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Applied Statistics Using SPSS, STATISTICA, MATLAB and R

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n [ X ] = ∑ i=<br />

1<br />

A.6 Expectation, Variance <strong>and</strong> Moments 417<br />

v ≡ V ( x − x)<br />

f<br />

ˆ . A. 28<br />

i<br />

2<br />

i<br />

This is the so-called sample variance. The square root of v, s = v , is the<br />

sample st<strong>and</strong>ard deviation. In Appendix C we present a better estimate of v.<br />

The variance can be computed using the second order moment, observing that:<br />

2 2<br />

2 2 2<br />

[ ] = Ε[<br />

( X − µ ) ] = Ε[<br />

X ] − 2µ<br />

Ε[<br />

X ] + µ = Ε[<br />

X ] −<br />

V X µ . A. 29<br />

4. Gauss’ approximation formulae:<br />

i. [ g( X ) ] ≈ g(<br />

Ε[<br />

X ])<br />

Ε ;<br />

2<br />

⎡dg<br />

⎤<br />

g ⎢ ⎥ .<br />

⎢⎣<br />

dx Ε[ X ] ⎥⎦<br />

ii. V[<br />

( X ) ] ≈ V[<br />

X ]<br />

A.6.2 Moment-Generating Function<br />

The moment-generating function of a r<strong>and</strong>om variable X, is defined as the<br />

tX<br />

expectation of e (when it exists), i.e.:<br />

tX<br />

ψ ( t) = Ε[<br />

e ] . A. 30<br />

X<br />

The importance of this function stems from the fact that one can derive all<br />

moments from it, using the result:<br />

k [ X ]<br />

n<br />

d ψ X ( t)<br />

Ε =<br />

. A. 31<br />

n<br />

dt<br />

t=<br />

0<br />

A distribution function is uniquely determined by its moments as well as by its<br />

moment-generating function.<br />

Example A. 15<br />

Q: Consider a r<strong>and</strong>om variable with the Poisson probability function<br />

−λ<br />

k<br />

P(<br />

X = k)<br />

= e λ / k!<br />

, k ≥ 0. Determine its mean <strong>and</strong> variance using the momentgenerating<br />

function approach.<br />

A: The moment-generating function is:<br />

tX ∞ tk −λ<br />

k −λ<br />

[ e ] = ∑ e e λ / k!<br />

= e ∑<br />

k = 0<br />

∞<br />

k = 0<br />

ψ ( t) = Ε<br />

( λe<br />

)<br />

X<br />

t<br />

k<br />

/ k!<br />

.<br />

Since the power series expansion of the exponential is ∑ ∞ x<br />

k<br />

e = x / k!<br />

one<br />

k = 0<br />

can write:<br />

−λ<br />

λe<br />

t<br />

λ(<br />

e<br />

t<br />

−1)<br />

ψ ( t)<br />

= e e = e .<br />

X

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