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Brownian

Actuary.Brownian History

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October 09, 2007, at 10:40 PM by 24.148.11.164 -
Changed line 58 from:
!!Useful computation
to:
!!Useful Computation
October 09, 2007, at 10:40 PM by 24.148.11.164 -
Changed lines 56-62 from:
Use multiplication rules to simplify the result.
to:
Use multiplication rules to simplify the result.

!!Useful computation
*Involving expectation

*Involving variance

October 09, 2007, at 10:38 PM by 24.148.11.164 -
Changed lines 24-25 from:
*{$ X(t+s)-X(t)  $} is normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $};  in particular,
{$ X(s)  $} is  normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $}.
to:
*{$ X(t+s)-X(t)  $} is normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $};  in particular, {$ X(s)  $} is  normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $}.
Changed line 26 from:
The term {$\alpha dt$} is called the ''drift'' in the process; it's the deterministic part of the equation and deals with the long term behavior of the process.  The term {$\sigma dZ(t)$} is the random part, which deals with the short term behavior.
to:
->The term {$\alpha dt$} is called the ''drift'' in the process; it's the deterministic part of the equation and deals with the long term behavior of the process.  The term {$\sigma dZ(t)$} is the random ''noise'' part, which deals with the short term behavior.
October 09, 2007, at 10:36 PM by 24.148.11.164 -
Changed lines 19-21 from:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n |Z(iT/n)-Z((i-1)T/n) = \infty. $$}

to:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n |Z(iT/n)-Z((i-1)T/n)| = \infty. $$}

October 09, 2007, at 10:36 PM by 24.148.11.164 -
Changed line 40 from:
{$$E(X(t)) = X(0)e^{\alpha t},$$
to:
{$$E(X(t)) = X(0)e^{\alpha t},$$}
October 09, 2007, at 10:35 PM by 24.148.11.164 -
Changed line 17 from:
The quadratic variation of a Brownian motion is deterministic.  This implies higher-ordered variations are zero.  This makes the multiplication rules below possible.
to:
The quadratic variation of a Brownian motion is deterministic.  This implies higher-ordered variations are zero. 
Added line 30:
Changed lines 36-37 from:
*{$$X(t)= X(0)e^{(\alpha - 0.5 \sigma^2)t+\sigma\sqrt{t}Z(t)}. $$}
to:
*We can verify that
{$$X(t)= X(0)e^{(\alpha - 0.5 \sigma^2)t+\sigma\sqrt{t}Z(t)}, $$}
using Ito's lemma below.
*A calculation shows
{$$E(X(t)) = X(0)e^{\alpha t},$$
which means {$ \alpha $} is the ''expected continuously compounded return'' on {$X$}.
Changed lines 43-44 from:

to:
!!Ito Processes
These are processes following
{$$ dX(t) = \alpha(X) dt + \sigma(X) dZ(t),$$}
where {$ \alpha $} and {$ \sigma $} are functions of {$X$}.  All processes above are examples.

!!Multiplication Rules
Basically approximations using 1st power of {$dt$}.
{$$ dt dZ = 0; $$}
{$$ (dt)^2 =0; $$}
{$$ (dZ)^2 = dt $$}

!!Ito's Lemma
Let {$C(a, b)$} be a {$C^{2,1}$} function, and let {$S(t)$} be an Ito process.  Then
{$$d C(S, t) = C_a(S,t) dS + \frac{1}{2} C_{aa}(S, t) (dS)^2 + C_b(S,t) dt. $$}
Use multiplication rules to simplify the result.
October 09, 2007, at 09:47 PM by 24.148.11.164 -
Changed line 25 from:
{$ Z(s)  $} is  normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $}.
to:
{$ X(s)  $} is  normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $}.
Changed lines 28-29 from:
to:
*We also have
{$$ X(t)= \alpha t + \sigma Z(t).$$}
Changed line 31 from:
Here the process follows
to:
*Here the process follows
Changed lines 34-36 from:
{$$ dX(t)/X(t) = = \alpha dt + \sigma dZ(t).$$}
to:
{$$ dX(t)/X(t) = d \ln X(t) = \alpha dt + \sigma dZ(t).$$}
*{$$X(t)= X(0)e^{(\alpha - 0.5 \sigma^2)t+\sigma\sqrt{t}Z(t)}. $$}

October 09, 2007, at 09:16 PM by 24.148.11.164 -
Changed lines 19-21 from:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n |Z(iT/n)-Z((i-1)T/n)}^2 = \infty. $$}

to:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n |Z(iT/n)-Z((i-1)T/n) = \infty. $$}

Changed lines 23-36 from:
to:
{$X(t)$} is modified from {$Z(t)$} by introducing two parameters, {$ \alpha $} and {$ \sigma $}.
*{$ X(t+s)-X(t)  $} is normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $};  in particular,
{$ Z(s)  $} is  normally distributed with mean {$\alpha s$} and variance {$ \sigma^2 s $}.
*For an incremental time {$dt$}, {$$ dX(t) = X(t+dt)-X(t) = \alpha dt + \sigma dZ(t).$$}
The term {$\alpha dt$} is called the ''drift'' in the process; it's the deterministic part of the equation and deals with the long term behavior of the process.  The term {$\sigma dZ(t)$} is the random part, which deals with the short term behavior.

!!Geometric Brownian Motion
Here the process follows
{$$ dX(t) = X(t+dt)-X(t) = \alpha X(t) dt + \sigma X(t) dZ(t),$$}
or
{$$ dX(t)/X(t) = = \alpha dt + \sigma dZ(t).$$}
 

October 09, 2007, at 07:09 PM by 24.148.11.164 -
Changed lines 5-7 from:
*{$ Z(0)=0 $};
*{$ Z(t+s)-Z(t)  $} is normally distributed with mean 0 and variance {$ s $};
*Nonoverlapping intervals are independently distributed.  That is, {$ Z(t+a)-Z(t) $} is dndependent of {$Z(t)- Z(t-b)$}, for all {$a, b >0$}.
to:
*{$ Z(0)=0 $}.
*{$ Z(t+s)-Z(t)  $} is normally distributed with mean 0 and variance {$ s $};  in particular,
{$ Z(s)
  $} is  normally distributed with mean 0 and variance {$s$}.
*Nonoverlapping intervals are independently distributed.  That is, {$ Z(t+a)-Z(t) $} is independent
of {$Z(t)- Z(t-b)$}, for all {$a, b >0$}.
Changed lines 13-16 from:
*Binomial approximation:  Assume that the change in {$ Z(t) $} is a binomial distribution, with a scale factor
*quadratic variation
*total
variation
to:
*Binomial approximation: 
->Let{$ dt $} be a short change in time.  Then {$$ dZ(t) = Z(t+dt)-Z(t) = Y(t)\sqrt{dt},$$} where {$ Y(t) $} is the binomial distribution taking values {$\pm 1$} with probability 0.5.
*quadratic
variation:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n (Z(iT/n)-Z((i-1)T/n))^2 = T. $$}
The quadratic variation of a Brownian motion is deterministic.  This implies higher-ordered variations are zero.  This makes the multiplication rules below possible.
*total variation:
->{$$\lim_{n\rightarrow \infty} \sum_{i=1}^n |Z(iT/n)-Z((i-1)T/n)}^2 = \infty. $$}


!!Arithmetic Brownian Motion

Added lines 1-15:
!!Brownian Motion
A ''stochastic process'' is  collection of random variables indexed by time {$t$}, or a function from the reals to a set of random variables.  A ''Brownian motion'' is a stochastic process that is a random walk occurring in continuous time, with continuous movement.  More precise definition:

Let {$ Z(t) $} be the value of a Brownian motion at time {$ t $}.  It has the following properties:
*{$ Z(0)=0 $};
*{$ Z(t+s)-Z(t)  $} is normally distributed with mean 0 and variance {$ s $};
*Nonoverlapping intervals are independently distributed.  That is, {$ Z(t+a)-Z(t) $} is dndependent of {$Z(t)- Z(t-b)$}, for all {$a, b >0$}.
*{$ Z(t) $} is continuous.

!!!Properties:
*{$ Z(t) $} is a ''martingale'', that is, {$ E[Z(t+s) | Z(t)] = Z(t) $}.  This is also called a ''diffusion process''.
*Binomial approximation:  Assume that the change in {$ Z(t) $} is a binomial distribution, with a scale factor
*quadratic variation
*total variation

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Page last modified on October 09, 2007, at 10:40 PM