WEBVTT
Kind: captions
Language: en
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Now we move into next example, example 6,
Let X1, X2, so on be a sequence of independent
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random variables each having probability mass
function, probability of Xi is equal to 1
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that is same as probability of Xi takes the
value minus 1, the probability is 1 by 2.
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This is valid for that means it is a sequence
of iid random variables and they are discrete
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type.
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Define M suffix k as the sum of first k Xi
random variables.
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So this running index is k is equal to 1,
2 and so on.
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So we are defining a sequence of a random
variable Mk by summing first k Xi random variables.
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For a fixed integer n, we define another sequence
of random variable that is denoted by W superscript
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n of t that is nothing but 1 divided by square
root of n M suffix n times t.
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This is for all t greater than or equal to
zero such that
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n times t is an integer.
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So we are defining another sequence of a random
variables W superscript n of t that is 1 divided
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by square root of n times Mn of t where nt
is an integer.
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So this is valid for all t greater than or
equal to zero.
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If you find out the mean and variance for
the difference of the random variable of W
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n of t minus W n of s for zero less than or
equal to s or less than or equal to t, this
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quantity will be zero.
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That means W n of t is a 1 divided by square
root of n Mn of t and the way we define the
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Mn of t that is the summation of Xi and the
probability of Xi is equal to 1 and the probability
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of X is equal to minus 1, minus 1 is 1 by
2.
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Therefore, the mean of Xi are going to be
zero because of that the expectation of or
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mean of W n of t minus W n of s that is equal
to zero.
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Also if you evaluate the variance of W n of
t minus W n of s by finding first variance
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of Xi’s using that you find out the variance
of Mn of t then find out the variance of W
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n of t minus W n of s that is going to be
t minus s.
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It need a calculation of expectation of a
Xi square then using expectation of a Xi square
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and the expectation of Xi, you can find out
the variance of Xi using variance of Xi, you
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can find out the variance of W n of t, then
you find out the variance of W n of t minus
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W n of s.
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By using mean and variance, for fix t greater
than or equal to zero as n tends to infinity
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we can conclude W n of t tends to a random
variable X and this converges takes place
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in distribution using CLT, one can conclude
W n of t converges to the random variable
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X and the converges in distribution where
X is normal distribution with the mean zero
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and the variance t.
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Using a central limit theorem, one can prove,
W n of t converges to X in distribution where
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X is normal distribution with a mean zero
and the variance t.
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This result is very useful in Brownian motion
and this same problem will be discussed in
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detail when we are discussing the module of
a Brownian motion.