Question

Theorem still holds of we replace it with Prove that the Central Limit Sn (defined in Problem, 2), i.e. Xu -M d N(0.1) Solin

Answer 1

ANSWERS:

Theorem 2.54. Let X, be i.i.d with finite third moment, and having zero mean and unit variance. Then, converges in distribution to N0,1). PROOF. By Lévy's continuity theorem, it suffices to show that the characteristic functions of niSn converge to the of N(0,1). Note that V n(t):=E (@its/vā] =v ( where y is the c.f of X1. Use Taylor expansion eitx = 1+it ** x3 for some r* [0,x] or [x,0). Apply this with X1 in place of x, tn-12 in place of t, take expectations and recall that E[X]]= 0 and E[X²]=1 to get "A)=-- +RC), where R.40 = 2* ["Xx]. Clearly, Rn(t) = Cn 32 for a constant C (that depends on t but not n). Hence nRn(t)- and by Exercise 2.53 we conclude that for each fixed tER, va(o)= (1+R2(0)* -e? which is the c.f of N(0,1).

OR -

X1, X2, ......, Xn are independent, identically distributed RVs (i.e. they all have the same PMF if discrete, or the same PDF if continuous), each with mean u and variance 6, and let Sn= X1 + X2 + ...... + Xn THEN limPlc, Solin S 0,) - Talou ?du. In words, when n gets large, the quantity Zn = behaves like a standard normal R.V., i.e. olyn oks normal with mean u and standard deviation or, equivalently (multiplying Zn by ), 2. Sn looks normal with mean nu and standard deviation no

Proof: Beyond the scope of our course. See Spiegel (p.131, example 4.25: Proof of CLT using moment generating functions) if you are interested. What is important is that, as n gets very large, although X1, X3, .... Xn may not be individually normal (or even continuous!), Sn always is. Also, since becomes a tiny number, n = Ã has a Gaussian distribution that is vn sharply peaked at u, as indicated in the following picture: 2.6 2. 8 3 3.2 3.4