# Formal proof for risk-neutral pricing formula

As you know, the key equation of risk neutral pricing is the following:

$\exp^{-rt} S_t = E_Q[\exp^{-rT} S_T | \mathcal{F}_t]$

That is, discounted prices are Q-martingales.

It makes real-sense for me from an economic point of view, but is there any "proof" of that?

I'm not sure my question makes real sense, and an answer could be "there is no need to prove anything, we create the RN measure such that this property holds"...

Is this sufficient to prove that, within this model, the risk-neutral measure exists?

EDIT:

Some answers might have been misled by my notation.

Here is a new one:

$\exp^{-rt} X_t = E_Q[\exp^{-rT} X_T | \mathcal{F}_t]$

where $X_t$ can be any financial asset. For example, a binary option on an underlying stock $S$.

In order to price the option, you would start with this equation and develop the right-hand side to finally solve for $X_t$.

My key question: what allows me to write the initial equation assuming I have no information about the dynamics of the option or its underlying.

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assume Laplace. like they said in the 90s "don't be normal". will the ideal probability distribution please stand up? –  user3232 Feb 15 '13 at 5:58
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## 5 Answers

Note first that this key equation is only assumed to hold true under some extra assumptions. Typically those assumptions are taken to be about absence of arbitrage, though it is possible to weaken them somewhat if you are willing to consider portfolio arguments or collectively agreeable objective function.

Anyway, the argument is this: if all the risk can be arbitraged away, then the price of any contingent claim should be equal to its price under the risk-neutral measure Q.

The mathematical proof can be grasped most easily by the old-school arguments where one shows delta-hedges eliminating stochastic terms from the SDE. More mathematically elegant arguments involving the Girsanov theorem and Feynman-Kac formula are less intuitive.

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Thanks for the answer, I updated the question, because I still can't see how to conclude.... –  SRKX Apr 14 '11 at 21:02
If you have absolutely no information, then the conditions of the Girsanov theorem fail to apply and you cannot write the equation. –  Brian B Aug 12 '11 at 16:58
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You can find a simple proof in the discrete time case at http://kalx.net/ftapd.pdf. I'm not sure what you are trying to derive with your Ito calculus, but here is a rigourous derivation of the Black-Sholes/Merton PDE: http://kalx.net/dsS2011/bms.pdf. The Black-Scholes '73 derivation is not mathematically correct.

The modern approach does not use so called real-world measures.

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Ok I'll have a look a it. Regarding my Ito calculus, I'm just trying to prove that the equation in my question holds. I can't find a way to prove that it holds regardless the dynamics of the asset. –  SRKX Apr 18 '11 at 14:35
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Since I did not get any comments to my latest update, and since I find it quite convincing, I hereby post my solution as an answer.

maybe I can prove that Q exists assuming a lognormal distribution of $S_t$.

Assuming $dS_t = \mu S_t dt + \sigma S_t dW_t$

By Itô, $d(e^{-rt} S_t) = -r e^{-rt} S_t dt + e^{-rt} S_t dS_t$.

Replacing with the definition of $dS_t$, I get:

$d(e^{-rt} S_t) = e^{-rt} S_t [(\mu - r)dt + \sigma dW_t] = e^{-rt} S_t \sigma [\theta dt + dW_t]$. (with $\theta=\frac{\mu-r}{\sigma}$)

By Girsanov, we know that there exists a measure s.t $dW_t = dW_t^* - \theta dt$

We thus get

$d(e^{-rt} S_t) = e^{-rt} S_t \sigma [\theta dt + dW_t] = e^{-rt} S_t \sigma dW_t^*$ which means that $e^{-rt} S_t$ is a Q-martingale.

If you think it's incorrect, feel free to comment, or even better, to correct!

The thing that worries me is that I need to know the dynamics of the asset, and I think I shouldn't have to...

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The price to set up a dynamic hedge portfolio absolutely depends on the dynamics of the asset's price process through the quadratic variance. If you change your assumption about the dynamics of the price process, say by using a jump-diffusion or variance gamma process instead of geometric Brownian motion, you change the value of the option. –  user3296 Jul 8 '11 at 6:44
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The only requirement if you are risk neutral is the property of martingale on your discounted stock price $M_t=e^{-rt} S_t$.

But if you apply Itô $d( S_t\cdot e^{-rt} e^{rt})=d(M_t\cdot e^{rt})=r_tM_te^{rt}dt + ..dW_t=r_tS_tdt+..dW_t$

you see see that under the risk free probability, the asset price must have $r_t$ as yield and to answer to your question, why the GB is so popular? simply because $dS_t=S_t(r dt+ \sigma dW_t)$ is the most simple SDE that hold this property.

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I'm not sure if my question was clear enough. I edited it because I don't think this is the answer I'm looking for. –  SRKX Jul 13 '11 at 9:43
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The first think you have to ask is ¿¿What price??? Monetary price or equity price?? All answers,the ones I read, related to monetary price, but are equity price really risk free???? One of the biggest problem with Black Scholes (personal opinion) is that they consider the behave of equity price as monetary price: Solve this ODE: S(t)'/dt= r*S(0), this tell you how money change in a deposit account (monetary price). All finance theory comes form this simple implementation. Now check Hull, and you will see that they use this expression to get the final result en the E(S(T)). Nice question, and I dont pretent to solve it, cause risk neutral in equity price does not exist........., is better to understand the model, limitation, etc.......

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I am sure you mean something interesting but what you are saying is far from obvious. –  nicolas Jul 9 '11 at 18:46
option theory does not say that equity prices are risk free. it says that you can offset the risk from the stock in the option using a correct amount of stocks. –  nicolas Jul 9 '11 at 18:47
"Monetary price or equity price?": not sure what you mean... –  SRKX Jul 13 '11 at 9:31
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