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Elementary Linear Algebra (MindTap Course List)

Elementary Linear Algebra (MindTap Course List)

8th Edition

ISBN:9781305658004

Author:Ron Larson

Publisher:Ron Larson

Chapter5: Inner Product Spaces

Section5.CR: Review Exercises

Problem 62CR

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Question

3
Let Y be a random variable with probability density function given by
2(1-y), Osys 1,
fly)
0,
elsewhere.
Use the method of transformation to find the densities of U₁, U., and U.
(a) U-2Y-3
(L)
elsewhere
(b) U₂-3-2
(3)-
(c) U₁₂ = y2
fu₂ (4) -
elsewhere
elsewhere
SUS
SUS
SUS

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Q4) The probability mass function of Y is f(y) = y/6 for y=1,2,3,4. Find the variance of Y.
Suppose Y1 and Y2 are random variables with joint pdf (6(1 — у2), 0, 0 < y1 < y2 fv, otherwise Y1 Let U1 and U2 = Y,. Use the transformation technique to show that U, follows a uniform Y2 distribution from 0 to 1.
Answer all three questions in this section 2. Let V and W be two random variables with joint probability density function given by fvw (v,w) = aw exp(-dv +(8-A)w²), w>0, v> w², where A, & are positive constants. (a) Find the value of a. (b) Consider the transformation X = WP, Y=V_W². Find the joint density of X and Y. Hence show that X and Y are independent exponential random variables. (c) State the distribution of AX + SY.
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The probability distribution for the random varaibles X and Y is shown in the table below. Find the covariance of X and Y x 0 1 2 0 11/73 18/73 14/73 1 10/73 8/73 0 2 12/73 0 0 ^ y on this side using coviaraince = E(XY)-E(X)E(Y)
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Let X1 and X2 be two continuous random variableshaving the joint probability density f(x1, x2) = 4x1x2 for 0 < x1 < 1, 0 < x2 < 10 elsewhereFind the joint probability density of Y1 = X21 and Y2 = X1X2.
(47) Let X z b(8,-) find E(5+6x) and distribution function.
Let Xand Y be two continuous random variables with joint probability density [3x function given by: f(x.y)%D 0sysxsl elsewhere with E(X) = ECX)- EC) - EC*)= ;and E(XY) = 10 3 E(Y*) = - and E(XY) =; %3D Then the value of the variance of 2X+Y is: O 3/80 O 91/320 43/320 7/20
Suppose that Y₁ and Y₂ are uniformly distributed over the triangle shaded in the accompanying figure. 3₂ (0, 1) (-1,0) (a) Find Cov(Y₁ Y₂). Cov(Y₁, Y₂) = (b) Are Y₁ and Y₂ independent? Yes O No (1, 0) (c) Find the coefficient of correlation for Y₁ and Y₂. P= y/₁ (d) Does your answer to part (b) lead you to doubt your answer to part (a)? Why or why not? O Even though Cov(Y₁Y₂) # 0, Y₁ and Y₂ are not necessarily dependent. Since Cov(Y₁ Y₂) # 0, we should expect Y₁ and Y₂ to be dependent. O Since Cov(Y₁, Y₂) = 0, we should expect Y₁ and Y₂ to be independent. O Even though Cov(Y₁Y₂) = 0, Y₁ and Y₂ are not necessarily independent.
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