We have two uncorrelated Stock price processes and the classical Money-Market (MM) account. Under the MM Numeraire, both stocks are Martingales when discounted by the MM, as usual.

Question: I would like to change the Numeraire to one of the stocks and apply the Fundamental Theorem of Asset pricing to show that the stock, when discounted by the other stock as Numeraire, is still a martingale.

I fail to show that, here goes my attempt:

Assume that we have two stocks, $X(t)$ and $Y(t)$ and the usual deterministic Money Market Account $M(t)$. Under the risk-neutral measure $Q$ associated with $M(t)$, let the processes for $X(t)$ and $Y(t)$ be as follows:

$$ X(t)=X(0)+\int^{t}_{0}r X(h)dh+\int^{t}_{0}\sigma_x X(h)dW_x(h) $$

$$ Y(t)=Y(0)+\int^{t}_{0}r Y(h)dh+\int^{t}_{0}\sigma_y X(h)dW_y(h) $$

Above, $W_x$ and $W_y$ are two uncorrelated standard Wiener processes under $Q$. Changing the Numeraire to $X(t)$ gives the following Radon-Nikodym:

$$ \frac{dQ_{X(t)}}{dQ}=\frac{M(t_0)}{M(t)}* \frac{X(t)}{X(t_0)} = \frac{e^{rt-0.5\sigma_x^2t+\sigma_xW_t}}{e^{rt}} = e^{-0.5\sigma_x^2t+\sigma_xW_t} $$

We see that the Radon-Nikodym can be directly applied via the Cameron-Martin-Girsanov theorem to both $W_x$ and $W_y$, whereby it will add drift of $\sigma_xt$ to both Brownians under the measure $Q_{X(t)}$. Consequently, the processes for $X_t$ and $Y_t$ under $Q_{X(t)}$ become:

$$ X_t = X_0exp{(rt+0.5\sigma_x^2t+ \sigma_x \tilde{W_x}(t)}) $$

$$ Y_t = Y_0exp{(rt-0.5\sigma_y^2t+ \sigma_x \sigma_y t+ \sigma_y \tilde{W_y}(t)}) $$

Above, $\tilde{W_y}(t)=W_y(t)-\sigma_xt$ is a standard Wiener process under the measure $Q_{X(t)}$ and also $\tilde{W_x}(t)=W_x(t)-\sigma_xt$ is another (uncorrelated) standard Wiener proces under the same measure $Q_{X(t)}$.

Now will the ratio of $X_t$ and $Y_t$ be a Martingale under $Q_{X(t)}$?

Let's see. First note that the ratio $\frac{Y(t)}{X(t)}$ itself will require the application of Ito's Lemma. Let $F(X_t,Y_t)=\frac{Y(t)}{X(t)}$, application of two-dimensional Ito's lemma yields:

$$ F(t)=F(0)+\int^{t}_{0}\left(r_y - r_x + \sigma_x^2-\rho \sigma_x\sigma_y\right) F(h)dh+\int^{t}_{0}\sigma_x F(h)d\tilde{W}_x(h)+\int^{t}_{0}\sigma_y F(h)d\tilde{W}_Y(h) =\\F(0)+\int^{t}_{0}\left(\sigma_x\sigma_y\right) F(h)dh+\int^{t}_{0}\sigma_x F(h)d\tilde{W}_x(h)+\int^{t}_{0}\sigma_y F(h)d\tilde{W}_Y(h) = \\ F_0exp\left( \left[\sigma_x\sigma_y-0.5\left(\sigma_x^2 + \sigma_y^2 \right) \right]t+ \sigma_x \tilde{W}_x + \sigma_y \tilde{W}_y \right) $$


$$\mathbb{E}^{Q_{X(t)}}{ \left[ \frac{Y(t)}{X(t)}|\mathcal{F}_0 \right] } = \mathbb{E}^{Q_{X(t)}}{ \left[ F(t)|\mathcal{F}_0 \right] } = F_0exp \left( \sigma_x \sigma_y \right) = \frac{Y(0)}{X(0)} exp \left(\sigma_x \sigma_y \right) $$

(above, I used the fact that: $\sigma_1 W_1(t)+\sigma_2W_2(t)=(distribution)={W_3}(t)\sqrt{\sigma_1^2 + \sigma_2^2}$ where $W_1$ and $W_2$ are two independent Brownian motions and $W_3$ is just another Standard Brownian motion. The equality holds in distribution, so can be used inside an expectation operator).

Clearly $\frac{Y(0)}{X(0)} exp \left(\sigma_x \sigma_y \right) \ne \frac{Y(0)}{X(0)} $ so the Martingale condition is NOT satisfied.

Any help on this would be hughly appreciated.

  • 1
    $\begingroup$ Your calculations for the process $X_t$ are fine. However, you say that the Brownian drivers are uncorrelated. So when applying Girsanov theorem you should actually write $$ \tilde{W}_y(t) = W_y(t) - \langle W_y, \sigma_x W_x \rangle_t = W_y(t) $$ is a Brownian motion under $\Bbb{Q}_X$. The SDE for $Y_t$ under $\Bbb{Q}_X$ is thus the same as under $\Bbb{Q}$. $\endgroup$ – Quantuple Jun 8 at 11:40
  • $\begingroup$ @Quantuple: thank you so much. Intuitively, shouldn't the drift of $Y(t)$ change under $\mathbb{Q_x}$ though? I am referring to this thread: quant.stackexchange.com/questions/50502/… $\endgroup$ – Jan Stuller Jun 8 at 12:08
  • $\begingroup$ The drift of $Y_t$ under the new measure only if the driver under the original measure and the ratio of the new versus old numéraire (or more precisely the RN derivative of the change of measure) are correlated. Here it is not the case. $\endgroup$ – Quantuple Jun 9 at 6:26
  • 1
    $\begingroup$ Bond SDE under its own forward measure deals with a similar problem but for zero-coupon bonds, it might be helpful. $\endgroup$ – Daneel Olivaw Jun 11 at 11:05

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