Independence of random variables
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Transcript Independence of random variables
Law of Large Numbers
• Toss a coin n times.
• Suppose
1
Xi
0
if i th toss cam eup H
if i th toss cam eup T
• Xi’s are Bernoulli random variables with p = ½ and E(Xi) = ½.
1 n
• The proportion of heads is X n X i .
n i 1
• Intuitively X n approaches ½ as n ∞ .
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Markov’s Inequality
• If X is a non-negative random variable with E(X) < ∞ and a >0 then,
P X a
EX
a
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Chebyshev’s Inequality
• For a random variable X with E(X) < ∞ and V(X) < ∞, for any a >0
P X E X a
V X
a2
• Proof:
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Back to the Law of Large Numbers
• Interested in sequence of random variables X1, X2, X3,… such that the
random variables are independent and identically distributed (i.i.d).
Let
1 n
Xn Xi
n i 1
Suppose E(Xi) = μ , V(Xi) = σ2, then
1 n
1 n
E X n E X i E X i
n i1 n i 1
and
1 n
1
V X n V X i 2
n i1 n
n
V X
i 1
i
2
n
• Intuitively, as n ∞, V X n 0 so X n EX n
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• Formally, the Weak Law of Large Numbers (WLLN) states the following:
• Suppose X1, X2, X3,…are i.i.d with E(Xi) = μ < ∞ , V(Xi) = σ2 < ∞, then for
any positive number a
P Xn a 0
as n ∞ .
This is called Convergence in Probability.
Proof:
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Example
• Flip a coin 10,000 times. Let
1
Xi
0
if i th toss cam eup H
if i th toss cam eup T
• E(Xi) = ½ and V(Xi) = ¼ .
• Take a = 0.01, then by Chebyshev’s Inequality
1
1
1
1
P X n 0.01
2
2
4
410,000 0.01
• Chebyshev Inequality gives a very weak upper bound.
• Chebyshev Inequality works regardless of the distribution of the Xi’s.
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Strong Law of Large Number
• Suppose X1, X2, X3,…are i.i.d with E(Xi) = μ < ∞ , then X n converges to μ
as n ∞ with probability 1. That is
1
P lim X 1 X 2 X n 1
n n
• This is called convergence almost surely.
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Continuity Theorem for MGFs
• Let X be a random variable such that for some t0 > 0 we have mX(t) < ∞ for
t t0 ,t0 . Further, if X1, X2,…is a sequence of random variables with
mX n t and lim m X t m X t for all t t0 ,t0
n
n
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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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).
X E X n X n n
• Let Z n n
then we have that E(Zn) = 0, V(Zn) = 1.
V X n
n
• 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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Example
• Suppose X is Poisson(900) random variable. Find P(X > 950).
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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
n
2
variance σ and , S X
n
i 1
i
then by the WLLN we have that
Sn
in probability.
n
• 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
S n n
Zn
and we have that E(Zn) = 0, V(Zn) = 1.
n
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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
n
Sn X i
i 1
Then, lim P S n n x x for - ∞ < x < ∞
n
n
where Ф(x) is the cdf for the standard normal distribution.
• This is equivalent to saying that Z n S n n converges in distribution to
n
Z ~ N(0,1).
•
Xn
P
x x
Also, lim
n
n
i.e. Z n X n converges in distribution to Z ~ N(0,1).
n
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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 P X1 X n 3n x 3n x as n ∞.
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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 P X 1 X n np x np1 p x as n ∞.
• Let Yn = X1 + … + Xn then Yn has a Binomial(n, p) distribution.
So for large n, PYn y P Yn np
np1 p
y np
y np
np1 p
np1 p
• 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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