Learning about change of numeraire, and came across this statement:

The price of any asset divided by a reference asset (called numeraire) is a martingale (no drift) under the measure associated with that numeraire.

This sounds intuitive, especially when we consider the reference asset as bank account then this would result the risk neutral measure. However, more rigorously, how to prove this, or which theorem (Fundamental theorem of Asset Pricing?) implies this ?


1 Answer 1


Proving the existence of a risk neutral measure is the difficult part. Once its existence is established, a simple calculation of conditional expectations allows to go from a numeraire to any other.

Write $\beta$ for the cash numeraire and $Q_\beta$ the corresponding risk neutral measure. Let $N$ be a numeraire (so $N$ is a positive process and $N/\beta$ is $Q_\beta$ martingale). Define a new measure by $$ \frac{dQ_N}{dQ_\beta}|_{\mathcal{F}_T} = \frac{N_T/\beta_T}{N_0/\beta_0} $$ Then, for any $P$ s.t. $P/\beta$ is a $Q_\beta$ martingale
\begin{eqnarray} \mathbb{E}^{Q_N}[ \frac{P_T}{N_T} | {\mathcal{F}_t} ] &=& \frac{\mathbb{E}^{Q_\beta}[ \frac{P_T}{N_T} \frac{N_T/\beta_T}{N_0/\beta_0} | {\mathcal{F}_t} ]}{\mathbb{E}^{Q_\beta}[ \frac{N_T/\beta_T}{N_0/\beta_0} | {\mathcal{F}_t} ]} = \frac{\mathbb{E}^{Q_\beta}[ \frac{P_T}{\beta_T} | {\mathcal{F}_t} ]}{\mathbb{E}^{Q_\beta}[ \frac{N_T}{\beta_T} | {\mathcal{F}_t} ]} = \frac{\frac{P_t}{\beta_t} }{\frac{N_t}{\beta_t} } = \frac{P_t}{N_t} \end{eqnarray} So $P/N$ is a $Q_N$ martingale.

If you assume that you have a Brownian market:
$$ \frac{dN_t}{N_t} = r_t dt + \sigma^N_t dW_t^\beta $$ $$ \frac{dP_t}{P_t} = r_t dt + \sigma^P_t dW_t^\beta $$ $$ \frac{dQ_N}{dQ_\beta}|_{\mathcal{F}_T} = \frac{N_T/\beta_T}{N_0/\beta_0} = \exp\left(\int_0^T \sigma^N_t dW^\beta_t - \frac{1}{2} \int_0^T |\sigma^N_t|^2 dt \right) $$ By Girsanov, under $Q_N$, $$ dW^N_t = dW^\beta_t - \sigma^N_t dt $$ is a Brownian motion and using Ito's lemma you can check that $$ \frac{d(P_t/N_t)}{P_t/N_t} = (\sigma^P_t - \sigma^N_t)dW^N_t $$ which also shows that it is a Brownian martingale under $Q_N$.

  • 3
    $\begingroup$ Concept cleared! $\endgroup$
    – kchpchan
    Apr 4, 2015 at 19:06
  • $\begingroup$ Proving existence? I think that's for what the Girsanov Theorrm is? $\endgroup$
    – BCLC
    Apr 5, 2015 at 2:47

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.