Kenneth Kuttler

59.18. THE CENTRAL LIMIT THEOREM 1925

which converges to 0 as n→ ∞ by the Dominated Convergence theorem. Therefore,

limn→∞

∣∣∣∣φ Zn(t)−

(1− t∗Σt

)n∣∣∣∣= 0

and solimn→∞

φ Zn(t) = e−

12 t∗Σt = φ Z (t)

where Z∼Np (0, ). Therefore, FZn (x)→ FZ (x) for all x because Rx ≡∏pk=1(−∞,xk] is a

set of λ Z continuity due to the assumption that λ Z≪mp which is implied by Z∼Np (0, ).This proves the theorem.

Here is the proof of the little inequality used above. The inequality is obviously trueif n = 1. Assume it is true for n. Then since all the numbers have absolute value no largerthan one, ∣∣∣∣∣n+1

∏i=1

zi−n+1

∏i=1

∣∣∣∣∣ ≤∣∣∣∣∣n+1

∏i=1

zi− zn+1

∏i=1

∣∣∣∣∣+

∣∣∣∣∣zn+1

∏i=1

wi−n+1

∏i=1

∣∣∣∣∣≤

∣∣∣∣∣ n

∏i=1

zi−n

∏i=1

∣∣∣∣∣+ |zn+1−wn+1|

≤n+1

∑k=1|zk−wk|

by induction.Suppose X is a random vector with covariance Σ and mean m, and suppose also that

Σ−1 exists. Consider Σ−(1/2) (X−m)≡ Y. Then E (Y) = 0 and

E (YY∗) = E(

Σ−(1/2) (X−m)(X∗−m)Σ

−(1/2))

= Σ−(1/2)E ((X−m)(X∗−m))Σ

−(1/2) = I.

Thus Y has zero mean and covariance I. This implies the following corollary to Theorem59.18.8.

Corollary 59.18.9 Let independent identically distributed random variables,{X j}∞

j=1

have mean m and positive definite covariance where −1 exists. Then if

Zn ≡n

∑j=1

−(1/2) (X j−m)√n

it follows that for Z∼Np (0,I) ,FZn (x)→ FZ (x)

for all x.

59.18. THE CENTRAL LIMIT THEOREM 1925which converges to 0 as n — co by the Dominated Convergence theorem. Therefore,txt \"t)h-({1-62,00 - (1-5)lim oy, (t) =e" 2!" = oy (t)noowhere Z ~N, (0,m). Therefore, Fz, (x) —> Fz (x) for all x because Rx = []}_, (—°, xx] is aset of Az continuity due to the assumption that Az, < mp which is implied by Z ~N, (0,m).This proves the theorem.Here is the proof of the little inequality used above. The inequality is obviously trueif n = 1. Assume it is true for n. Then since all the numbers have absolute value no largerthan one,lim =0noand son+1 n[[s-2[]i=li=ln+l n+l[l«-I]™i=1 i=1<n+lnzntt [[wi—[] wii=1 i=1+IA+ [Zn-t1 —Wn+1 |n n[He-T™i=] i=]n+l< DY lee —wei=by induction.Suppose X is a random vector with covariance © and mean m, and suppose also that©! exists. Consider £~(!/2) (X—m) = Y. Then E (Y) = 0 andE(vY) = & (2°?) (X—m)(X*—m)E('))mV E ((X—m)(X*—m))r/) = 1.Thus Y has zero mean and covariance J. This implies the following corollary to Theorem59.18.8.Corollary 59.18.9 Let independent identically distributed random variables,Jn ?J{Xj } j=lhave mean m and positive definite covariance w where w! exists. Then ifn _Z,=Y wl?) (Xj; —m)=it follows that for Z ~N, (0,1) ,Fy, (x) => Fz (x)for all x.