The Black Scholes formula gives the formula for European calls, for stock with no dividends, $$ c = S N(d_1) - K e^{- r (T-t)}N(d_2) $$ with $S$ the price of stock at time $t$, $T$ is the maturity date, $K$ is the strike price, $r$ is the risk free interest rate.
I have learned that change in stock price is normally distributed is a 'good' assumption. I think that this is playing some role in this Black Scholes formula, seeing that there is a normal distribution popping out in the formula.
Could someone please clarify if this is indeed the case or not? If so, what is the technical assumption equivalent of change in stock price being normally distributed? thank you
As @Snoop points out, the assumption is not that changes in $S$ are normally distributed, which would allow the possibility of $S < 0$ but that changes are lognormally distributed, i.e., changes in $\ln S$ is normally distrbuted. This makes sense since the magnitude of a change $S$ depends on the currency you use but the ratio of a change in $S$ over $S$ itself does not.
The normal distribution is the most fundamental distribution on the real line. In particular, it satisfies the central limit theorem. This makes it easy to sum up normally distributed changes in $S$ with respect to a small change in time to get the distribution of $S$ over a larger time interval. You can then take a limit to get a stochastic process known as Brownian motion. So it is mathematically far easier to work with than other distributions. In particular, if you assume that $\ln S$ follows a Brownian motion with constant volatility, it leads to the Black-Scholes formula.
The problem with a normal distribution is that its density function decays far too fast and underestimates the probability of "rare" events. So there has been an effort to develop stochastic processes using so-called fat-tailed distributions. This has so far been met with limited success.