My question concerns the Black-Scholes formula for the value of a European option, namely
\begin{align} C(S_t, t) &= N(d_1)S_t - N(d_2) Ke^{-r(T - t)} \\ d_1 &= \frac{1}{\sigma\sqrt{T - t}}\left[\ln\left(\frac{S_t}{K}\right) + \left(r + \frac{\sigma^2}{2}\right)(T - t)\right] \\ d_2 &= d_1 - \sigma\sqrt{T - t} \\ \end{align}First, I want to ignore for purposes of this question the derivation of this formula using the Black-Scholes PDE dynamic hedging argument, which I haven't been through in detail, but I understand in principal. Rather, I'd like someone to explain to me why the following argument, which leads me to a (slightly) different answer than the one above, is wrong.
The Black-Scholes model of stock movements posits that the change $\Delta S$ in a stock price over a small time interval $\Delta t$ behaves as
$\Delta S = \mu S \Delta t + \sigma \sqrt{\Delta t} \varepsilon S$
where $\mu = \text{drift rate}$, $\sigma = \text{volatility}$ (constant), and $\varepsilon$ is a fair coin flip resulting in $1$ and $-1$ (I prefer this incremental equation to a stochastic one, I'm not up on Ito's lemma and all that). $S_T$, the stock price at time $T$, is then (for fixed $\Delta t$) the random variable
$S_T = S_0 \left(1+\mu \Delta t + \sigma \sqrt{\Delta t} \right)^X \left(1+\mu \Delta t - \sigma \sqrt{\Delta t}\right)^{N-X}$
where $X$ is a binomial R.V. counting the number of $1$'s from the coin flips and $N = T/\Delta t$. Using the normal approximation for $X$ and letting $\Delta t \rightarrow 0$ gets us
$S_T = S_0 e^{(\mu-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$
where $Z$ is standard normal (or we could replace $\sqrt{T} Z$ with brownian motion $W$ for a dynamic model).
Now, what is the fair value of a European option on this stock at strike price $K$ and time to expiry $T$? Well, the seller of such an option would expect to have to pay at expiry, on average
$E[\text{max}(S_T - K,0)]$
That is, the expected value of the payout of the option. And so to price it today, one would discount this expected payout by the risk-free rate:
$e^{-rT}E[\text{max}(S_T - K,0)]$
But, this is wrong, according to the Black-Scholes formula given a the beginning. I'll spare you the computations, but in fact, this computation returns the correct Black-Scholes formula only if we change our model $S_T$ to
$S_T = S_0 e^{(r-\sigma^2/2)T}e^{\sigma \sqrt{T} Z}$
That is, by replacing $\mu$, the drift rate of the stock, to $r$, the risk-free rate. And so my question is, why are we justified in doing this? Why is it okay, for purposes of this computation, to "pretend" that the drift rate of the stock is $r$, when it very well may not be in real life?
Two objections come to my mind regarding my computation. First, it assumes that the person valuing the option only cares about its expected payout, and not the volatility of the payout, due to the volatility of the stock itself (i.e. it assumes risk-neutrality, which may not be the case). Still, if someone were to sell enough of these options (on identical but independent stocks) for the law of averages to win out, then this would be a correct valuation. And secondly, this doesn't seem to be actually computing the market value of the option, but only the expected cost to the seller. That is, expected cost to the seller might not be exactly the same thing as value to the buyer. But still, if someone were to offer options at a price other than this one, obvious (statistical) arbitrage opportunities would exist (for a market participant with enough reserves to weather the volatility in the payouts).
I'm interested in your thoughts on these objections. But my big question is still: why does replacing $\mu$ with $r$ lead us to the correct formula, when $\mu$ and $r$ are almost never equal? Thanks you for any help.