Let Z = XY. 1. Find the expectation E[Z]. E[Z] =18.6667
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How was the correct answer of 18.6667 attained?
![Suppose the joint PMF of the random variables X and Y is
P(X = x, Y = y)
Let Z = XY.
1. Find the expectation E[Z].
E[Z] =
= 18.6667
=
x+y
81
if(x, y) {5,6,7} × {2,3,4},
0 Otherwise.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1467f162-7185-45ab-925a-2589d3c8cce7%2F6b5a6dd5-0d01-4d5e-8ad0-2bd332695e99%2Frwq9xio_processed.png&w=3840&q=75)
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- Let X1,X2,... be a sequence of identically distributed random variables with E|X1|<∞ and let Yn = n−1max1≤i≤n|Xi|. Show that limnE(Yn) = 0Consider two securities, the first having μ1 = 1 and σ1 = 0.1, and the secondhaving μ2 = 0.8 and σ2 = 0.12. Suppose that they are negatively correlated,with ρ = −0.8. Denote the expected return and its standard deviation as functions of π byμ(π ) and σ (π ). The pair (μ(π ), σ (π )) trace out a curve in the plane as πvaries from 0 to 1. Plot this curve in R.Let X be a random variable with pdff(x) = 4x^3 if 0 < x < 1 and zero otherwise. Use thecumulative (CDF) technique to determine the pdf of each of the following random variables: 1) Y=X^4, 2) W=e^(-x) 3) Z=1-e^(-x) 4) U=X(1-X)
- If Z is a discrete random variable with possible values of -1, 0, and 1, and the probability mass function is given by P(Z = -1) = 0.4, P(Z = 0) = 0.3, and P(Z = 1) = 0.3, what is the standard deviation of Z?Consider a function F (x ) = 0, if x < 0 F (x ) = 1 − e^(−x) , if x ≥ 0 Is the corresponding random variable continuous?Let X be a random variable with probability mass function P ( X = 1 ) = 1/2 , P ( X = 2 ) = 1/3 , a n d P ( X = 5 ) = 1/6 . Then E[1/x]=?
