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MAYbe
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How can i derive this first passage probability formula

April 19th, 2017, 4:06 pm

In this article the authors present this first passage probability formula 

$$Z = {1 \over \sigma }\left[ {\log S/{S_t} + (r - {1 \over 2}{\sigma ^2})t} \right]$$

${{S_t}}$ value of the stock at time t , r ror on the stock, $\sigma $ standard deviation.

The authors ref. Feller (1971) "An Introduction to Probability Theory and Its Applications, Vol 2. " But i have been unable to find the formula , does anyone know where this comes from and any literature where this formula is presented?


Link to feller : http://ruangbacafmipa.staff.ub.ac.id/fi ... Feller.pdf


Link to article : download/file.php?id=7750
 
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list1
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Re: How can i derive this first passage probability formula

April 19th, 2017, 5:31 pm

The idea to get expectation of the 1st passage time is following. Right Ito formula up to this moment, take expectation- eliminate stoch integral. Under integral you get parabolic operator Lu. Solve eq Lu = -1. Solve parabolic eq Lu = -1 , you need specify the boundary condition. Using solution for eq Lu = -1 we arrive at the formula for E[$]\tau[$]. I read in the book I. Gikhman, A. Skorohod Stochastic differential equation 1ed. It might other books too.

 
 
MAYbe
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Re: How can i derive this first passage probability formula

April 19th, 2017, 5:34 pm

can you write it more clearly?
 
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list1
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Re: How can i derive this first passage probability formula

April 19th, 2017, 10:00 pm

Sorry, my response does not relate to your question. In the paper they introduced 'Z' as a random variable (2). Then they introduced a constant 'a' , (3). Then they assumed that 

[$]\frac{r}{2 \sigma^2}[$] = const             (1)

In this case the second term in Z and a equal to zero. Then they stated that "Equation (1) therefore requires an expression for the first-passage time of a standardized Brownian motion to the “position” a " . After that they referred to Feller's result about first passage time for Brown process. 
They did not specified whether their process S ( t ) has a drift r or [$]\mu[$]. In (1) [$]\sigma \not= 0[$] and it is not clear why they assumed that they deal with Brown motion. After that point they used known formulas. Sorry, it is not everything clear for me though I might be wrong.
 
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Cuchulainn
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Re: How can i derive this first passage probability formula

April 20th, 2017, 9:37 am

can you write it more clearly?
I think he means that the 1st passage time is the solution of a parabolic PDE.  Not exactly a closed solution but very computable and universal.
https://tel.archives-ouvertes.fr/tel-00009074/document
 
MAYbe
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Re: How can i derive this first passage probability formula

April 20th, 2017, 11:47 am

I wonder if it is derived from $$Z = \exp \left\{ {\sigma X(t) + (\sigma r - {1 \over 2}{\sigma ^2})t} \right\}$$ where $$X(t) = \log S/Sc$$
 
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outrun
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Re: How can i derive this first passage probability formula

April 20th, 2017, 11:53 am

I started to understand the underlying intuitive aspects of passage time formulas once I understood the reflection principle.
https://galton.uchicago.edu/~yibi/teach ... 23_6up.pdf
 
MAYbe
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Re: How can i derive this first passage probability formula

April 20th, 2017, 4:59 pm

Possible typo in first passage probability formula?

The first passage probability given by Bunch and Johnson is given as
$$Z = {1 \over \sigma }\left[ {\log {S \over {Sc}} + (r - {1 \over 2}{\sigma ^2})t} \right]$$

after reading some literature should it not be the case that the first passage probability should be
$$Z = {1 \over {\sigma \sqrt t }}\left[ {\log {S \over {Sc}} + (r - {1 \over 2}{\sigma ^2})t} \right]$$
 
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list1
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Joined: July 22nd, 2015, 2:12 pm

Re: How can i derive this first passage probability formula

April 20th, 2017, 9:19 pm

Possible typo in first passage probability formula?

The first passage probability given by Bunch and Johnson is given as
$$Z = {1 \over \sigma }\left[ {\log {S \over {Sc}} + (r - {1 \over 2}{\sigma ^2})t} \right]$$

after reading some literature should it not be the case that the first passage probability should be
$$Z = {1 \over {\sigma \sqrt t }}\left[ {\log {S \over {Sc}} + (r - {1 \over 2}{\sigma ^2})t} \right]$$
It was said before (3) that "Under the standard assumptions, Z is normal with a zero mean and a variance of t" , ie it looks Z does not normalized by sq root of t. If you look through (3-12) you could not find where Z is used. More interesting is an assumption $$\gamma = \frac{r}{\sigma^2} = 1$$ and the next statement that "Equation (1) therefore requires an expression for the first-passage time of a (standardized) Brownian motion to the “position” a. It was not written but it is very possible that  $$dS_t  = r S_t dt  +  \sigma  S_t dw ( t )$$
though it is not a fact because this is underlying of the option that guarantees perfect hedging. If perfect hedging does not mention then it might be real stock with drift [$]\mu[$]. Why does assumption [$]\gamma = 1 [$] is sufficient that S is a Brownian process does not clear it looks that under this condition S is log-Brownian. Though, who knows but they have a proof. If this problem is clarified they do not use Z for P at least up to(12)