The second example, this stochastic process is a continuous-time discrete state stochastic
process. So here let N(t) where t varies from 0 to infinity be a Poisson process with the
intensity ? or parameter ? and the F(t) is its natural filtration. The natural filtration
means it has the information till time t and each random variable N(t) for fixed t is F(t)
measurable. That is the meaning of a natural filtration.
We know that for fixed t, N(t) is a Poisson distributed random variable with the parameter
?. Therefore, the mean of N(t) that is same as the parameter ?. So the first condition
is satisfied. The second condition, for fixed t, N(t) has
to be F(t) measurable. Since it is a natural filtration, the N(t) is a F(t) measurable.
The third condition we are going to verify. The conditional expectation of N(t) given
F(s) where F(s) is the filtration at time s or the information till s. Obviously, here
s is less than t. So to find out this conditional expectation,
what we do? We add and subtract N(s) with the N(t). So instead of a conditional expectation
of N(t) given F(s), we subtract N(s) and add N(s).
Expectation is a linear operator. Therefore, you can split this, these three terms into
two different expectations, conditional expectations. Therefore, the first one you can keep N(t)
- N(s). The second one you can keep it separately N(s). Hence, you have conditional expectation
of N(t) - N(s) given the filtration F(s) plus conditional expectation of N(s) given F(s).
In the first term, the first term, this conditional expectation is nothing but you know the information
till time s and we are asking the conditional expectation of N(t) - (s) given F(s). That
means this t minus s and s, this is a non-overlapping intervals. S is the point. T minus s is the
non-overlapping intervals. So this is the random variable corresponding to the non overlapping
interval with respect to s. Therefore, non -- you know the property of
Poisson process. For s less than t, N(t) - N(s) is nothing but the increments. The increments
are stationary and independent. Therefore, the N(t) - N(s) is independent of F(s). If
it is independent, the conditional expectation is nothing but expectation of N(t) - N(s).
You know the Poisson process, the properties. Since N(t) is a Poisson process, N(t) - N(s)
for a fixed t and s, this is a Poisson distributed random variable with the mean ? times (t-s).
Therefore, this conditional expectation will be ? times (t-s) based on the N(t) - N(s)
is independent of F(s) and for fixed s and t, N(t) - N(s) is a Poisson distributed random
variable with the mean ? times (t-s) whereas the second term, conditional expectation of
N(s) given F(s) that means for information till time s, what is the expectation of N(s)
at the same time s? So since you know the information till or up to the time s, N(s)
is constant. N(s) is a constant. Therefore, expectation of constant is a constant.
Therefore, it is N(s). It's not a ? times s because you know the information till time
s. Once you know the information till time s, that means you know the value of N(s) also.
Once you know the value of N(s), therefore, N(s) is no more a random variable. So it's
a constant. So expectation of a constant is a constant. So it is N(s).
Hence, conditional expectation of N(t) given F(s) is same as you can take ?t in this side,
so conditional expectation of N(t) - ?t given F(s) is same as N(s) minus ? times s. You
see this is the expectation of N(t) - ?s given F(s). That is same expression. Therefore,
as such expectation of N(t) given F(s) is not N(s). It has the some positive value.
(t - s) is always greater than 0. Therefore, ? times (t - s) will be greater than 0. Therefore,
this conditional expectation is always greater than or equal to N(s).
Hence, N(t) is not a Martingale whereas if you treat N(t) - ?t as a stochastic process
over the t ranges from 0 to infinity, then this stochastic process is a Martingale. The
N(t) is not satisfying is same as N(s). The conditional expectation is not equal to N(s),
but it is greater than or equal to N(s). Therefore, N(t) is not a Martingale whereas if you make
another stochastic process, that is nothing but N(t) - ? times t for t greater than or
equal to 0, then this stochastic process satisfies the third condition and also satisfies the
other two conditions. Therefore, the N(t) - ?t is a Martingale.
Third example. This is related to the application of finance. Consider
the binomial tree model. Let Sn be a stochastic process and the Fn be the -- its natural filtration.
Define the probability of Sn+1 given u times Sn given Fn that is same as -- that is P and
the probability of Sn+1 is equal to d times Sn given Fn is equal to 1 - p where u and
d are the next value of Sn with the probability p and q respectively.
Therefore, suppose the previous, the nth value was Sn, then the next value will be, it is
decremented with d, therefore, d times Sn or it would have been incremented with u.
Therefore, u times Sn will be the Sn+1th value. Therefore, this -- this stochastic process
is called the binomial tree model. Now I am considering the discounted stochastic
process that is nothing but e power minus r times, basically e power rt St. Since it
is a discrete-time stochastic process, the first S1 is multiplied by e power minus r
whereas the second random variable S2 is multiplied by e power minus 2 times r and so on. Therefore,
the nth random variable Sn will be e power minus n times r where r is the riskless interest
rate. So whenever you multiply the e power minus r times t, the corresponding stochastic
process is called the discounted stochastic process.
The discounted stochastic process is a martingale if -- only if the right hand side is equal
to e power -n times r times s because here it is a conditional expectation of e-(n+1)r
Sn+1 given Fn. If this quantity is same as e-nr multiplied by Sn, then this discounted
stochastic process will be a martingale or it has the martingale property.
So this is the case only if -- if the p value takes er times -d and divided by u minus d
if the p is the probability of incremented by u, if the p is equal to er -d divided by
u-d, then the discounted stochastic process is a Martingale. So that is possible because
since p lies between 0 to 1, since the -- whenever the r is the riskless interest rate, er is
also lies between d to u. Therefore, the u - d and er -d, these values is going to be
lies between 0 to 1. So, therefore, with the proper value of r, d and u, the p can be -- if
the p is of this form, then the discounted stochastic process is a martingale.