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Let $(W^{(1)},W^{(2)})$ be a two-dimensional standard Brownian motion and let $$dV_t = \kappa(\theta - V_t)dt+ \sigma \sqrt{V_t}dW^{(1)}_t$$ where $\kappa, \theta$ and $\sigma$ are constants such that $$2 \kappa \theta > \sigma^2.$$ Let $\lambda$ and $a$ be constants and define \begin{align*} L^{(1)}_t &:= \exp \bigg\{-\int_0^t \lambda \sqrt{V_u}dW_u^{(1)}-\frac{1}{2}\int_0^t(\lambda \sqrt{V_u})^2du \bigg\}; \\ L^{(2)}_t &:=\exp \Bigg\{-\int_0^t \frac{1}{\sqrt{1-\rho^2}}\bigg(\frac{\mu -r}{\sqrt{V_u}}-\lambda \rho \sqrt{V_u}\bigg)dW_u^{(2)} \\ &-\frac{1}{2}\int_0^t \bigg[\frac{1}{\sqrt{1-\rho^2}}\bigg(\frac{\mu -r}{\sqrt{V_u}}-\lambda \rho \sqrt{V_u}\bigg) \bigg]^2du \Bigg\}; \\ L_T &:= L^{(1)}_TL^{(2)}_T. \end{align*}

I have to prove that $L$ is a martingale. The process $L$ is clearly a positive local martingale with $L_0=1,$ so it's a supermartingale. Thus, we can prove that $L$ is a true martingale by showing that $E[L_T]=1.$ As a part of the proof, in this paper: https://www.hindawi.com/journals/ijsa/2006/018130/
(in page 5) they say that:

"as $W^{(2)}$ and $V$ are independent, and $0<V_t< \infty$ for every $t \leq T$ with probability 1, by conditional expectations we have $$E[L_T]=E[L_T^{(1)}] \tag*{($\star$)}."$$ I don't know how to use conditional expectation and independence between $V_t$ and $W_t^{(2)}$ to prove $(\star).$ There are $V_t$ terms in $L^{(1)}_T$ and $L^{(2)}_T$ so they are not independent, I don't understand why $L^{(2)}_T$ goes away. Any ideas?

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  • $\begingroup$ Hi only some guesses I don't know if it's useful at all. They might be conditioning on the whole path $V_s$ for s in (0,t), so they could split the product but I'm not if this is licit. Then (still guessing) as $L_2$ looks to me like Radon-Nikodym term in the measure change of Girsanov, this might give $E[L_2(t)|(V_s)_{s \in(0,t)}]=1$ integrating over V would give (*). $\endgroup$
    – TheBridge
    Commented Jan 29, 2020 at 20:56
  • $\begingroup$ @TheBridge: Thank you for the suggestion. I'm not sure if I understand. Do you a know a book or paper where that idea is applied (even if it is in a different context)? $\endgroup$
    – UBM
    Commented Jan 31, 2020 at 17:47

1 Answer 1

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Based on @user159517 proof and @TheBridge suggestion.

First, an observation.

Observation 1. It's well known that if $\eta_t$ is a deterministic function, the stochastic exponential $\{M_t; 0 \leq t \leq T\},$ where $$M_t:= \exp \bigg\{-\int_0^t \eta_u dW_u-\frac{1}{2} \int_0^t \eta^2_u du \bigg\}$$ is a martingale and since $E[M_0]=1$ we must have $E[M_T]=1.$

Let $\mathcal F_T^{W^{(1)}}$ be the $\sigma$-algebra generated by $W^{(1)}.$ In order to prove ($\star$), the key point is to realize that $$E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]=1 \quad \text{a.s.} \tag{1}$$
Then \begin{align*} E[L_T]&=E[E[L_T|\mathcal F_T^{W^{(1)}}]] \\ &=E[E[L^{(1)}_TL^{(2)}_T|\mathcal F_T^{W^{(1)}}]] \\ &= E[L^{(1)}_T E[L^{(2)}_T|\mathcal F_T^{W^{(1)}}]] \\ &= E[L^{(1)}_T] \end{align*} and condition ($\star$) is proved.

Proof of (1): $\quad$ (from @user159517)

Let $(\Omega_i,\mathcal F_i, \mathbb P_i),i=1,2$ and take as a probability space $\Omega := \Omega_1 \times \Omega_2.$ Also, let $\mathbb P = \mathbb P_1 \otimes \mathbb P_2$ be the product measure on $\Omega.$ Note that because $\mathcal F_T^{W^{(1)}}$ is generated by $W^{(1)}$ which is not a function of $\omega_2$, the same holds for any $\mathcal F_T^{W^{(1)}}$-measurable random variable. As the coefficients in the SDE for $V$ are locally Lipschitz, by the Ito theorem we may take $V$ to be a strong solution, hence $V$ is $\mathcal F_T^{W^{(1)}}$-measurable and therefore a function only of $\omega_1$.

By the definition of conditional expectation, (1) is proved if we show that for all $A \in \mathcal F_T^{W^{(1)}}$ we have $E[1_A E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]=E[1_A].$ So let $A \in \mathcal F_T^{W^{(1)}}.$ Then

\begin{align}\int_{\Omega} E[L^{(2)}_T | \mathcal F_T^{W^{(1)}}]1_{A} ~d\mathbb{P} &= \int_{\Omega} L^{(2)}_T 1_{A} ~d\mathbb{P} = \int_{\Omega_1} 1_{A}(\omega_1)\left(\int_{\Omega_2}^{} L_{T}^{(2)}(\omega_1,\omega_2)d\mathbb{P}_2(\omega_2)\right)d\mathbb{P}_1(\omega_1) \\ &= \int_{\Omega_1} 1_{A}(\omega_1)~d\mathbb{P}_1(\omega_1) = \int_{\Omega} 1_{A}~d\mathbb{P}, \end{align}

where in the third equality above, we have used Observation 1 and the fact that $V = V(\omega_1)$ to conclude that $\int_{\Omega_2}^{} L_{T}^{(2)}(\omega_1,\omega_2)d\mathbb{P}_2(\omega_2) = 1$ for any $\omega_1 \in \Omega_1$.

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  • $\begingroup$ @user159517: Thank you, I edited the answer. $\endgroup$
    – UBM
    Commented Feb 14, 2020 at 19:57
  • $\begingroup$ very good, deleted my old answer. $\endgroup$
    – user159517
    Commented Dec 20, 2020 at 11:34
  • $\begingroup$ user159517 : Thank you! $\endgroup$
    – UBM
    Commented Dec 20, 2020 at 12:18

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