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I'm reading Leif Andersen's "Interest Rate Modeling, vol 2 Term Structure Models" and met a problem on Chapter 14 LM Dynamics and Measures, $\S$ 14.2.5 Stochastic Volatility, Lemma 14.2.6, on page 602.

Before the lemma the stochastic volatility LMM model was defined in spot measure $Q^B$ (numeraire is the discrete banking account $B(t)=P(t,T_{q(t)}) \prod \limits _{n=0} ^{q(t)-1} \left(1+\tau_nL_n(T_n)\right)$).

$$dz(t) = \theta (z_0-z(t)) dt + \eta \psi(z(t)) dZ(t), \quad z(0) = z_0 \tag{14.15}$$ $$dL_n(t) = \sqrt{z(t)} \varphi(L_n(t)) \lambda_n(t)^\top \left(\sqrt{z(t)} \mu_n(t) dt + dW^B(t) \right) \tag{14.16}$$ where $$\mu_n(t) = \sum_{j=q(t)}^n \frac{\tau_j \varphi(L_j(t)) \lambda_j(t)}{1+\tau_j L_j(t)}$$ , and Z(t) a Brownian motion under the spot measure $Q^B$.

Now Lemma 14.2.6 says, the SDE for $z(t)$ in measure $Q^{T_{n+1}}$, $n\geq q(t) -1$ is

$$\begin{align} dz(t) &= \theta(z_0-z(t)) dt + \eta \psi(z(t)) \\ &\times \left( -\sqrt{z(t)} \mu_n(t)^\top \langle dZ(t), dW^B(t) \rangle + dZ^{n+1}(t) \right) \tag{14.17} \end{align} $$

, where $Z^{n+1}(t)$ is a Brownian motion in measure $Q^{T_{n+1}}$.

The proof on the book is as follows:

From earlier results, we have

$$dW^{n+1}(t) = \sqrt{z(t)} \mu_n(t) dt + dW^B(t)$$

Let us introduce the $m$-dimensional vector $$a(t) = \langle dZ(t), dW^B(t) \rangle / dt \tag{1}$$

so that we can write

$$dZ(t) = a(t)^\top dW^B(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t) \tag{2}$$

where $\widetilde W(t)$ is a scalar Brownian motion independent of $W^B(t)$. In the measure $Q^{T_{n+1}}$, we then have

$$\begin{align} dZ(t) &= a(t)^\top \left(dW^{n+1}(t) - \sqrt{z(t)}\mu_n(t) dt \right) + \sqrt{1-\|a(t)\|^2} d\widetilde W(t) \tag{3}\\ &=dZ^{n+1}(t) - a(t)^\top \sqrt{z(t)} \mu_n(t) dt \tag{4} \end{align} $$ and the result follows.

My questions are:

Q1. Why with $(1)$ one can write $(2)$? is this a property for any 2 Brownian motions?

Q2. After $(3)$ I can only get $$dZ(t) = a(t)^\top dW^{n+1}(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t) - a \sqrt{z(t)}\mu_n(t) dt$$

To get $(4)$ it means one can define Brownian motion $Z^{n+1}(t)$ in measure $Q^{T_{n+1}}$ as: $$dZ^{n+1} (t) = a(t)^\top dW^{n+1}(t) + \sqrt{1-\|a(t)\|^2}d\widetilde W(t)$$

How could one do so? notice $\widetilde W(t)$ is a Brownian motion in measure $Q^B$, but $ W^{n+1}(t)$ is a Brownian motion in measure $Q^{T_{n+1}}$, isn't here a bit messy?

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Can you please clarify notation "$<dz_t,dW_t>$" is Quadratic Variation? –  emcor Aug 9 at 21:59
    
@emcor yes, it's the quadratic variation –  athos Aug 10 at 0:54

2 Answers 2

up vote 3 down vote accepted
+200

For Q1, the function $a(t)$ is the instantaneous correlation. The form given by (2) is basically the Cholesky decomposition. Of course, you may directly show, uisng Levy's characterization, that $$ \widetilde{W}(t) = \int_0^t\bigg[\frac{1}{\sqrt{1-||a(t)||^2}} dZ(t) -\frac{a(t)^T}{\sqrt{1-||a(t)||^2}} dW^B(t) \bigg] $$ is a standard scalar Brownian motion that is independent of $W^B(t)$.

For Q2, note that as $\widetilde{W}(t)$ is a Brownian motion independent of $W^B(t)$, it is also a Brownian motion under the measure $Q^{T_{n+1}}$ -- the Radon-Nykodim derivative $$ \frac{dQ^{T_{n+1}}}{dQ^B}|_{t} = \exp\bigg(-\int_0^t\sqrt{z(s)}\mu_n(s)dW^B_s - \frac{1}{2} \int_0^t z(s)\mu_n^2(s) ds\bigg),$$ employed in the Girsanov transformation, is independent of $\widetilde{W}(t)$.

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Thanks for Q2! I thought I understand Girsanov but not. It actually says, in $Q$, $W_t$ Brownian, $X_t$ adapted, $Y_t$ martingale, define $Z_t := \mathcal E (X_t)$, $\tilde Q: d \tilde P / dP = Z_t$, then $\tilde Y_t = Y_t - [Y,X]_t$ is martingale. Normally we use $X_t = \int_0^t u(s) ds$ and $Y_t = W_t$ but that's not necessary. Here we can use $Y_t = Z_t$ the other Brownian motion independent to $W_t$. –  athos Aug 13 at 4:36
    
notess for Q1 and Cholesky decomposition: math.stackexchange.com/questions/163470/… –  athos Aug 13 at 6:37

Q1: $$(1)\rightarrow(2)$$

(1): $a(t)$ is the instantaneous correlation of $\rho(Z_t,W_t)$ because:

$$\rho(dZ_t,dW_t)=\dfrac{Cov(dZ_t,dW_t)}{\sigma_{dZ_t}\sigma_{dW_t}}=\dfrac{E(dZ_t\cdot dW_t)}{\sqrt{dt} \sqrt{dt}}=\dfrac{\langle dZ_t, dW_t\rangle}{t}=a(t)$$

$\Rightarrow$ (2) holds as following, in the 1-dim case:

  • $dZ_t\sim N(0,dt),$

  • $dW_t,\tilde{dW_t}\stackrel{iid}{\sim}\,N(0,dt),$

$$\Rightarrow a_tdW_t+\sqrt{1-a_t^2}d\tilde{W_t}\sim N\left(0\cdot a_t+\sqrt{1-a_t^2}\cdot 0,\,a^2dt+\sqrt{1-a_t^2}^2dt\right)=N(0,dt)$$

because we know the sum of independent normal variates is again normal with $N(\mu=\mu_1+\mu_2,\sigma^2=a^2\sigma_1^2+b^2\sigma_2^2)$.

So: $$a_tdW_t+\sqrt{1-a_t^2}d\tilde{W_t} \stackrel{d}{\sim}dZ_t$$ q.e.d.


Hence this holds for Q2 aswell, and you can plug it in.

q.e.d.

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