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Suppose a process for a stock price of a US-based company traded in the USA is, under the USD money-market numeraire:

$$dS_t=S_tr_{USD}dt+S_t\sigma_SdW_1(t)$$

Using fundamental theorem of asset pricing, we must have that:

$$\frac{S_0}{B_{USD}(t_0)}=\frac{S_0}{1}\stackrel{?}{=}\mathbb{E}^Q_{USD}\left[\frac{S_t}{B_{USD}(t)}\right]=\mathbb{E}^Q_{USD}\left[\frac{S_0e^{r_{USD}t-0.5\sigma_S^2t+\sigma_SW_1(t)}}{e^{r_{USD}t}}\right]=S_0$$

So clearly, the discounted process for $S_t$ is a martingale under $B_{USD}(t)$ as numeraire.

Suppose I am interested in the exchange rate between USD and EUR, and I denote the process that describes how many units of USD I need to pay for 1 unit of EUR as $X_t$ (i.e. analogously to the process $S_t$, which tells me how many units of USD I need to pay for 1 unit of $S_t$).

Let the process of $X_t$ be as follows:

$$dX_{EUR\rightarrow USD}(t)=(r_{USD}-r_{EUR})X_{EUR\rightarrow USD}(t)dt+\sigma_XX_{EUR\rightarrow USD}(t)dW_2(t)$$

The Forward on $X_t$ is denoted as $F(X_t)=\mathbb{E}_{USD}^Q[X_t|X_0]$.

The no-arbitrage condition on the forward is trivially: $$F(X_{EUR\rightarrow USD}(t))=\frac{e^{r_{USD}t}}{e^{r_{EUR}t}}X_{EUR\rightarrow USD}(t_0)$$

Clearly, this condition is satisfied because $$\mathbb{E}_{USD}^Q[X_t|X_0]=\mathbb{E}_{USD}^Q[X_0e^{r_{USD}t-r_{EUR}t-0.5\sigma_X^2t+\sigma_XW_2(t)}]=X_0e^{r_{USD}t-r_{EUR}t}=\frac{e^{r_{USD}t}}{e^{r_{EUR}t}}X_{EUR\rightarrow USD}(t_0)$$

But clearly, under the USD numeraire, the discounted process for $X_t$ is NOT a martingale, since:

$$\frac{X_0}{1}\stackrel{?}{=}\mathbb{E}^Q_{USD}\left[\frac{X_t}{B_{USD}(t)}\right]=\mathbb{E}^Q_{USD}\left[\frac{X_0e^{r_{USD}t-r_{EUR}t-0.5\sigma_X^2t+\sigma_XW_2(t)}}{e^{r_{USD}t}}\right]=X_0e^{-r_{EUR}t}\neq X_0$$

So the process for $X_t$ cannot be a valid process under the $B_{USD}(t)$ numeraire. It would not be a valid process under the $B_{EUR}(t)$ numeraire either, because again, if discounted by the EUR numeraire, it would not be a martingale.

Where is the catch here?

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The catch is that when a stock pays dividends, say, by a continuously compounded dividend yield $q$ then $$ \frac{dS_t}{S_t}=r_{USD}\,dt\color{red}{-q\,dt}+\sigma\,dW_t\,, $$ and $S_te^{-r_{USD}t}$ is not longer a martingale either. Thank god this can be fixed because when you add to this the present value of the paid dividends then the process $$ M_t=S_t\,e^{-r_{USD}\,t}+\int_0^tq\,S_u\,e^{-r_{USD}\,u}\,du $$ is a martingale. See this answer.

Now to the FX rate. The well-known SDE $$ \frac{dX_t}{X_t}=r_{USD}\,dt\color{red}{-r_{EUR}\,dt}+\sigma\,dW_t $$ suggests that $r_{EUR}$ is the continuous compounded dividend yield that you receive when you hold one unit of $X_t$ which is the price of one EUR in USD.

This makes a lot of sense because when we have 1 EUR in cash we can say that we hold $X_t$ USD and we do receive $r_{EUR}$ per time and share in dividends from holding that cash.

To make a long story short: $X_te^{-r_{USD}t}$ is not a martingale but $$ M_t=X_t\,e^{-r_{USD}\,t}+\int_0^tr_{EUR}\,X_u\,e^{-r_{USD}\,u}\,du $$ is one.

As a sanity check to see directly that this is a martingale write $r_{USD}=r$ and $r_{EUR}=q$ for brevity and note that \begin{align} dM_t&=e^{-r\,t}\,dX_t-r\,e^{-r\,t}X_t\,dt+q\,X_t\,e^{-r\,t}\,dt\\ &=e^{-r\,t}\Big\{dX_t-r\,X_t\,dt+q\,X_t\,dt\Big\}\\ &=e^{-r\,t}\,\sigma\,X_t\,dW_t \end{align} by the SDE $dX_t=(r-q)\,X_t\,dt+\sigma\,X_t\,dW_t\,.$

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    $\begingroup$ Nice answer (+1) but aren't you missing a bunch of $\text{d}$'s in your SDEs for $\text{d}t$ and $\text{d}W_t$? $\endgroup$
    – Kevin
    Nov 4, 2022 at 22:18
  • $\begingroup$ Absolutely. Thanks! Corrected now. $\endgroup$
    – Kurt G.
    Nov 5, 2022 at 5:53
  • $\begingroup$ The process for $X_t$ is: $$X_t=X_0+\int_{h=0}^{h=t}r_{USD}X_hdh-\int_{h=0}^{h=t}r_{EUR}X_hdh+\int_{h=0}^{h=t}\sigma X_hdW_h$$ Therefore, we should have that: $$M_t=e^{-r_{USD}t}\left\{X_0+\int_{h=0}^{h=t}r_{USD}X_hdh-\int_{h=0}^{h=t}r_{EUR}X_hdh\color{red}{+\int_{h=0}^{h=t}r_{EUR}X_hdh}+\int_{h=0}^{h=t}\sigma X_hdW_h\right\}$$ In other words: $$M_t=e^{-r_{USD}t}\left\{X_t+\int_{h=0}^{h=t}r_{EUR}X_hdh\right\}=e^{-r_{USD}t}X_t+\int_{h=0}^{h=t}r_{EUR}X_he^{-r_{USD}t}dh$$ instead of: $$M_t=e^{-r_{USD}t}X_t+\int_{h=0}^{h=t}r_{EUR}X_he^{\color{orange}{+}r_{USD}(t\color{orange}{-h})}dh$$ $\endgroup$ Nov 14, 2022 at 10:55
  • $\begingroup$ @JanStuller . Thanks for reading this carefully. There were typos in my answer. I agree with your equation for $X_t$ but do not see why your $M_t$ should be a martingale. $\endgroup$
    – Kurt G.
    Nov 14, 2022 at 11:17
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    $\begingroup$ I doubt that $X_t$ and your $\bar{X}_t$ are related in that way. Most likely a hick-up inside the integrals (factor $X_t$ is different from factor $\bar{X}_t$). To the end of the answer I just added a very direct proof that the $M_t$ there is without doubt a martingale. $\endgroup$
    – Kurt G.
    Nov 14, 2022 at 14:38

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