Probability and Statistics

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Probability and Statistics

1.) f (x; 0,1) = 1/π (1 x2)

Y = aX b

E (Y) = E (aX b) = a E (X) = 0, since E (x) = 0

Var (Y) = a2 Var (X) = a2, since Var (X) = 1

Hence;

f (Y: 0, a2) = a/π (a2 y2)

Z = log (|X|)

Z = log (|X|) = log (|X|)

=1/xπ [δ/ (lnx – μ)2 δ2]

But X=ez, therefore the distribution of Z is;

f (z;0, 1)= 1/ezπ [δ/ (z – μ)2 δ2]

2.) Let Y = 2(X – 1)2- 1, where X is uniformly distributed over the interval [0, 2].

Determine the pdf of Y and the expected value of Y.

75438064579500X is in a closed interval of 0 and 2. Substitute the values of x’s in equation Y given to establish the pdf of Y as follows:

Pdf of Y =1, x =0, 2

-1, x =1

Getting expected value of Y, we first expand its expression knowing very well that since X is uniformly distributed, then its distribution is:

X~ N (0, 1)

Y =2X2-4X 1

E(Y) = E (2X2- 4X 1) = 2E (X2)-4E (X)

But E (X) = 0, and Var (X) =E (X2), Thus E (Y) = (2*1) – 4*0 =2

3.) Y1 = X12 and Y2 = X1/X2

We solve for X1 and X2 in terms of Y1 and Y2.

Thus

X1 = (Y1)1/2 ………………………. (1)

X2 = X1/Y2…………………………. (2)

Let’s substitute the value of X1 in equation (2) above,

X2 = (Y1)1/2/Y2

Furthermore let the determinant be 1/(2)1/2, hence the joint pdf of Y1 and Y2 is:

fy1,y2(Y1,Y2) = 1/(2)1/2 fx1,x2 ( (Y1)1/2, (Y1)1/2/Y2)

= 1/ (2)1/2 {2 * (Y1)1/2[(Y1)1/2, (Y1)1/2/Y2]}

Marginal distributions of Y1 is

f (Y1) = ∫- ∞∞ fy1, y2 (Y1,Y2) dY2 = ∫- ∞∞ 1/ (2)1/2 {2 * (Y1)1/2[(Y1)1/2, (Y1)1/2/Y2]} dY2

Marginal distributions of Y2 is

F (Y2) = ∫- ∞∞ fy1, y2 (Y1, Y2) dY1 = ∫- ∞∞ 1/ (2)1/2 {2 * (Y1)1/2[(Y1)1/2, (Y1)1/2/Y2]} dY1

Y1 and Y2 are not independent due to the fact that the product of their respective marginal pdf’s does not result to their joint pdf.

Postgraduate problem

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:

f (x; 0,1) = 1/π (1 x2)

Let α = aX and β = bY

f (x; 0,1) = 1/π (1 x2)

E (α) = E (aX b) = a E (X) = 0, since E (x) = 0

Var (α) = a2 Var (X) = a2, since Var (X) = 1

Hence;

f (α: 0, a2) = a/π (a2 x2)

f (y; 0,1) = 1/π (1 y2)

E (β) = E (bY) = b E (Y) = 0, since E (Y) = 0

Var (β) = b2 Var (β) = b2, since Var (Y) = 1

Hence;

f (β; 0, b2) = b/π (b2 y2)

Therefore,

α β = aX bY= a/π (a2 x2) b/π (b2 y2)

= 1/ π { a(a2 x2) b(b2 y2)}

Reference

DeGroot, M. H.,

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