Re: Probability of exceeding a specific value
- From: Robert Israel <israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx>
- Date: Tue, 18 Sep 2007 13:36:48 -0500
Randy Poe <poespam-trap@xxxxxxxxx> writes:
On Sep 18, 11:31 am, grosu <gr...@xxxxxxxxxxxxx> wrote:
The price is not normally distributed.
Not by assumption. But it turns out that it is.
The initial price is known to be 10$.
Then, the price changes every time step, with the change in the price
being normally distributed, that is the Delta of the price. But, the
price can go down as well as up.
And the price at step k is 10 + a sum of k normals with zero
mean and variance 1. Thus, the price at step k is normally
distributed with mean 10 and variance k.
I am not interested in the price after k time stpes. I am
interested in the probability of the price exceeding 13$,
There's no single answer to that. That probability increases
in time. It is P(Z > 3/sqrt(k)). The probability that $13 will
be exceeded eventually is 100%.
and the time expected for this to hapen.
I can't see an easy way to approach this, but it could be
approximated numerically by evaluated P(Z > 3/sqrt(k))
for a large finite number of values of k.
If u(p) is the expected number of steps for the price to exceed 10 + p
(starting with price 10), we have
u(p) = { 0 for p <= 0
{ 1 + int_0^infty f(p-x) u(x) dx for p > 0
where f(x) is the probability density function for the normal distribution
with mean 0 and variance 1. I don't know if this integral equation has
a closed-form solution, but you could probably do fairly well using finite
sums to approximate the integral.
--
Robert Israel israel@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
.
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