9
$\begingroup$

Or simply: why do we call equivalent martingale measures as risk-neutral measures?

In the utility or game theory, when we consider a person's preferences to certain outcomes, we often deal with the utility functions. For example, if we consider an investor with a utility function $U$ whose return on a portfolio $\Pi$ is $x_\Pi$, we assume that he choose a portfolio that maximizes his expected utility $$ \Pi^* \quad\text{ such that }\quad\mathsf E U(x_{\Pi^*}) = \sup_{\Pi}\mathsf E U(x_\Pi). $$ In particular, we say that an investor is risk-averse (risk-seeking) whenever $U$ is concave (convex). We say that an investor is risk-neutral when $U(x) = ax + b$ is an affine function.

The risk-neutral valuation - taking expectations w.r.t. martingale measures equivalent to the real-world ones - is used in quant finance a lot for the pricing purposes. I do understand the theory behind this method, and the relation with non-arbitrage arguments. I wonder though, whether there is any relation with the risk-neutrality as in the paragraph above.


I thought of the following idea: let us think of a fair price for a contract (when we write it) as the highest one at which the agent will buy it. The agent $A$ with utility $U_A$ and expectation of prices $\mathsf E_A$ has to make a choice between the zero utility (when he does not buy contract) and $$ \mathsf E_AU_A(\mathrm e^{-rT}C_T - C_0) $$ where $T$ is maturity of the contract, $r$ is a rate used to compute present value of future cashflow, $C_T$ is the payoff of the contract, $C_0$ is the price of the contract. Hence, we need to solve the equation $$ \mathsf E_AU_A(\mathrm e^{-rT}C_T - C_0) = 0 $$ with unknown $C_0$. Assuming that agent is risk-neutral, we obtain $$ C_0 = \mathrm e^{-rT}\cdot\mathsf E_A(C_T). $$ At the same time, pricing using the $\Delta$-hedging in the Black \& Scholes framework gives us $$ C_0 = \mathrm e^{-rT}\cdot\mathsf E_Q(C_T) $$ where $Q$ is a risk-neutral measure. Hence, if we assume that our agent is risk neutral, then his expectations (at least at any given time $T$) have to be given exactly by the measure $Q$.

$\endgroup$
3
  • $\begingroup$ your notation is confusing me a bit do you mean: The investor looks for a $\Pi^*$ so that $$\mathbb{E}[ U(x_{\Pi^*})]=\max_{\Pi} \mathbb{E}[U(x_\Pi)]$$ $\endgroup$ Mar 26, 2014 at 17:57
  • 1
    $\begingroup$ @Probilitator: indeed, I've modified this part - better now? $\endgroup$
    – SBF
    Mar 26, 2014 at 18:46
  • $\begingroup$ yep everything fine now :D $\endgroup$ Mar 26, 2014 at 21:29

1 Answer 1

6
$\begingroup$

An agent with utility function $U$ values a final position $X_T$ by $E\left[U(X_T)\right]$. You can think of this as a function mapping random variables to $\mathbb{R}$, $X_T \mapsto E \left[U(X_T)\right]$.

A risk-neutral mapping should be a linear mapping of the kind above. In other words, $f$ should map some space of random variables to $\mathbb{R}$, and satisfy $$a \cdot f(X_T) + b \cdot f(Y_T) = f(a \cdot X_T + b \cdot Y_T)$$ for scalar $a,b$.

Using a martingale measure, the valuation rule is $X_T \mapsto E \left[ \frac{dQ}{dP} X_T \right]$. Note that this is a linear map. Therefore, it's like a risk-neutral agent is pricing the positions.

$\endgroup$
2
  • $\begingroup$ Thanks, I'm assuming in your case $E = E_P$: the expectation over a market measure. In such case, the map $X_T \mapsto \mathsf E_P[X_T]$ is linear as well, so nothing distincts it from the case of a martingale measure. $\endgroup$
    – SBF
    Mar 27, 2014 at 7:45
  • $\begingroup$ Yes, there are many other linear mappings on this space, but only one of them is arbitrage free! $\endgroup$
    – quasi
    Aug 10, 2019 at 2:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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