Chapter 10: Generating Functions

402 CHAPTER 10. GENERATING FUNCTIONS
This is hard to check directly, but easy to check by using characteristic functions.
Note first that
∫ +∞
µ 2 = E(X 2 x 2
)=
−∞ π(1 + x 2 ) dx = ∞
so that µ 2 is infinite. Nevertheless, we can define the characteristic function k X (τ)
of x by the formula
∫ +∞
k X (τ) = e iτx 1
−∞ π(1 + x 2 ) dx .
This integral is easy to do by contour methods, and gives us
Hence,
and since
we have
k X (τ) =k Y (τ) =e −|τ| .
k X+Y (τ) =(e −|τ| ) 2 = e −2|τ| ,
k Z (τ) =k X+Y (τ/2) ,
k Z (τ) =e −2|τ/2| = e −|τ| .
This shows that k Z = k X = k Y , and leads to the conclusions that f Z = f X = f Y .
It follows from this that if X 1 , X 2 , ...,X n is an independent trials process with
common Cauchy density, and if
A n = X 1 + X 2 + ···+ X n
n
is the average of the X i , then A n has the same density as do the X i . This means
that the Law of Large Numbers fails for this process; the distribution of the average
A n is exactly the same as for the individual terms. Our proof of the Law of Large
Numbers fails in this case because the variance of X i is not finite.
Exercises
1 Let X be a continuous random variable with values in [ 0, 2] and density f X .
Find the moment generating function g(t) for X if
(a) f X (x) =1/2.
(b) f X (x) =(1/2)x.
(c) f X (x) =1− (1/2)x.
(d) f X (x) =|1 − x|.
(e) f X (x) =(3/8)x 2 .
Hint: Use the integral definition, as in Examples 10.15 and 10.16.
2 For each of the densities in Exercise 1 calculate the first and second moments,
µ 1 and µ 2 , directly from their definition and verify that g(0)=1,g ′ (0) = µ 1 ,
and g ′′ (0) = µ 2 .