week 11 1
Convergence in Distribution• Recall: in probability if
• Definition
Let X1, X2,…be a sequence of random variables with cumulative distribution functions F1, F2,… and let X be a random variable with cdf FX(x). We say that the sequence {Xn} converges in distribution to X if
at every point x in which F is continuous.
• This can also be stated as: {Xn} converges in distribution to X if for all such that P(X = x) = 0
• Convergence in distribution is also called “weak convergence”. It is weaker then convergence in probability. We can show that convergence in probability implies convergence in distribution.
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week 11 2
Simple Example
• Assume n is a positive integer. Further, suppose that the probability mass function of Xn is:
Note that this is a valid p.m.f for n ≥ 2.
• For n ≥ 2, {Xn} convergence in distribution to X which has p.m.f
P(X = 0) = P(X = 1) = ½ i.e. X ~ Bernoulli(1/2)
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week 11 3
Example
• X1, X2,…is a sequence of i.i.d random variables with E(Xi) = μ < ∞.
• Let . Then, by the WLLN for any a > 0
as n ∞.
• So…
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week 11 4
Continuity Theorem for MGFs• Let X be a random variable such that for some t0 > 0 we have mX(t) < ∞ for
. Further, if X1, X2,…is a sequence of random variables with
and for all
then {Xn} converges in distribution to X.
• This theorem can also be stated as follows:Let Fn be a sequence of cdfs with corresponding mgf mn. Let F be a cdf with mgf m. If mn(t) m(t) for all t in an open interval containing zero, then Fn(x) F(x) at all continuity points of F.
• Example:Poisson distribution can be approximated by a Normal distribution for large λ.
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week 11 5
Example to illustrate the Continuity Theorem
• Let λ1, λ2,…be an increasing sequence with λn ∞ as n ∞ and let {Xi} be a sequence of Poisson random variables with the corresponding parameters.
We know that E(Xn) = λn = V(Xn).
• Let then we have that E(Zn) = 0, V(Zn) = 1.
• We can show that the mgf of Zn is the mgf of a Standard Normal random variable.
• We say that Zn convergence in distribution to Z ~ N(0,1).
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week 11 6
Example
• Suppose X is Poisson(900) random variable. Find P(X > 950).
week 11 7
Central Limit Theorem
• The central limit theorem is concerned with the limiting property of sums of random variables.
• If X1, X2,…is a sequence of i.i.d random variables with mean μ and variance σ2 and ,
then by the WLLN we have that in probability.
• The CLT concerned not just with the fact of convergence but how Sn /n fluctuates around μ.
• Note that E(Sn) = nμ and V(Sn) = nσ2. The standardized version of Sn is
and we have that E(Zn) = 0, V(Zn) = 1.
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week 11 8
The Central Limit Theorem• Let X1, X2,…be a sequence of i.i.d random variables with E(Xi) = μ < ∞
and Var(Xi) = σ2 < ∞. Suppose the common distribution function FX(x) and the common moment generating function mX(t) are defined in a neighborhood of 0. Let
Then, for - ∞ < x < ∞
where Ф(x) is the cdf for the standard normal distribution.
• This is equivalent to saying that converges in distribution to Z ~ N(0,1).
• Also,
i.e. converges in distribution to Z ~ N(0,1).
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week 11 9
Example
• Suppose X1, X2,…are i.i.d random variables and each has the Poisson(3) distribution. So E(Xi) = V(Xi) = 3.
• The CLT says that as n ∞. xnxnXXP n 331
week 11 10
Examples• A very common application of the CLT is the Normal approximation to the
Binomial distribution.
• Suppose X1, X2,…are i.i.d random variables and each has the Bernoulli(p) distribution. So E(Xi) = p and V(Xi) = p(1- p).
• The CLT says that as n ∞.
• Let Yn = X1 + … + Xn then Yn has a Binomial(n, p) distribution.
So for large n,
• Suppose we flip a biased coin 1000 times and the probability of heads on any one toss is 0.6. Find the probability of getting at least 550 heads.
• Suppose we toss a coin 100 times and observed 60 heads. Is the coin fair?
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