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 dSt=μStdt+σStdWt, where Wt is the Wiener process.
It means that (for simplicity, let's talk about European call) lnST is distributed as N(ln(S0)+(μ−12σ2)T,σ2T)
Correct me if I'm wrong, you'd like to find EP(C)=e−rTEP[max(ST−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 S0N(d1(μ))e(μ−r)T−KN(d2(μ))e−rT
the terms d1 d2 are from the B-S formula, with the adjustment to replace risk-free rate r there with "risky" μ
Now, I write down some derivation steps, please check them.
Let's rewrite expectation as follows, EP[...]=EP[I(ST≥K)(ST−K)], where I(.) is the indicator function.
Notice that the inequality ST≥K is equivalent to lnST≥lnK
Then, ...=EP[STI(lnST≥lnK)]−EP[KI(lnST≥lnK)]
=EP[elnSTI(lnST≥)lnK)]−KN(d2(μ))
To calculate the first term, use the following lemma: if X distributed as N(a,s2) then $E(e^X\textbf{I}(l
Take lnST as X and l as lnK, obtain EP[STI(lnST≥lnK)]=elnS0+(μ−12σ2)T+12σ2TN(lnS0+(μ−12σ2)T+σ2T−lnKσ√T)=S0eμN(d1(μ))
Finally, discount it with the risk-free rate r and we get the result.
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