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Foundations of Data Science

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Replacing e tx by its power series expansion 1 + tx + (tx)2 · · · gives<br />

2!<br />

∫ ∞ (<br />

)<br />

Ψ(t) = 1 + tx + (tx)2 + · · · p(x)dx<br />

2!<br />

−∞<br />

Thus, the k th moment <strong>of</strong> x about the origin is k! times the coefficient <strong>of</strong> t k in the power<br />

series expansion <strong>of</strong> the moment generating function. Hence, the moment generating function<br />

is the exponential generating function for the sequence <strong>of</strong> moments about the origin.<br />

The moment generating function transforms the probability distribution p(x) into a<br />

function Ψ (t) <strong>of</strong> t. Note Ψ(0) = 1 and is the area or integral <strong>of</strong> p(x). The moment<br />

generating function is closely related to the characteristic function which is obtained by<br />

replacing e tx by e itx in the above integral where i = √ −1 and is related to the Fourier<br />

transform which is obtained by replacing e tx by e −itx .<br />

Ψ(t) is closely related to the Fourier transform and its properties are essentially the<br />

same. In particular, p(x) can be uniquely recovered by an inverse transform from Ψ(t).<br />

∑<br />

More specifically, if all the moments m i are finite and the sum ∞ m i<br />

i! ti converges absolutely<br />

in a region around the origin, then p(x) is uniquely determined.<br />

The Gaussian probability distribution with zero mean and unit variance is given by<br />

p (x) = √ 1<br />

2π<br />

e − x2<br />

2 . Its moments are given by<br />

u n = 1 √<br />

2π<br />

=<br />

∫∞<br />

−∞<br />

x n e − x2<br />

2 dx<br />

{ n!<br />

n even<br />

2 n 2 (<br />

n<br />

2 )!<br />

0 n odd<br />

i=0<br />

To derive the above, use integration by parts to get u n = (n − 1) u n−2 and combine<br />

this with u 0 = 1 and u 1 = 0. The steps are as follows. Let u = e − x2<br />

2 and v = x n−1 . Then<br />

u ′ = −xe − x2<br />

2 and v ′ = (n − 1) x n−2 . Now uv = ∫ u ′ v+ ∫ uv ′ or<br />

∫<br />

e − x2<br />

2 x n−1 =<br />

x n e − x2<br />

2 dx +<br />

∫<br />

(n − 1) x n−2 e − x2<br />

2 dx.<br />

From which<br />

∫<br />

x n e − x2<br />

2 dx = (n − 1) ∫ x n−2 e − x2<br />

2 dx − e − x2<br />

2 x n−1<br />

∞∫<br />

∞∫<br />

x n e − x2<br />

2 dx = (n − 1) x n−2 e − x2<br />

2 dx<br />

−∞<br />

−∞<br />

422

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