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