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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?

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  • $\begingroup$ 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. $\endgroup$ – Gordon Aug 25 '16 at 17:46
  • $\begingroup$ @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. $\endgroup$ – Medan Aug 25 '16 at 17:51
  • $\begingroup$ 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. $\endgroup$ – Gordon Aug 25 '16 at 17:53
  • $\begingroup$ @Gordon: Thanks, edited all of the terms. $\endgroup$ – Medan Aug 25 '16 at 17:59
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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 $f(t, S)=g(t, S)$.

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  • $\begingroup$ Crystal clear answer, +1 $\endgroup$ – Quantuple Aug 25 '16 at 20:10

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