2
$\begingroup$

I am modeling a stock price that follows Geometric Brownian Motion and have the following:

$E(X)$ = .16 (16%)

$\sigma$ = .24 (24%)

$X_0$ = 95

$T$ = 1 (12 months)

I am trying to find the probability that the price of this stock will be below 93 at the end of this time period. I am calculating this analytically, using the Log Normal Distribution given as the following:

$P(X,t)$ = $1\over X $$ \cdot$$1\over {\sigma \sqrt{2 \pi t}}$$\cdot$$e^{-(ln(x)- ln(x_0)-(\mu- \sigma^2 /2)t)^2}\over 2\sigma^2t$

I can plug in the values as the following:

$P(X,t)$ = $1\over X $$ \cdot$$1\over {(.24) \sqrt{2 \pi (1)}}$$\cdot$$e^{-(ln(x)- ln(95)-((.16)- (.24)^2 /2)(1))^2}\over 2(.24)^2(1)$

But then I am still left with the X. My question, is this just the 93 value that should be plugged in? Would this represent the probability of the price being below 93 after this time period? What if we wanted to find the probability that the price would close above this 93 (just 1 - this probability)?

$\endgroup$

2 Answers 2

5
$\begingroup$

knowing that the log of the prices in a GBM follows the following normal distribution:

$$\operatorname{ln}(S_t) \sim N\left(\operatorname{ln}S_0 + T*\left( \mu - \frac{\sigma^2}{2} \right), \sigma^2 T \right) $$

You can create a normal distribution with these values and then check the CDF. Here is the python code:

from scipy.stats import norm; 
mu=0.16; sigma=0.24;S_0=95;T=1
my_var=sigma**2*T
my_norm=norm(np.log(S_0) + (mu-sigma**2/2)*T,np.sqrt(my_var))
my_norm.cdf(np.log(93))

from this normal distribution you get the CDF value for log(93) since you want to know the probability of values below 93, it is 0.26260905311083976

and this probability is time dependent if instead of 1 year it was for 6 months then $T$ would be 0.5.

And yes, the probability of the price being above 93 is the complementary of that, i.e. 1-0.26.

$\endgroup$
2
$\begingroup$

Just like the normal density, this will give the probability density of x=93. So to find the probability of $P\left[ S\le 93\right]$, you will need to calculate the cumulative probability. See some discussion here. https://math.stackexchange.com/questions/2445900/probability-from-log-normal-distribution

Also try the Matlab free page here: https://uk.mathworks.com/help/stats/logncdf.html to get an understanding of the log normal probabilities, and then just look up the equivalent in whatever software you are using. Excel has a function as well.

$\endgroup$
1
  • $\begingroup$ Thank you for your note. $\endgroup$
    – QFII
    May 12, 2019 at 14:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.