Deriving Numeraire Change with Itô's Lemma
Summary
The document works through a numeraire-change derivation for the ratio of an asset value to a numeraire. It applies Itô's lemma to the reciprocal of the numeraire, then uses the product rule to obtain the ratio's drift and diffusion terms from the dynamics of both processes. The cross-variation term contributes to the drift.
The derivation uses the fact that the asset-to-numeraire ratio is a martingale under the numeraire measure: its drift must therefore vanish. Equating the derived drift to zero yields the relationship sought in the referenced proposition. The presentation assumes one-dimensional Brownian motion to keep notation simple; the author notes that this simplifies the calculation without changing its basic logic. The document is a symbolic derivation and gives no numerical example or discussion of the conditions needed for the martingale property.
Key ideas
- Apply Itô's lemma to the reciprocal of the numeraire before differentiating the asset-to-numeraire ratio.
- The product rule introduces a cross-variation term that affects the ratio's drift.
- Under the numeraire measure, a martingale ratio has zero drift.
- The displayed derivation assumes one-dimensional Brownian motion for simpler notation.
Tags
Full text
# numéraires change
# numéraires change
I am currently reading the Brigo and Mercurio's book about interest rate models. I am stuck at understanding the numéraire change proof for the 2.3.1 proposition. Could someone be kind enough to enlighten me about this? Thanks a lot! Kind regards.
Edit, here the part:
## Answer by NC520 (score 1, accepted)
https://quant.stackexchange.com/a/79353
By definition from the previous page, the dynamics of $S_t/U_t$ are assumed to be given by: \begin{align*} d \frac{S_t}{U_t} = \sigma_t^{S/U} dW_t^U \end{align*} Moreover, by Ito's lemma we have that: \begin{align*} d \frac{1}{U_t} = - \frac{1}{U_t^2} dU_t + \frac{1}{U_t^3} dU_t dU_t \end{align*} By the Leibnitz rule, substituting in the previous result, as well as the dynamics for $S_t$ and $U_t$: \begin{align*} d \frac{S_t}{U_t} &= \frac{1}{U_t} dS_t + S_t d \frac{1}{U_t} + dS_t d \frac{1}{U_t} \\ &= \frac{1}{U_t} dS_t + S_t \left( - \frac{1}{U_t^2} dU_t + \frac{1}{U_t^3} dU_t dU_t \right) + dS_t \left( - \frac{1}{U_t^2} dU_t + \frac{1}{U_t^3} dU_t dU_t \right) \\ &= \frac{1}{U_t} \left( \mu_t^S dt + \sigma_t^S dW_t^U \right) + S_t \left( - \frac{1}{U_t^2} \left( \mu_t^U dt + \sigma_t^U dW_t^U \right) + \frac{1}{U_t^3} (\sigma_t^U)^2 dt \right) - \frac{1}{U_t^2} \sigma_t^S \sigma_t^U dt + o(dt) \\ &= \left[ \frac{1}{U_t} \mu_t^S - \frac{S_t}{U_t^2} \mu_t^U + \frac{S_t}{U_t^3} (\sigma_t^U)^2 - \frac{1}{U_t^2} \sigma_t^S \sigma_t^U \right] dt + \left[ \frac{1}{U_t} \sigma_t^S - \frac{S_t}{U_t^2} \sigma_t^U \right] dW_t^U + o(dt) \end{align*} Since $S_t/U_t$ is a martingale under $U$, it must be that the drift rate is zero. Comparing with the first equation you can see the result.
Note: In the derivation above I have assumed that the Brownian motion is one dimensional, so that $C=1$. This simplifies the notation, but does not impact the derivation.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.