Re: Probability of exceeding a specific value

On Sep 18, 10:31 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
matt271829-n...@xxxxxxxxxxx writes:
On Sep 18, 4:31 pm, grosu <gr...@xxxxxxxxxxxxx> wrote:
The price is not normally distributed.
The initial price is known to be 10$.
Then, the price changes every time step, with the change in the price
being normally distributed, that is the Delta of the price. But, the
price can go down as well as up.
I am not interested in the price after k time stpes. I am interested in
the probability of the price exceeding 13$, and the time expected for
this to hapen.

Although, as has been pointed out, the probability of the price
eventually exceeding $13 is 1, the expected time for this to happen is
infinite. Isn't it?

Oops, yes... I should have known that. In fact, the process is just a
Brownian motion observed at discrete times, so it's sufficient to note
that the expected time for Brownian motion to hit 3 is infinite.

On Sep 18, 10:31 pm, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
matt271829-n...@xxxxxxxxxxx writes:
On Sep 18, 4:31 pm, grosu <gr...@xxxxxxxxxxxxx> wrote:
The price is not normally distributed.
The initial price is known to be 10$.
Then, the price changes every time step, with the change in the price
being normally distributed, that is the Delta of the price. But, the
price can go down as well as up.
I am not interested in the price after k time stpes. I am interested in
the probability of the price exceeding 13$, and the time expected for
this to hapen.

Although, as has been pointed out, the probability of the price
eventually exceeding $13 is 1, the expected time for this to happen is
infinite. Isn't it?

Oops, yes... I should have known that. In fact, the process is just a
Brownian motion observed at discrete times, so it's sufficient to note
that the expected time for Brownian motion to hit 3 is infinite.

Given that the expectation is infinite, I imagine that the OP might
find the distribution of the first time to hit $13 more useful. Then I
think the discrete problem (discrete jumps of N(0,1)) is different
from the continuous case, even though the price distributions at some
given time are the same. Basically because with the continuous case
you've got the chance of hitting $13 and then falling back between
timesteps.

Failing to see how to find a closed form for the discrete case, I
looked at the continuous case, translated to start at zero and looking
to hit k (so in the original problem k = 3). I got the probability
density function of the first time to hit k to be

1/Sqrt(2*pi) * k * t^(-3/2) * Exp(-k^2/(2*t))

with the cumulative distribution being

1 - Erf(k/Sqr(2*t))

Ans, as, erm, expected the expected value of t looks to be infinite.

Comparing this with simulations of the discrete case there is a fair
discrepancy.

.

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