# Why the Black-Scholes formula can be used in the real world?

The BS formula is deduced using the risk neutral measure. Why can it be used in the real world?

• A very warm welcome to Quant.SE and thank you for your question. – vonjd Mar 22 '15 at 8:55
• Doesn't this question also illustrate we can still have a hard time even one of the most fundamental questions of option pricing. I can also contribute by quoting one of the trading company managers I encountered during an academic presentation: "If you are not hedging, why don't you go with with $\mu$?". I think this also summarizes the most important difference between risk neutral and real world measures. – berkorbay Mar 22 '15 at 23:39
• Why would you not care about hedging? The whole point of the risk neutral measure is that it goes hand in hand with the no-arbitrage theorem. If options are priced in such a way that there is no arbitrage possible, then they are priced with respect to a risk neutral measure. This also goes the other way: if they are not priced by a risk neutral measure (but e.g. directly by the physical measure) then there is an arbitrage opportunity. That's why we don't go with $\mu$. – Olaf Mar 24 '15 at 22:40
• @Olaf That's not true IMHO. Because Q and P are equivalent measures, no arbitrage under Q means no arbitrage under P. We use Q for pricing simply because it is also a martingale measure and hence has nicer properties than P since we can play with expectations. That's all. – Quantuple Apr 7 '16 at 16:03
• Yes, you're right, but that's not what I meant. What I meant with "directly by the physical measure" is: pricing the option as a martingale with respect to the physical measure. So if we say that the price of the option follows from its expected discounted value ($V(t) = E[D(t, T)V(T)]$) with respect to the physical measure, then we have an arbitrage opportunity. – Olaf Apr 8 '16 at 8:33

## 5 Answers

The short answer is:

As long as a derivative can be perfectly replicated via hedging in the underlying asset then the price of the derivative should be independent of investors' risk aversion and hence the application of risk-neutral probabilities and discounting of the future expected payoff under risk neutral probability leads to the same price of the derivative as an application of real-world probabilities would.

Longer answer:

The above has been proven and shown by Black and Scholes' seminal work (along with Robert Merton). Note that several conditions have to be met in order to assure one can move into a risk-neutral pricing framework:

• Derivative can be replicated by trading in the underlying asset(s) and trading in the money market.
• No arbitrage requirement
• Complete Market
• Law of one price, among others

Please note that contrary to the statement of another user one can perfectly price derivatives via real-world probabilities. But the problem is that one would have to have knowledge of how to discount each expected future payoff, meaning, one would have to know investors' risk preferences. That is virtually impossible or at least very difficult, hence, the whole application of risk-neutral pricing is in order to simplify the work of pricing derivative securities.

Note of caution: One cannot use risk-neutral pricing to assess the future value of a derivative. The reason for that is that the underlying assets do not grow at a risk-free rate, generally a risk-premium has to be applied.

• I think your answer would be even better if you made the difference between "price" and "value" more precise. Otherwise some people could be confused. – vonjd Mar 25 '15 at 10:13

The risk-neutral probability is used as a convenient mathematical tool but, strictly speaking, it is not a necessary ingredient for the BS formula.

In fact, the formula can be derived by computing the expectation (under the physical probability) of the option payoff, properly discounted with the right stochastic discount factor:

$$p_t = E_t[\;m_{t,T}\; f(S_T)\,]$$

where $f$ is the payoff function of the derivative, a function of the realisation at maturity time $T$ of the underlying, i.e. $S_T$. The BS formula for a call option is obtained by choosing the corresponding payoff function (exercise: what is it?).

The stochastic discount factor (SDF) $m_{t, T}$ is a positive random variable which discounts for both time and risk. In every possible state of the world encompassed by the expectation operator, the value of the $SDF$ in that state gives a weight to the payoff. For instance, a good payoff in a very bad state is highly valuable: the SDF will give a very strong weight to that state, which ultimately leads to an higher price.

The effect of the SDF may as well be described by means of the risk-neutral probability: after all, it all boils down to give the right weights (probabilities) to each state. Hence we can write

$$p_t = E_t[\;m_{t,T}\; f(S_T)\,] = E^Q[\, e^{-r(T-t)} f(S_T) \,]$$

where $Q$ is the risk-neutral probability, which accounts for the risk discounting.

• But the factor $m_{t,T}$ is defined as the measure change from the physical measure to the risk neutral measure, right? So saying that the risk neutral measure is just a convienent tool is a bit misleading, I would say. After all, the risk neutral measure determines the measure change $m_{t,T}$. Or am I missing something in your argument? – Olaf Mar 24 '15 at 22:35
• I completely agree with you! Maybe I should rephrase it in a better way. My point is that one can get the BS formula reasoning only with the physical probability, without even defining the concept of "risk-neutral probability". – Abramo Mar 24 '15 at 23:55

If we are talking about BS world, one plausible explanation is every move in the risk-free world is "perfectly hedgeable". In other terms if you consider the drift term $\mu = r + \lambda\sigma$ BS formula allows you to reduce $\lambda$ to zero by constantly hedging with the underlying. That is why you can get a correct price with $r$ under BS world.

However "real" real world is different and even in theoretical incomplete markets there are more than one risk-free measure. With some models, there is no perfect hedge. Good luck with that.

Risk-neutral probabilities and physical probabilities agree on what is possible and what is impossible. Also, a hedge under Risk-neutral probability works almost surely and so does the hedge under physical probability.

• Just a quick note to anybody who wants to read more about it on wikipedia, it's known as equivalent probability measure. – SmallChess Mar 22 '15 at 2:34
• I think this answer is unclear and doesn't address the question. – vonjd Mar 22 '15 at 10:10

Because BS is about derivatives and not about the underlying. In a way if you priced derivatives with real world measures (all else being equal) you would double count risk preferences because these are already included in the underlying - think about it this way (beware, oversimplification ahead):

You want to price a derivative on gold, a gold certificate. The product just pays the current price of an ounce in \$.

Now, how would you price it? Would you think about your risk preferences? No, you won't, you would just take the current gold price and perhaps add some spread. Therefore the risk preferences did not matter (=risk neutrality) because this product is derived (= derivative) from an underlying product (=underlying).

This is because all of the different risk preferences of the market participants are already included in the price of the underlying and the derivative can be hedged with the underlying continuously (at least this is what is taken for granted in a BS world). As soon as the price of the gold certificate diverges from the original price traders would just buy/sell the underlying and sell/buy the certificate to pocket a risk free profit - and the price will soon come back again.

So, you see, the basic concept of risk neutrality is quite natural and easy to grasp.

• @Downvoter: Why did you downvote my answer? – vonjd Mar 22 '15 at 10:49
• Because I find your explanation less clear and less to the point than the one you downvoted. Talking about "double counting risk preference" is very vague and you did not even mention probability measures which is what the question was about. – AFK Mar 22 '15 at 10:55
• @AFK: The question was why you could use risk neutral pricing for derivatives in a world that is not risk neutral. I answered that. Did you only read my first paragraph? (btw way I mention "measure" there!) Because "double counting" is vague I wrote a few paragraphs of explanation! – vonjd Mar 22 '15 at 11:03
• If your point is that you can hedge the payoff, it was already in the answer you downvoted. Except that it also explained that the risk neutral and real word measures agree on what has probability 1 which is why almost sure replication under risk neutral measure is equivalent to almost sure replication under real world measure which is crucial and would apply with any other equivalent measure (=any numeraire). You think your answer is better, fine. Let's agree to disagree. – AFK Mar 22 '15 at 11:36
• @AFK: 1 I don't like your tone, it seems inappropriate and overly aggressive. This should be a professional technical discussion and not some religious debate. 2 The OP asks a very basic question from which we might conclude that his/her background is not prepared to understand sentences like "a hedge under Risk-neutral probability works almost surely" thrown at him/her. This is what I meant by "the answer is unclear. 3 You implied that you downvoted my answer because I downvoted another answer. This is an act of revenge which is generally frowned upon here. I flagged your comment accordingly. – vonjd Mar 22 '15 at 11:48