Calculating GBM Price Threshold Probabilities with the Lognormal CDF

# Geometric Brownian Motion - Price Probabilities


# Geometric Brownian Motion - Price Probabilities












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)?

## Answer by ricmarchao (score 5)

https://quant.stackexchange.com/a/47284

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.

## Answer by Magic is in the chain (score 2)

https://quant.stackexchange.com/a/45564

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.