Option pricing with dependent risk factors

Question

I'm a bit stuck with the pricing of an option where the underlying stock is correlated to an additional process.

Setting: Assume that we have a probability space where under $Q$ the dynamics of the stock and an additional process are given by

$$ \begin{align} dS(t) &= S(t)(rdt+\sigma dW_1 (t)) \\[6pt] d\lambda(t) &= c\lambda(t)dt+\xi dW_2 (t) \end{align}$$

where: $$ W_2 (t)=\rho W_1 (t)+ \sqrt{1-\rho^2}Z(t) $$

$W_1 (t)$ and $Z(t)$ are independent Brownian motions.

The question is now how to determine the following conditional risk-neutral valuation:

$$ E^Q [e^{-\int_0^T\lambda(v)dv} \max(S(T),K) \ | \ e^{-\int_0^T\lambda(v)dv} =x] $$

The last expression can be rewritten as:

$$ x E^Q [\max(S(T),K) \ | \ e^{-\int_0^T\lambda(v)dv} =x] $$

but then I'm stuck how we can deal with the dependence between $S$ and $\lambda$.

Thanks a lot in advance for your help!

N.B: $\lambda$ is not the interest rate but just a stochastic discount factor.

Answer 1

Note that \begin{align*} d\left(e^{-ct} \lambda_t \right) &= -ce^{-ct} \lambda_t dt + e^{-ct}d\lambda_t\\ &=\xi e^{-ct}dW_2(t). \end{align*} Then \begin{align*} \lambda_t = \lambda_0 e^{ct} + \xi\int_0^t e^{-c(u-t)} dW_2(u). \end{align*} Moreover, \begin{align*} \int_0^T \lambda_s ds &= \frac{\lambda_0}{c}\left(e^{cT}-1 \right) + \xi\int_0^T \int_0^s e^{-c(u-s)} dW_2(u)ds\\ &=\frac{\lambda_0}{c}\left(e^{cT}-1 \right) + \xi\int_0^T ds \int_u^T e^{-c(u-s)} ds\, dW_2(u)\\ &=\frac{\lambda_0}{c}\left(e^{cT}-1 \right) +\frac{\xi}{c}\int_0^T\left(e^{-c(u-T)}-1 \right)dW_2(u)\\ &\equiv \frac{\lambda_0}{c}\left(e^{cT}-1 \right) + \frac{\xi}{c} X_T, \end{align*} where \begin{align*} X_T = \int_0^T\left(e^{-c(u-T)}-1 \right)dW_2(u) \sim N\left(0,\ \frac{1}{2c}e^{2cT}-\frac{2}{c}e^{cT}+T+\frac{3}{2c} \right). \end{align*} Furthermore, note that \begin{align*} E\left(W_1(T) X_T \right) &= E\left(\int_0^T dW_1(u) \int_0^T\left(e^{-c(u-T)}-1 \right)dW_2(u) \right)\\ &=\rho \left(\frac{1}{c}e^{cT}-\frac{1}{c}-T \right). \end{align*} Then \begin{align*} corr(W_1(T), \, X_T) = \frac{\rho \left(\frac{1}{c}e^{cT}-\frac{1}{c}-T \right)}{\sqrt{T \left( \frac{1}{2c}e^{2cT}-\frac{2}{c}e^{cT}+T+\frac{3}{2c}\right)}}\equiv \rho_0. \end{align*} We can then assume that \begin{align*} W_1(T) = \frac{\rho_0\sqrt{T}}{\sqrt{\frac{1}{2c}e^{2cT}-\frac{2}{c}e^{cT}+T+\frac{3}{2c}}}X_T + \sqrt{T\left(1-\rho_0^2\right)}Z, \end{align*} where $Z$ is a standard normal random variable that is independent of $X_T$.

From \begin{align*} e^{-\int_0^T \lambda_sds} = x, \end{align*} we obtain that \begin{align*} X_T = -\frac{c}{\xi}\ln x - \frac{\lambda_0}{\xi}\left(e^{cT}-1 \right). \end{align*} Then \begin{align*} S(T) &= S(0) e^{(r-\frac{1}{2}\sigma^2)T + \sigma W_1(T)}\\ &=S(0) e^{\frac{-\sigma \rho_0\sqrt{T}}{\sqrt{\frac{1}{2c}e^{2cT}-\frac{2}{c}e^{cT}+T+\frac{3}{2c}}}\left(\frac{c}{\xi}\ln x + \frac{\lambda_0}{\xi}\left(e^{cT}-1 \right) \right)} e^{(r-\frac{1}{2}\sigma^2)T + \sigma \sqrt{1-\rho_0^2} \sqrt{T} Z}\\ &\equiv \tilde{S}(0) e^{-\frac{1}{2}\sigma^2 (1-\rho_0^2)T + \sigma \sqrt{1-\rho_0^2} \sqrt{T} Z}, \end{align*} where \begin{align*} \tilde{S}(0) = S(0) e^{\frac{-\sigma \rho_0\sqrt{T}}{\sqrt{\frac{1}{2c}e^{2cT}-\frac{2}{c}e^{cT}+T+\frac{3}{2c}}}\left(\frac{c}{\xi}\ln x + \frac{\lambda_0}{\xi}\left(e^{cT}-1 \right) \right)} e^{(r-\frac{1}{2}\sigma^2\rho_0^2)T}. \end{align*} Therefore, \begin{align*} &\ E\left(e^{-\int_0^T \lambda_s ds} \max(S(T), K) \,|\, e^{-\int_0^T \lambda_s ds} =x\right) \\ =&\ xE\left(K+\max(S(T)-K, 0) \,|\, e^{-\int_0^T \lambda_s ds} = x\right)\\ =&\ x\Big(K + \tilde{S}(0)\Phi(d_1)-K\Phi(d_2) \Big), \end{align*} where $\Phi$ is the cumulative distribution function of a standard normal random variable, \begin{align*} d_1 = \frac{\ln \frac{\tilde{S}(0)}{K} + \frac{1}{2}\sigma^2 (1-\rho_0^2)T}{\sigma \sqrt{(1-\rho_0^2)T}}, \end{align*} and \begin{align*} d_2 = d_1 - \sigma \sqrt{(1-\rho_0^2)T}. \end{align*}