# How to obtain true probabilities from Black-Scholes?

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.

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## 3 Answers

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 $E(e^X\textbf{I}(l<X))=e^{s+\frac{1}{2}s^2} N (\frac{\mu+s^2-l}{s})$

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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You cannot get "true probabilities" (empirical distribution) from the BS model. Option price is required initial investment, which is risk neutral expectation of payout. “True probabilities” are irrelevant in Black-Scholes. However, you can estimate the risk-neutral probability distribution (i.e. implied risk-neutral density) of the stock returns through Breeden-Litzenberger formula.

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You cannot deduce the real-world probabilities from the option prices.

It may seem strange, but here is a simple example which might help you to understand.

Suppose that everyone in the market agrees on the real-world probabilities, and that they are not changing for any external reason.

Then suppose that the investment board of a large pension fund decides that they need to increase the amount of options they have bought because they get a feeling that they would like to hold more protection against an adverse move (and since most pension funds are net long equities, this is likely to mean that they want to buy out-of-the-money equity put options to protect against a sell off in the equity market).

The pension fund will come to the dealers (investment banks probably) and will buy a whole load of put options, say. Naturally the price in the market will go up (simple law of supply/demand, and demand has increased), which implies that the implied vols will go up.

In summary: no change in the real-world probabilities, but a big change in the implied volatilities which will in turn lead to a change in the implied underlying probability distribution.

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