two discrete random variables A and B is Pab(a,b) = {ca2b, a = -2,2 and b = 1,2 otherwise Clearly stating your reasons, answer the following two (1) Are A and B are uncorrelated? (ii) Are A and B independent? [2] X is continuous uniform (1,7) while Y is exponential with mean 2. If the variance of (X+2Y) is 20, find the correlation coefficient of X and Y.
The joint probability mass function of two discrete random variables A and B is (i) Are A and B are uncorrelated? (ii) Are A and B independent? Sca²b, a=-2,2 and b = 1,2 PA,(a,b) = 0, otherwise
Please answer all parts of the question. Thank you [1] The joint probability density function of two continuous random variables X and Y is fx,x(x,y) = {6. sc, 0 Sy s 2.y = x < 4-y otherwise Find the value of c and the correlation of X and Y.
Please show how did you came up with the answer, show formulas and work. Also, please do Parts e to i. Thank you so much 1. Consider the following probability mass function for the discrete joint probability distribution for random variables X and Y where the possible values for X are 0, 1, 2, and 3; and the possible values for Y are 0, 1, 2, 3, and 4. p(x,y) <0 3 0 4 0.01 0 0 0.10 0.05 0.15...
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