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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:

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  • $\begingroup$ Perhaps you could screenshot the part of the book you are confused about, it would make it easier for the relevant personnel to answer your question. $\endgroup$
    – KaiSqDist
    Commented May 10 at 19:25
  • $\begingroup$ I edited with the screenshot, thanks for your suggestion. $\endgroup$
    – JohnGalt
    Commented May 10 at 19:30
  • $\begingroup$ Do you have any specific questions? This proposition tells you how the dynamics of an asset change when changing the numeraire. The more well known formula is the one in terms of the Brownian motion $dW$, but it is equivalent to the one here in terms of drift rate. $\endgroup$
    – NC520
    Commented May 10 at 19:35
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    $\begingroup$ I am trying to understand the proof, I don't see how the Leibnitz/ito part connect togheter, when I inject d(1/Ut) on the first equation, I don't get how to find the s/u sigma $\endgroup$
    – JohnGalt
    Commented May 10 at 19:45

1 Answer 1

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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.

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    $\begingroup$ Many thanks! I should have find it myself but don't why I was stuck on this, thanks again for your help! $\endgroup$
    – JohnGalt
    Commented May 11 at 16:33
  • $\begingroup$ Can you please close the question if the answer was satisfactory? $\endgroup$
    – NC520
    Commented May 15 at 9:50

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