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Homework answers / question archive / 5) Geometry basics: L2 = {X €R| EX2 < ∞}, ||X|| = √EX2, d(X,Y)=||Y-X|| Let X ~ exp (1), Y = e-x, and consider the simple linear model Y= α + βx + W w

5) Geometry basics: L2 = {X €R| EX2 < ∞}, ||X|| = √EX2, d(X,Y)=||Y-X|| Let X ~ exp (1), Y = e-x, and consider the simple linear model Y= α + βx + W w

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5) Geometry basics: L2 = {X €R| EX2 < ∞}, ||X|| = √EX2, d(X,Y)=||Y-X||

Let X ~ exp (1), Y = e-x, and consider the simple linear model

Y= α + βx + W w. EW = 0 = p(X, W).

a) Evaluate the constants α and β.

b) Determine the relative proximity of Y to its closest linear predictor

|Y — (α + βx)|/|Y-EY|

4) Random variable basics: R= (F)={X:Ω → R|(X < x) ? F V x ? R}

L = L1 = {X €R|E|X| < ∞}

a) Given that E : L —> R is normed, non-negative & linear

Verify that the continuity property: 0 < Zn ↑ Z = 0 < Zn ↑ Z

Is equivalent to o-linearity: Zn > 0, n=1, 2, ... = E Σn=1 Zn = Σn=1 EZn. proof:

b) If EX is not undefined, verify that |EX| < E|X|, and describe the circumstances for equality.

c) X =d Y, X < Y = x w P1= y. proof:

3. Probability basics: Suppose P: F — R is any set-function, on the non-empty domain F ‹ P (Ω) closed wrt countable unions and complementation, that is

i) Normed: P(Ω)=1

ii) Non-negative: A ? F => P(A) > 0

iii) F-additive: P(A+B) = P(A)+P(B)

And show that the following three versions of an additional property iv) are entirely equivalent to each other:

iv) σ -additive: P(Σn=1 An) =Σn=1 P(An)

iv)’ continuous: An→ A => P(An) → P(A)

iv)” continuous at 0: An ↓ Ø  => P(An) → 0.

2) Quantile basics 2: g(u) =inf F-1 [u,1] & h(u) =supF-1[0,u]

a) F(h(p)-) < p < F(h(p)) proof:

b) F(x-) < p < F(x) = g(p) < x < h(p) proof:

c) Just as was the case with g, we find that h(U) =d X. proof:

Quantile basics 1: g(u) =infF-1 {u,1] & h(u) =supF-1 [0, u]

a) Let c(u) = sup F-1 (0,u) and verify that g(u) = c(u).

b) [g(u) < x = u < F(x)] = [u < Fg(u) & gF(x) < x]. Proof:

c) Unfortunately, the inequalities for h are a bit different than those for g.

Demonstrate that u < Fh(u) & x < hF(x).

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