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$$ p(S,t;S',t') = \frac{1}{\sigma S'\sqrt{2\pi (t'-t)}} \exp\left(-\frac{(\log(S/S') + (\mu-1/2\sigma^2)(t'-t))^2}{2\sigma^2(t'-t)}\right) $$

I found this equation when I was reading "Paul Wilmots on Quantitative Finance" which calculates the probability that of a stock price ending/landing on a particular price (S'). So if the stock price is 100 USD and the volatility is 30%, the probability the stock closes at 105 USD, exactly 30 days from now, according to the formula, is 3.850%.

When I tried to use the formula for different strikes S' (95 96 97 98 99 100 101 102 103..), integer wide, and a fixed 30 days, the probabilities at those respective strikes summed to exactly 1, as it should. However when I used a fractional spacing/mesh (100 100.1 100.2 100.3 ...) I the probabilities summed more than 1. And that makes sense I guess? If the probability that a stock worth a 100 USD now will close at 105 USD, 30 days from now, is 3.85% than it shouldn't vary too far from the the probability of a 100 USD stock closing at 105.10 USD 30days, the two numbers are within the vicinity of one another by 10 cents.

But is there a way to normalize the formula such that if I wanted to know the probability that a stock will land within the interval [105.00, 106.00], I would not be getting probabilities that sum to more than 1?

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That formula actually does not make much sense, given that for a continuous random variable the probability of any given point is zero.

Assuming a Black-Scholes world you are better of by doing:

$P(S_T>S_T^*)=N(d_2)$ where $d_2$ is the standard black-scholes term.

From this it is straightforward to compute $P(S_T^{-}<S_T<S_T^+)$.

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Actually, the probabilities in the first case will not sum to exactly 1, since you are truncating the distribution (S' is unbounded above), but will be arbitrarily close.

To get the 'right' cumulative probability, you have to adjust for the step size; so, in the first case, you were assigning a weight of 1. In the second case, using a weight of 0.1 will yield a cumulative probability arbitrarily close to 1, given the truncation.

Alternatively (as pointed out by volcompt), since this formula assumes a log-normal distribution, it seems much simpler to use the difference between the CDF on the returns: $P(S' \leq 106) - P(S' \lt 105)$.

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