Change of Numeraire: Why Asset Prices Become Martingales
Summary
The document explains the change-of-numeraire result in asset pricing: when a positive asset is selected as numeraire, the price of another asset divided by that numeraire is a martingale under the associated probability measure. It starts with the cash account as numeraire and its risk-neutral measure, then defines a new measure by weighting outcomes according to the numeraire’s discounted value at the horizon.
A conditional-expectation calculation shows that the ratio of the two asset prices has no drift under the new measure. The answer also gives a Brownian-market illustration: the measure change adjusts the Brownian motion by the numeraire’s volatility, after which Itô’s lemma yields a zero-drift process for the price ratio. The argument assumes the relevant risk-neutral measure already exists and uses positive numeraires; the existence of the measure is identified as the harder foundational step. The stochastic calculus illustration applies under its stated Brownian model assumptions.
Key ideas
- A numeraire change defines a probability measure using the numeraire’s discounted terminal value.
- Under the resulting measure, the price of another asset divided by the numeraire is a martingale.
- The conditional-expectation identity provides a general proof once the cash risk-neutral measure exists.
- In a Brownian model, Girsanov’s theorem adjusts the Brownian motion and Itô’s lemma verifies the zero drift.
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# Change of numeraire and reference asset
# Change of numeraire and reference asset
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 ?
## Answer by AFK (score 7, accepted)
https://quant.stackexchange.com/a/17240
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 $Q_N$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.