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  • Probability Theory and Examples Solutions PDF 下载

    Probability Theory and Examples Solutions PDF 下载

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    2 Chapter 1 Laws of Large Numbers
    let G = {A : A × A2 ×···× An ∈ F}. Since F is a σ-field it is easy to see that
    if Ω ∈ G then G is a σ-field so if G⊃A then G ⊃ σ(A). From the last result it
    follows that if A1 ∈ R, A1 × (a2, b2) ×···× (an, bn) ∈ F. Repeating the last
    argument n 1 more times proves (ii).
    1.4. It is clear that if A ∈ F then Ac ∈ F. Now let Ai be a countable collection
    of sets. If Aci is countable for some i then (∪iAi)c is countable. On the other
    hand if Ai is countable for each i then ∪iAi is countable. To check additivity
    of P now, suppose the Ai are disjoint. If Aci is countable for some i then Aj is
    countable for all j = i so Pk P(Ak)=1= P(∪kAk). On the other hand if Ai
    is countable for each i then ∪iAi is and Pk P(Ak)=0= P(∪kAk).
    1.5. The sets of the form (a1, b1) ×···× (ad, bd) where ai, bi ∈ Q is a countable
    collection that generates Rd.
    1.6. If B ∈ R then {Z ∈ B} = ({X ∈ B} ∩ A) ∪ ({Y ∈ B} ∩ Ac) ∈ F
    1.7.
    P(χ ≥ 4) ≤ (2π) 1/24 1e 8 = 3.3345 × 100 5
    The lower bound is 15/16’s of the upper bound, i.e., 3.126 × 100 5
    1.8. The intervals (F(x ), F(x)), x ∈ R are disjoint and each one that is
    nonempty contains a rational number.
    1.9. Let Fˆ 1(x) = sup{y : F(y) ≤ x} and note that F(Fˆ 1(x)) = x when F is
    continuous. This inverse wears a hat since it is different from the one defined
    in the proof of (1.2). To prove the result now note that
    P(F(X) ≤ x) = P(X ≤ Fˆ 1(x)) = F(Fˆ 1(x)) = x
    1.10. If y ∈ (g(α), g(β)) then P(g(X) ≤ y) = P(X ≤ g 1(y)) = F(g 1(y)).
    Differentiating with respect to y gives the desired result.
    1.11. If g(x) = ex then g 1(x) = log x and g0(g 1(x)) = x so using the formula
    in the previous exercise gives (2π) 1/2e (log x)2/2/x.
    1.12. (i) Let F(x) = P(X ≤ x). P(X2 ≤ y) = F(√y) F( √y) for y > 0.
    Differentiating we see that X2 has density function
    (f(√y) + f( √y))/2√y
    (ii) In the case of the normal this reduces to (2πy) 1/2e y/2

     

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