Question: Question 1: Problem 1(a) 1 point possible (graded, results hidden) Suppose that X, Y, and Z are independent, with E [X] = E [Y] =

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![[Y] = E [Z] = 2, and E [X?] = E [Y?]](https://s3.amazonaws.com/si.experts.images/answers/2024/06/667624ac5aff2_668667624ac288e8.jpg)
![= E [Z3] = 5. Find cov (XY, XZ). (Enter a numerical](https://s3.amazonaws.com/si.experts.images/answers/2024/06/667624acb636d_668667624ac8903d.jpg)





Question 1:

![{X= Y]. Cov (X, Y) = Are X and Y independent? 0](https://s3.amazonaws.com/si.experts.images/answers/2024/06/667624b01b44a_671667624afd42c5.jpg)







Problem 1(a) 1 point possible (graded, results hidden) Suppose that X, Y, and Z are independent, with E [X] = E [Y] = E [Z] = 2, and E [X?] = E [Y?] = E [Z3] = 5. Find cov (XY, XZ). (Enter a numerical answer.) cov (XY, XZ) =Problem 2. Problem 1(b) 2.0 points possible (graded, results hidden) Let X be a standard normal random variable. Another random variable is determined as follows. We flip a fair coin {independent from X}. In case of Heads, we let Y = X. In case of Tails, we let Y = ~X. 1. Is Y normal? Justify your answer. Ono 0 not enough information to determine 2. Compute COV {X= Y]. Cov (X, Y) = Are X and Y independent? 0 yes 0 no 0 not enough information to determine Problem 3. Problem 1(c) 2.0 points possible (graded, results hidden) Find P (X + Y a. What is a?Problem 5 3.0 points possible (graded, results hidden) Let X and Y be independent positive random variables. Let Z = X/Y. In what follows, all occurrences of :c, y. ,z are assumed to be positive numbers. 1. Suppose that X and Y are discrete, with known PMFs, pg and pg. Then, my (Zly) =1: (1?)- What is the argument in the place of the question mark? 2. Suppose that X and Y are continuous, with known PDFs, f3: and fy. Provide a formula, analogous to the one in part (a)I for ley (2| y) in terms of f1. That is, find A and Bin the formula below. My (Ely) = Afx (3] - 3. Which of the following is a formula for fz (z)? fz ( z ) =... (Choose all that apply.) fz (z) = fr,z (y, z) dy fz (2 ) = / fy,z (y, z) dz fz ( 2 ) = / fr (y) fz,x (z,y) dy Ofz ( z ) = fx (y) fzx ( zly ) dy fz ( z ) = fy (y) fx (yz) dy $2 (2) = / ufx (3) fx (yz) dy
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