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