12
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
7
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

It is true that you cannot infer the real World probabilities from the BSM formula directly. It is also equally true that the "right value" of the option in the real world is obtained by replacing the risk free rate with the expected return of the stock.

Another example of this is simply to look at the real world price of a forward on the stock. If arbitrage, or replication, is not allowed would you systematically buy or sell a ten year forward on a stock where the forward is priced using the risk free rate?

As long as the expected return on the stock is above the risk free rate, which is consistent with most classical financial assumptions, you would chose to buy the forward. The equilibrium price for the forward when arbitrage is not allowed is obtained when the forward is priced using the expected return of the stock as the rate. Then you would be indifferent to buying or selling the forward.

The simple example of the forward pricing did not use any real world probabilities in pricing the forward. The probabilities might be behind how you came up with the expected rate of return, or the drift, of the stock if you wish but all you need to know to price the forward is actually the expected return.

Derivatives people, like myself, might complain that this is truoblesome because you need to make an assumption of the drift of the stock to arrive at the answear. Well tough luck, that is what equities people need to do every day when pricing stocks.

$\endgroup$
0
$\begingroup$

I think the simplest answer is just to understand that Black and Scholes uses risk neutral probabilities, vis-a-vis "d1" and "d2" in commonly cited derivations. These are very similar to "Z-scores" in statistics, but they are derived with the assumption that a dynamically risk-free portfolio can be constructed -- this allows us to derive a closed form solution for the differential equation. Just as how a standard Z-score can be converted to a probability, the Black and Scholes converts d1 and d2 to a "risk neutral" probabilities through the cumulative distribution function (CDF) of the standard normal ("i.e., Gaussian) variety.

If you want to find the raw probability of an options expiring above or below a given strike, it would be so much easier to just utilize the Z-score formula, as in:

$$Z = \frac{\ln(S / K) - m(T-t)}{s \sqrt{T-t}}$$

$$p = \phi(Z)$$

where $S$ is the stock price, $K$ is the arbitrary strike price, $m$ is expected annualized drift, $s$ is the annualized volatility, and $T - t$ is the time till expiration in years, $p$ is the probability that the underlying will expire at or above the strike at time $T - t$, and $\phi$ is the CDF of the Gaussian distribution.

$\endgroup$
  • $\begingroup$ I think you are missing a $(T - t)$-factor multiplying $m$ in the nominator of $Z$. Also - $m$ is the drift of the logarithmic stock price. I.e. if $\mathbb{E} \left[ S_t \right] = S_0 e^{\mu t}$, then $m = \mu - s^2 / 2$. $\endgroup$ – LocalVolatility Feb 21 '17 at 22:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.