moment generating function

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圖片參考：http://www.upload-jpg.com/images.php/941691707/MGF.jpg

1) For mean = μ and variance = 1, the cont. function of X is given by:

f(x) = [1/√(2π)] e-(x - μ)^2/2 Hence the distribution of y is:

x2 = {[1/√(2π)] e-(x - μ)^2/2}2 = e-(x - μ)^2/(2π)

So the moment generating function of Y is:

M(t) = ∫ (x = - ∞ → ∞) etx [e-(x - μ)^2/(2π)] dx

= ∫ (x = - ∞ → ∞) [etx-(x - μ)^2/(2π)] dx

2) Finding E[Y] by M'(0)

M'(t) = (d/dt)∫ (x = - ∞ → ∞) [etx-(x - μ)^2/(2π)] dx

= ∫ (x = - ∞ → ∞) d(etx)/dt [-(x - μ)^2/(2π)] dx

= ∫ (x = - ∞ → ∞) x etx [-(x - μ)^2/(2π)] dx

So M'(0) = ∫ (x = - ∞ → ∞) x [-(x - μ)^2/(2π)] dx

= ∫ (x = - ∞ → ∞) [e-(x - μ)^2/(2π)] d[(x - μ)2] + 2μ∫ (x = - ∞ → ∞) [e-(x - μ)^2/(2π)] dx

= [1/(2π)] [- e-(x - μ)^2] (x = - ∞ → ∞) + 2μ

= 2μ

Finding E[Y] by y = x2:

E[Y] = ∫ (x = - ∞ → ∞) {[1/√(2π)] e-(x - μ)^2/2}2 dx

= ∫ (x = - ∞ → ∞) [e-(x - μ)^2/(2π)] dx

= 2μ