# Option pricing with dependent risk factors

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 $\lambda$.

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

• The dynamics of $S_t$ is not directly linked to that of $\lambda_t$ in what you wrote above, is that normal? Also is $\lambda_t$ a traded asset? – Quantuple Apr 6 '17 at 12:07
• The dynamics of $S_t$ and $\lambda_t$ are indirectly linked via the correlated Brownian motions $W_1$ and $W_2$. $\lambda_t$ is in fact the force of mortality in variable annuities (insurance products). This plays the role of a discount factor which is not traded. – Wiles01 Apr 6 '17 at 12:43
• Ok just wanted to make sure. Also I assume you're looking for a closed form solution not a numerical evaluation of the expectation? – Quantuple Apr 6 '17 at 13:01
• Yes, I would like, maybe via an adequate measure of change, to come back to the B&S call price but I'm stuck with the conditional expectation... – Wiles01 Apr 6 '17 at 13:04
• You can probably obtain a closed form solution for the variable annuity option price $E^Q[\exp(-\int_0^T \lambda(v)dv) \max(S(T),K))]$ by introducing the "maturity $T$ mortality zero coupon" $X(t,T) = E^Q[\exp(-\int_t^T \lambda(v)dv)]$ and then using $X(t,T)$ as your numeraire instead of $\exp(\int_0^t \lambda(v)dv)$. – Antoine Conze Oct 5 '17 at 11:56

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*}
• Great answer @Gordon. Just a remark, maybe I have missed something but shouldn't we have: $E\left(W_1(T) X_T \right)=\rho \left(e^{cT}/c\color{red}{-1/c}-T \right)?$ – Daneel Olivaw Oct 6 '17 at 10:23
• @Gordon Thanks for your great answer. I still have a small remark in terms of interpretation: In the B&S model, we know that the risk-neutral expectation gives the value to perfectly replicate the call option (through continuous trading). Given your calculus, does this mean that the value to perfectly replicate the call option will change depending on the value of $x$ ? Thanks in advance. – Wiles01 Feb 15 '18 at 10:19
• @Wiles01: Yes, your hedge will depend on the realization of $x$. As there are two random sources, to completely hedge, you may need to introduce another asset, but this will be worthy for discussing as another question. – Gordon Feb 15 '18 at 18:11
• @Gordon A last remark: Shouldn't we have $T$ under the square root in $d_1$ and $d_2$? The sigma has changed but not the time length. Thanks for the reply. – Wiles01 Mar 14 '18 at 15:26