A model for the movement of a stock supposes that if the present price of the stock is s, then after one time period it will be either (1.012)s with probability 0.52, or (0.99)s with probability 0.48. Assuming that successive movements are independent, approximate the probability that the stock s price will be up at least 30% after the next 1000 time periods.
2026-04-06 14:31:35.1775485895
probability normal distribution
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The following is an analysis that requires logarithms. We will use base $e$ ("natural") logarithms. But if you are more comfortable with base $10$, with minor adjustments it can be done that way. Let $X_i$ be the fractional price "increase" from period $i$ to period $i+1$. Let $Y_i=\ln(X_i)$.
Note that with probability $0.52$, we have $Y_i=\ln(1.012)\approx 0.119286$, and with probability $0.48$, we have $Y_i=\ln(0.99)\approx -0.0100503$.
We assume, as directed, that the $X_i$, and hence the $Y_i$, are independent, despite the fact that this is quite implausible.
Calculate the mean $\mu$ and the variance $\sigma^2$ of the $Y_i$. This is a routine calculation.
Let $W$ be the sum $Y_1+Y_2+\cdots+Y_{1000}$. The $W$ is the sum of $1000$ independent identically distributed random variables. The cumulative distribution function of $W$ is well-approximated by the cumulative distribution function of a normal mean $1000\mu$, variance $1000\mu^2$.
We are interested in the probability that the stock price climbs by a factor of at least $30\%$, so by a factor of at least $1.3$.
The required probability is $1$ minus the probability that the price climbs by a factor of less than $1.3$. We concentrate on finding that.
The stock climbs by less than $30\%$ precisely if $W\lt \ln(1.3)\approx 0.2623643$.
So we want to find the probability that a certain normal with known mean and known variance is $\lt 0.2623643$. That's a standard calculation.