# 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:

$$P(X,t)$$ = $$1\over X \cdot1\over {\sigma \sqrt{2 \pi t}}\cdote^{-(ln(x)- ln(x_0)-(\mu- \sigma^2 /2)t)^2}\over 2\sigma^2t$$

I can plug in the values as the following:

$$P(X,t)$$ = $$1\over X \cdot1\over {(.24) \sqrt{2 \pi (1)}}\cdote^{-(ln(x)- ln(95)-((.16)- (.24)^2 /2)(1))^2}\over 2(.24)^2(1)$$

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)?

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