How to price a stock under Q and stochastic interest rates?

Question

I am interested in pricing a stock under $\mathbb{Q}$ when I assume that

$$dS(t) = \mu(S(t))dt + \sigma(S(t))dW(t)$$

where $W(t)$ is a Wiener process under $\mathbb{P}$ and

$$dr(t) = a(b-r(t))dt + \sigma(r(t))dZ(t)$$

where $Z(t)$ is a Wiener process under $\mathbb{P}$. So I have real-world observations of interest rates and stock prices and want to use them to price the stock under $\mathbb{Q}$. I found in one of the papers that the differential equation for stock price will look like:

$$dS(t) = r(t)dt + \sigma(S(t))dB(t)$$

and

$$dr(t) = a(b-r(t))dt + \sigma(r(t))dZ(t),$$

where $B(t)$ is a Wiener process under $\mathbb{Q}$. But I don't understand, why the Wiener process for the Interest rate is the same as under $\mathbb{P}$. Does it mean that if I want to price the stock under $\mathbb{Q}$ is doesn't matter if my interest rates are priced under $\mathbb{Q}$ or not? Does it mean that if I price my stock under $\mathbb{Q}$ with real-world interest rates it and it is martingale, it is risk-neutral? Could you please help me?

Thanks!

Answer 1

The derivation in Appendix A of the paper Valuation of Equity-Indexed Annuities under Stochastic Interest Rates that you mentioned is Wrong: the Girsanov transformation is applied to an $n$-dimensional Brownian motion, where the components are independent. However, for the case here with $n=2$, the Brownian motions are dependent, we can not naively combine them together to form a two-dimensional Brownian motion and then apply the Girsanov transformation.

For your case, we assume that, under the real-world probability measure $\mathbb{P}$, \begin{align*} dS(t) &= \mu(S(t)) dt + \sigma(S(t)) dW(t)\\ dr(t) &= a(b-r(t)) dt + \sigma(r(t)) dZ(t),\\ \end{align*} where $\{W(t), t \ge0\}$ and $\{Z(t), t \ge0\}$ are two standard Brownian motions with instantaneous correlation $\rho$. Based on Cholesky decomposition, we can re-write the above dynamics as \begin{align*} dS(t) &= \mu(S(t)) dt + \sigma(S(t)) dW(t)\\ dr(t) &= a(b-r(t)) dt + \sigma(r(t)) d\big(\rho W(t) + \sqrt{1-\rho^2} B(t)\big), \end{align*} where $\{W(t), t \ge0\}$ and $\{B(t), t \ge0\}$ are two independent standard Brownian motions.

To obtain the dynamics under the risk-neutral probability measure $\mathbb{Q}$, let \begin{align*} \lambda(t) = \frac{r(t) - \mu(S(t))}{\sigma(S(t))}. \end{align*} Then, \begin{align*} \frac{d\mathbb{Q}}{d\mathbb{P}}\big|_t = \exp\left(-\frac{1}{2} \int_0^t \lambda^2(s)ds + \int_0^t \lambda(s)dW_s \right). \end{align*} Moreover, under the measure $\mathbb{Q}$, \begin{align*} \widehat{W}(t) &= W(t) - \int_0^t \lambda(s)ds, \mbox{ and}\\ \widehat{B}(t) &= B(t), \end{align*} are two independent standard Brownian motions. Consequently, \begin{align*} dS(t) &= r(t) dt + \sigma(S(t)) d\widehat{W}(t)\\ dr(t) &= \big[a(b-r(t)) + \rho \lambda(t) \sigma(r(t)) \big]dt + \sigma(r(t)) d\big(\rho \widehat{W}(t) + \sqrt{1-\rho^2} \widehat{B}(t)\big). \end{align*} Note the extra term $\rho \lambda(t) \sigma(r(t))$ in this dynamics under the risk-neutral measure $\mathbb{Q}$.

Answer 2

If you model the spot price of the stock, then it is just the spot price (what else could be more accurate?).

If you model the forward price of a stock, then you most probably want to apply cost-of-carry (in order to avoid arbitrge). If there are no dividends in your spot, then the forward price for time $T$ is $$ F_T = S_0 rT $$ where $r$ is a rate that applies for the time period that you analyze.

If you have some interest rate model, then this should give the same factor for the same period (if it is calibrated to the rate). Thus it should give the same.

By the way: you look at short-rate models. The continuous short rate does not exist - it can not be traded by itself. Just objects similar to $$ E_Q[\exp(\int_0^T r_u du)] $$ can be traded (FRAs). So usually the SDE for $r_t$ the short rate is under Q. And Under Q the stock price grows by the risk-free rate.