Probability and Statistics<\/p>\n
Your Name<\/p>\n
Institutional Affiliation <\/p>\n
Probability and Statistics<\/p>\n
1.) f (x; 0,1) = 1\/\u03c0 (1 + x2)<\/p>\n
Y = aX + b<\/p>\n
E (Y) = E (aX + b) = a E (X) = 0, since E (x) = 0<\/p>\n
Var (Y) = a2 Var (X) = a2, since Var (X) = 1<\/p>\n
Hence;<\/p>\n
f (Y: 0, a2) = a\/\u03c0 (a2+ y2)<\/p>\n
Z = log (|X|)<\/p>\n
Z = log (|X|) = log (|X|)<\/p>\n
=1\/x\u03c0 [\u03b4\/ (lnx – \u03bc)2 + \u03b42]<\/p>\n
But X=ez, therefore the distribution of Z is;<\/p>\n
f (z;0, 1)= 1\/ez\u03c0 [\u03b4\/ (z – \u03bc)2 + \u03b42]<\/p>\n
2.) Let Y = 2(X \u2013 1)2- 1, where X is uniformly distributed over the interval [0, 2].<\/p>\n
Determine the pdf of Y and the expected value of Y.<\/p>\n
75438064579500X is in a closed interval of 0 and 2. Substitute the values of x\u2019s in equation Y given to establish the pdf of Y as follows:<\/p>\n
Pdf of Y =1, x =0, 2<\/p>\n
-1, x =1<\/p>\n
Getting expected value of Y, we first expand its expression knowing very well that since X is uniformly distributed, then its distribution is:<\/p>\n
X~ N (0, 1)<\/p>\n
Y =2X2-4X+1<\/p>\n
E(Y) = E (2X2- 4X+1) = 2E (X2)-4E (X)<\/p>\n
But E (X) = 0, and Var (X) =E (X2), Thus E (Y) = (2*1) \u2013 4*0 =2<\/p>\n
3.) Y1 = X12 and Y2 = X1\/X2<\/p>\n
We solve for X1 and X2 in terms of Y1 and Y2.<\/p>\n
Thus<\/p>\n
X1 = (Y1)1\/2 ………………………. (1)<\/p>\n
X2 = X1\/Y2…………………………. (2)<\/p>\n
Let\u2019s substitute the value of X1 in equation (2) above,<\/p>\n
X2 = (Y1)1\/2\/Y2<\/p>\n
Furthermore let the determinant be 1\/(2)1\/2, hence the joint pdf of Y1 and Y2 is:<\/p>\n
fy1,y2(Y1,Y2) = 1\/(2)1\/2 fx1,x2 ( (Y1)1\/2, (Y1)1\/2\/Y2)<\/p>\n
= 1\/ (2)1\/2 {2 * (Y1)1\/2[(Y1)1\/2, (Y1)1\/2\/Y2]}<\/p>\n
Marginal distributions of Y1 is<\/p>\n
f (Y1) = \u222b- \u221e\u221e fy1, y2 (Y1,Y2) dY2 = \u222b- \u221e\u221e 1\/ (2)1\/2 {2 * (Y1)1\/2[(Y1)1\/2, (Y1)1\/2\/Y2]} dY2<\/p>\n
Marginal distributions of Y2 is<\/p>\n
F (Y2) = \u222b- \u221e\u221e fy1, y2 (Y1, Y2) dY1 = \u222b- \u221e\u221e 1\/ (2)1\/2 {2 * (Y1)1\/2[(Y1)1\/2, (Y1)1\/2\/Y2]} dY1<\/p>\n
Y1 and Y2 are not independent due to the fact that the product of their respective marginal pdf\u2019s does not result to their joint pdf.<\/p>\n
Postgraduate problem<\/p>\n
Standard Cauchy distribution is the distribution of a random variable that is the ratio of two independent standard normal variables and has the probability density function:<\/p>\n
f (x; 0,1) = 1\/\u03c0 (1 + x2)<\/p>\n
Let \u03b1 = aX and \u03b2 = bY<\/p>\n
f (x; 0,1) = 1\/\u03c0 (1 + x2)<\/p>\n
E (\u03b1) = E (aX + b) = a E (X) = 0, since E (x) = 0<\/p>\n
Var (\u03b1) = a2 Var (X) = a2, since Var (X) = 1<\/p>\n
Hence;<\/p>\n
f (\u03b1: 0, a2) = a\/\u03c0 (a2+ x2)<\/p>\n
f (y; 0,1) = 1\/\u03c0 (1 + y2)<\/p>\n
E (\u03b2) = E (bY) = b E (Y) = 0, since E (Y) = 0<\/p>\n
Var (\u03b2) = b2 Var (\u03b2) = b2, since Var (Y) = 1<\/p>\n
Hence;<\/p>\n
f (\u03b2; 0, b2) = b\/\u03c0 (b2+ y2)<\/p>\n
Therefore, <\/p>\n
\u03b1 + \u03b2 = aX + bY= a\/\u03c0 (a2+ x2) + b\/\u03c0 (b2+ y2)<\/p>\n
= 1\/ \u03c0 { a(a2+ x2) + b(b2+ y2)}<\/p>\n
Reference<\/p>\n
DeGroot, M. H., & Schervish, M. J. (2012). Probability and Statistics. Boston: Addison-Wesley.<\/p>\n","protected":false},"excerpt":{"rendered":"
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