# discounted price is a martingale under any measure?

Assume I have some dynamics for the stock price under 2 different measures: risk-neutral and forward measures: $$dS_t=r S_tdt+\sigma S_td\tilde{W_t}$$ $$dS_t=\alpha S_tdt+\sigma S_td\hat{W_t}$$

now I define 2 functions: $$g(t,s)=\mathbb{\tilde{E}}[e^{-\int_t^Tr_udu}h(S(T))|\mathbb{F}_t]$$ $$f(t,s)=\mathbb{\hat{E}}[e^{-\int_t^Tr_udu}h(S(T))|\mathbb{F}_t]$$ Then in both cases $e^{-\int_0^tr_udu}g$ and $e^{-\int_0^tr_udu}f$ are martingales under $\tilde{\mathbb{P}}$ and $\hat{\mathbb{P}}$ respectively. In the first case, it is basically discounting with a bank account, which(using tower property) is RN measure associated with. In the second though, the measure is associated with the bond of maturity $T$ so I would expectt $f/B(t,T)$ be a martingale under $\hat{\mathbb{P}}$, but I see $e^{-\int_0^tr_udu}f$ is. So they both are?

• For constant, or deterministic, interest rate, the risk-neutral measure and the forward measure are the same, and any martingality property is also the same. They will be different under the stochastic interest rate assumption. Aug 25, 2016 at 17:46
• @Gordon: I edited the question as I did not mean the trivial case. I sort of guess that $g$ is the price of a derivative while $f$ is not, and rather just a function" and doesn't have to have the property of $f/B$ be a martingale, but would be nice if someone can clarify this precisely. Aug 25, 2016 at 17:51
• The notations for $e^{-rt}f$ and $e^{-rt}$ are still confusing unless $r$ is a constant. You are right that $g$ is a pricing function, while $f$ is not. Aug 25, 2016 at 17:53
• @Gordon: Thanks, edited all of the terms. Aug 25, 2016 at 17:59

Indeed, based on your definition, $e^{-\int_0^tr_s ds}f$ is a martingale under the forward measure $\hat{\Bbb{P}}$. Note that, since $f$ is not an asset price process, $f/B(t, T)$ is not a martingale process under $\hat{\Bbb{P}}$.
Since $g$ is an asset price process, then $e^{-\int_0^tr_s ds}g$ is a martingale under the risk-neutral measure $\tilde{\Bbb{P}}$. Moreover, $g/B(t, T)$ is a martingale under the forward measure $\hat{\Bbb{P}}$. In fact, note that \begin{align*} \eta_t =: \frac{d\tilde{\Bbb{P}}}{d\hat{\Bbb{P}}}|_t = \frac{B(0, T) e^{\int_0^t r_s ds}}{B(t, T)}. \end{align*} Then, \begin{align*} g(t, S) &= \tilde{\Bbb{E}}\left(e^{-\int_t^T r_s ds} h(S_T) \mid \mathscr{F}_t \right)\\ &=\hat{\Bbb{E}}\left(\frac{\eta_T}{\eta_t} e^{-\int_t^T r_s ds} h(S_T) \mid \mathscr{F}_t \right)\\ &= B(t, T)\hat{\Bbb{E}}\left(h(S_T) \mid \mathscr{F}_t \right). \end{align*} Therefore, \begin{align*} \frac{g(t, S)}{B(t, T)} = \hat{\Bbb{E}}\left(h(S_T) \mid \mathscr{F}_t \right) \end{align*} is a martingale.
Your definition of $f$ is just a conditional expectation, which does not have any financial meaning. Alternatively, you can define $f$ by \begin{align*} f(t, S) =B(t, T)\hat{\Bbb{E}}\left(h(S_T) \mid \mathscr{F}_t \right), \end{align*} but then $f(t, S)=g(t, S)$.