WEBVTT
Kind: captions
Language: en
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Now we will move into the theorem 5 which
discuss the differential equation corresponding
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to the probability generating function for
Pij of t. Let that u of s is equal to summation
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ak s power k summation over k. Then the probability
generating function for Pij of t satisfies
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partial derivative of Psi of t, s with respect
to t is equal to partial derivative of the
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Psi of t, s with respect to s multiplied us
and partial derivative of Psi of t,s with
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respect to t is same as u of Psi of t, s with
the initial condition Psi of 0,s is same as
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a summation over j Pij of 0 s power j but
that is nothing but s. So the theorem 5 gives
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the partial differential equation and the
ordinary differential equation satisfied by
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the partial by probability generating function
of a Pij of t.
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Let us see the proof. We start with Psi of
h of s, Psi of h,s that is nothing but the
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summation over j P of i,j of s, P of i,j of
h s power j; substitute the Pij of h and simplify
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you will get the first term will be s. The
second term will be h times summation over
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j aj, sj the second term will be order of
h. You know that u of s is same as summation
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over j, a suffix j of s power j therefore
the probability generating function for Pij
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of h, s that is nothing but s plus h u of
s plus order of h. We know that by the theorem
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four, psi of t plus h,s will be Psi of t,
psi of h of s, h,s. So substitute psi of h,
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s with s plus h of us plus order of h therefore
this will be psi of t, s plus h of us plus
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order of h. By Taylor's theorem we expand
the right hand side with respect to the second
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variable. Therefore, the right hand side will
be psi of t,s. The second term will be partial
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derivative of psi with respect to s times
h of us h times us plus order of h. All the
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other term vanishes throughout divide by h
and take psi of t, s in the left side.
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Therefore, the left-hand side becomes psi
of t plus h,s minus psi of t, s divided by
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h whereas in the right hand side will be partial
derivative of psi with respect to s times
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u of s order of h divided by h.
Taking h tends to 0 positive we get to the
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partial differential equation dou psi divided
by dou t is equal to dou psi by dou s time
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us. This is a partial differential equation
for the function of two variables psi t,s
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with the initial condition psi of 0,s is s.
So we have proved the first part of theorem
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five. Similarly one can prove the second part
of theorem five. The proof of a part two.
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You start with the probability generating
function psi of v plus h,s is same as psi
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of v, psi of h,s. By Taylor's theorem the
right hand side becomes psi of v, s plus h
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of u psi of v,s plus order of h. Then psi
of v,s in the left hand side divide throughout
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by h we will get this equation. Now limit
h tends to 0 plus and then substitute v is
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equal to t in this equation limit meta h tends
to 0 plus and substitute the v is equal to
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h, v is equal to t we get partial derivative
of psi with respect to t is equal to u of
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psi of t,s. This is ordinary differential
equation with the initial condition psi of
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0, s equal to s. So in the theorem 5 we conclude
the probability generating function satisfies
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the partial differential equation and the
initial and ordinary differential equation
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with the initial conditions psi of 0,s equal
to s.
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Now we will find out the mean of Z of t. You
start with the partial differential equation
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satisfied by probability generating function.
By differentiating with respect to s and the
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interchanging the order of differentiation
on the left-hand side we get the left-hand
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side is the second order partial derivative
of psi with respect to t and with respect
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to s. The right hand side u of s second order
partial derivative of psi with respect to
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s u dash of s partial derivative of psi with
respect to s. If s equal to 1 you know that
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u of 1 will be 0. Suppose the m of t will
be the mean of Z of t that is nothing but
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the partial derivative of psi with respect
to s then substitute s is equal to 1. Therefore
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this equation becomes partial derivative of
m of t with respect to t is equal to u dash
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of 1 m of t since m is the single variable
so this is the ordinary differential equation.
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So dmt by dt is equal to u dash of 1 times
m of t where m of t is a mean of Z of t. But
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since Z of 0 is equal to 1 m of 0 also 1.
Therefore you can solve this ordinary differential
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equation. The initial condition m of 0 is
equal to 1 hence the solution will be m of
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t is equal to e power t times u dash of 1.
Now we can discuss the mean of Z of t based
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on the value of u dash of 1. Before that
we discuss the probability of extinction.
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That is defined by q that is nothing but limit
t tends to infinity probability of 1,0 of
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t. This is called a probability of extension
that is denoted by the letter quarter.
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Now we will try to find out the probability
of extinction q. Assume that a naught is strictly
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greater than 0 otherwise extinction is impossible.
It is an enough to consider the case where
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the process starts with a single individual
at time 0. It means Z of 0 is equal to 1.
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With s equal to 0 you will get a Pi of 0 of
t that is nothing but a P10 of t power i in
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the probability generating function of Pij
of t.
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Now we will prove that Pij0 of t is a non
decreasing in t. You start with Pi0 of t plus
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v that is nothing but psi of t plus v,0 of
power i that is same as psi of t, psi of v,0
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power i that will be greater than or equal
to psi of 0,t power i but that is same as
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Pi0 of t. Hence we proved Pi0 of t is a non
decreasing in t.
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Let t be the fixed positive number consider
a discrete time branching process Z of 0,
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Z of t naught, Z of 2 times Z naught and so
on, Z of n times t naught where Z of t is
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population size at time t. Assume that the
population size at time 0 is 1, Z of 0 is
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equal to 1. Since Z of t is assumed to be
Markov process the discrete process Yn Y suffix
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n that is nothing but Z of n of t naught will
be a discrete time Markov chain which is also
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a discreet time branching process because
Z of t is a continuous time branching process
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therefore Y of n will form a discrete-time
branching process which is also a discrete
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time Markov chain. By the hypothesis of homogeneity
of probability function of Z of t and the
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probability generating function of Pij of
t that is nothing but probability generating
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function of Pij of t power P1 of 1 j of t
power i.
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This we have proved it in the earlier. We
have proved it in earlier therefore using
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these two we are finding summation over k
the conditional probability of Yn plus 1 is
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equal to k given Yn is equal to i multiplied
by s power k that is nothing but expectation
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of s power Yn plus 1 given Yn is equal to
i. That is same as because Yn is nothing but
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Z of n times t naught therefore Yn plus 1
is nothing but Z of Yn plus 1 times t naught.
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So it implies Yn by Zn n times t naught and
Yn plus 1 by Z of n plus 1 times t naught.
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This is true for all n therefore that the
same as expectation of s power Z of t naught
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given Z of 0 is equal to i but that is nothing
but psi of t naught,s power i but that can
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be written as expectation of s power Z of
t naught given Z naught is equal to 1 whole
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power i that is same as expectation of s power
Y1 given Y naught is equal to 1 the whole
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power i. This shows that Yn is a branching
process. Yn is a discrete-time branching process.
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So using this we have proved the Yn is a discrete
time branching process. The probability generating
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function for the number of offspring of a
single individual in this process is psi of
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t naught,s. By theorem 3 we know that the
probability of extension for Yn that is a
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discrete time branching process is the smallest
non-negative root of the equation psi of t
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naught,s equal to s.
So by using the theorem 3 we conclude the
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probability of extinction for the Yn process
is the smallest non-negative root of the equation
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psi of Z naught, s equal to s. But we know
that probability of Yn is equal to 0 for some
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n that is same as limit n tends to infinity
of probability of Yn is equal to 0 that is
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same as a limit n tends to infinity of a probability
of n times Z naught is equal to 0 but that
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is same as limit t tends to infinity of probability
Zt is equal to 0. By definition this is nothing
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but q that is a probability of extinction.
Hence the probability of extinction q of a
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continuous time branching process Z of t is
the smallest non-negative root of the equation
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psi of t naught,s is equal to s. Here we have
concluded by theorem 3 the probability of
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extinction for the discrete-time branching
process Yn is the smallest non-negative root
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of the equation psi of t naught,s equal to
s because of this we conclude the probability
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of extinction of the continuous-time branching
process Z of t is a smallest non-negative
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root of the equation psi of t naught,s equal
to s where t naught is any positive number.
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Hence we expect that we should be - we should
also be able to calculate q from equation
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that does not depend on time. From this equation
can able to calculate q from the equation
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that does not depends on time.