SLIDE 1
Formal connection to moments: MX(t) =
∞
- k=0
E[(tX)k]/k! =
∞
- k=0
Moments and independence Theorem : If X 1 , . . . , X p are - - PDF document
Moments and independence Theorem : If X 1 , . . . , X p are - - PDF document
Moments and independence Theorem : If X 1 , . . . , X p are independent and each X i is integrable then X = X 1 X p is in- tegrable and E ( X 1 X p ) = E ( X 1 ) E ( X p ) . Moment Generating Functions Defn : The moment
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SLIDE 3
For α = 1 get exponential distribution. Have power series expansion MX(t) = 1/(1 − t) =
∞
- tk
Write tk = k!tk/k!. Coeff of tk/k! is E(Xk) = k! Example: : Z ∼ N(0, 1). MZ(t) =
∞
−∞ etz−z2/2dz/
√ 2π =
∞
−∞ e−(z−t)2/2+t2/2dz/
√ 2π = et2/2 because rest of integrand is N(t, 1) density.
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Theorem: If M is finite for all t ∈ [−ǫ, ǫ] for some ǫ > 0 then
- 1. Every moment of X is finite.
- 2. M is C∞ (in fact M is analytic).
- 3. µ′
k = dk dtkMX(0).
Note: C∞ means has continuous derivatives
- f all orders.
Analytic means has convergent power series expansion in neighbourhood of each t ∈ (−ǫ, ǫ). Theorem: Suppose X and Y have mgfs MX and MY which are finite for all t ∈ [−ǫ, ǫ]. If MX(t) = MY (t) for all t ∈ [−ǫ, ǫ] then X and Y have the same distribution. The proofs, and many other facts about mgfs, rely on techniques of complex variables.
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MGFs and Sums If X1, . . . , Xp are independent and Y = Xi then the moment generating function of Y is the product of those of the individual Xi: E(etY ) =
- i
E(etXi)
- r MY = MXi.
Note: also true for multivariate Xi. Example: : If Xi ∼ N(µi, σ2
i ) then
MXi(t) =E
- et(σiZi+µi)
=etµiMZi(tσi) =etµiet2σ2
i /2
=eσ2
i t2/2+tµi .
Moment generating function for Y = Xi is MY (t) =
- exp{σ2
i t2/2 + tµi}
= exp{
- σ2
i t2/2 +
- µit}
which is mgf of N( µi, σ2
i ).
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Example: Now suppose Z1, . . . , Zν independent N(0, 1) rvs. By definition: Sν = ν
1 Z2 i has χ2 ν distribution.
It is easy to check S1 = Z2
1 has density
(u/2)−1/2e−u/2/(2√π) and then the mgf of S1 is (1 − 2t)−1/2 . It follows that MSν(t) = (1 − 2t)−ν/2 which is mgf of Gamma(ν/2, 2) rv. So: χ2
ν dstbn has Gamma(ν/2, 2) density:
(u/2)(ν−2)/2e−u/2/(2Γ(ν/2)) .
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