How to obtain true probabilities from Black-Scholes option pricing equation? Suppose, that we know risk adjusted discount rate for the underlying asset (the drift term in the physical measure) and risk free rate. The task is to find a real (not risk neutral) expected payoff for a call option.
Answer
The true probabilities underlying the B-S equation are actually postulated. The pricing process is assumed to follow the stochastic process $d S_t =\mu S_t d t + \sigma S_t dW_t$, where $W_t$ is the Wiener process.
It means that (for simplicity, let's talk about European call) $\ln S_T$ is distributed as $N(ln(S_0)+(\mu-\frac{1}{2}\sigma^2)T, \sigma^2 T)$
Correct me if I'm wrong, you'd like to find $E_P(C) = e^{-rT} E_P[max(S_T-K,0)] $, where P is a "physical" probability measure. Just to make sure, this expected value won't represent the fair price of the option.
If my calculations are correct, this expected value is equal to $S_0 N(d_1(\mu)) e^{(\mu-r)T} - K N(d_2(\mu))e^{-rT}$
the terms $d_1$ $d_2$ are from the B-S formula, with the adjustment to replace risk-free rate $r$ there with "risky" $\mu$
Now, I write down some derivation steps, please check them.
Let's rewrite expectation as follows, $E_P[...]=E_P[\textbf{I}(S_T\geq K)(S_T-K)]$, where $\textbf{I}(.)$ is the indicator function.
Notice that the inequality $S_T\geq K$ is equivalent to $\ln S_T \geq \ln K$
Then, $... = E_P[S_T \textbf{I}(\ln S_T \geq \ln K)]-E_P[K \textbf{I}(\ln S_T \geq \ln K)] $
$= E_P[e^{\ln S_T} \textbf{I}(\ln S_T \geq) \ln K)]- K N(d_2(\mu))$
To calculate the first term, use the following lemma: if X distributed as $N(a,s^2)$ then $E(e^X\textbf{I}(l Take $\ln S_T$ as $X$ and $l$ as $\ln K$, obtain $E_P[S_T \textbf{I}(\ln S_T \geq \ln K)]=e^{\ln S_0 + (\mu - \frac{1}{2}\sigma^2)T + \frac{1}{2}\sigma^2 T}N(\frac{\ln S_0 + (\mu-\frac{1}{2}\sigma^2)T + \sigma^2 T - \ln K}{\sigma\sqrt T}) = S_0 e^\mu N(d_1(\mu))$ Finally, discount it with the risk-free rate $r$ and we get the result.
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