# How is an exchange rate process a martingale under any measure?

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?

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 $$dX_t=(r-q)\,X_t\,dt+\sigma\,X_t\,dW_t\,.$$

• Nice answer (+1) but aren't you missing a bunch of $\text{d}$'s in your SDEs for $\text{d}t$ and $\text{d}W_t$? Nov 4, 2022 at 22:18
• Absolutely. Thanks! Corrected now. Nov 5, 2022 at 5:53
• 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: $$M_t=e^{-r_{USD}t}X_t+\int_{h=0}^{h=t}r_{EUR}X_he^{\color{orange}{+}r_{USD}(t\color{orange}{-h})}dh$$ Nov 14, 2022 at 10:55
• @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. Nov 14, 2022 at 11:17
• 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. Nov 14, 2022 at 14:38