11 - STAT 350 Fall 2008 Homework #11 - Solution covers...

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STAT 350 – Fall 2008 Homework #11 - Solution covers through Lecture #17 1. X is a discrete random variable with the following PMF: x -2 -1 0 1 2 p X ( x ) 0.1 0.15 0.5 0.15 0.1 a. Let Y = X 2 . Make a table giving the Joint PMF of X and Y . X Y -2 -1 0 1 2 0 0 0 0.5 0 0 0.5 1 0 0.15 0 0.15 0 0.3 4 0.1 0 0 0 0.1 0.2 0.1 0.15 0.5 0.15 0.1 1 b. Find the Correlation of X and Y . E(X) = -2(0.1) + (-1)(0.2) + 0(0.4) + 1(0.2) + 2(0.1) = 0 E(Y) = 0(0.4) + 1(0.3) + 4(0.2) = 1.1 Cov(X,Y) = 0 + 0 + (0-0)(0-1.1)(0.4) + 0 + 0 = 0 + (-1-0)(1-1.1)(0.2) + 0 + (1-0)(1-1.1)(0.2) + 0 = (-2-0)(4-1.1)(0.1) + 0 + 0 + 0 + (2-0)(4-1.1)(0.1) = 0 Because Cov(X,Y) = 0, Corr(X,Y) = 0 c. Are X and Y independent random variables? (Hint: what would the joint PMF look like if they were independent?) No. X and Y are clearly not independent. If they were independent, the joint PMF would be a product of the marginal PMFs and would look like this: X Y -2 -1 0 1 2 0 0.05 0.075 0.25 0.075 0.05 0.5 1 0.03 0.045 0.15 0.045 0.03 0.3 4 0.02 0.030 0.10 0.030 0.02 0.2 0.1 0.15 0.5 0.15 0.1 1 d. Find P(Y = 1 | X = 1) Hint: this is a conditional probability. Recall the definition of P(A|B). () 11 0.15 10 . PY X PX =∩ = = = 1 5 = 1 e. Find P(X = 1 | Y = 1) 0.15 0.5 . 3 0 X == = Homework #11 - Solution Page 1 of 6
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2. X and Y are discrete random variables with the following joint PMF. X Y 1 2 3 4 5 6 0 0.007 0.019 0.025 0.026 0.013 0.027 1 0.018 0.005 0.029 0.011 0.028 0.003 2 0.017 0.020 0.030 0.037 0.009 0.038 3 0.002 0.021 0.010 0.036 0.015 0.039 4 0.014 0.023 0.031 0.001 0.016 0.040 5 0.022 0.004 0.032 0.035 0.041 0.006 6 0.008 0.024 0.033 0.034 0.012 0.139 a. Find the marginal PMF of X X Y 1 2 3 4 5 6 0 0.007 0.019 0.025 0.026 0.013 0.027 1 0.018 0.005 0.029 0.011 0.028 0.003 2 0.017 0.020 0.030 0.037 0.009 0.038 3 0.002 0.021 0.010 0.036 0.015 0.039 4 0.014 0.023 0.031 0.001 0.016 0.040 5 0.022 0.004 0.032 0.035 0.041 0.006
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11 - STAT 350 Fall 2008 Homework #11 - Solution covers...

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