Numeraire correlated to the traded asset

Question

The Fundamental Theorem of Asset Pricing states that:

\begin{align*} \frac{X_0}{N_0} &= \mathbb{E}^N{ \left[ \frac{X(t)}{N(t)}|\mathcal{F}_0 \right] } \end{align*}

The usual conditions apply (both $ N(t) $ and $ X(t) $ are traded assets, markets are complete, etc.)

Question: does the equation above still hold if $N(t)$ is correlated to $X(t)$ ?

Mathematically, one could suppose that (under the real-world measure):

$$X(t)=X(0)+\int^{t}_{0}\mu_1 X(h)dh+\int^{t}_{0}\sigma_{1} k_{1,1} X(h)dW_1(h)+\int^{t}_{0}\sigma_{1} k_{1,2} X(h)dW_2(h)$$

$$N(t)=N(0)+\int^{t}_{0}\mu_2 N(h)dh+\int^{t}_{0}\sigma_{2} k_{2,1} N(h)dW_1(h)+\int^{t}_{0}\sigma_{2} k_{2,2} N(h)dW_2(h)$$

In other words, there are two Brownian motions that are the sources of risk. Asset $X(t)$ has linear loadings ($K_{1,1}$) onto $W_1$ and ($K_{1,2}$) onto $W_2$, whilst the Numeraire has linear loadings ($K_{2,1}$) onto $W_1$ and ($K_{2,2}$) onto $W_2$, which makes $N(t)$ and $X(t)$ correlated.

If you'd like to answer the question generally, without taking the specific process equations for $X(t)$ and $N(t)$ into account, that is also fine.

Thank you so much, I highly appreciate any inputs on this.

Answer 1

As @ilovevolatility explains, the seminal reference for this matter is Geman, El Karoui & Rochet (1995). We assume none of the assets are dividend paying, and they are strictly positive. There are two potential options.

• You are considering a market with only assets $X$ and $N$. Then Assumption 1 of their paper would apply, which is related to the two Fundamental Theorems of Asset Pricing: "there exists a probability measure $\mathcal{N}$ associated to the numéraire $N$ such that the asset $X$ is a martingale in measure $\mathcal{N}$".
  This is a required assumption in your model. The First Fundamental Theorem implies that this assumption is equivalent to assuming your market is arbitrage-free. If $\mathcal{N}$ is unique, then by the Second Fundamental Theorem the market is also complete. Hence correlation does not matter, because you are assuming the process is martingale (of course, your dynamics need to be specified in such a way that this actually holds!).

• You are considering a market with assets $X$, $N$ and $M$, where $M$ is for example the risk-free money market account. Your assumption is that $X/M$ and $N/M$ are martingales under the risk-neutral measure $\mathcal{Q}$ induced by $M$. Then Theorem 1 in Geman, El Karoui & Rochet (1995) states that there exists a probability measure $\mathcal{N}$ induced by $N$ under which $X/N$ and $M/N$ are martingales. This should hold independently on whether $X$ and $N$ are correlated $-$ their paper contains a nice proof which is independent of the specific dynamics of these processes.

For a practical example of the second case in a typical Brownian Motion setting, we require Girsanov theorem (see for example these notes). Let us assume the following dynamics under $\mathcal{Q}$, with $M_0$ equal to $1$: $$\begin{align} dX(t) & = r X(t)dt+\sigma X(t)dW^\mathcal{Q}(t) \\ dN(t) & = rN(t)dt + \varsigma N(t)dB^\mathcal{Q}(t) \end{align}$$ where $dW^\mathcal{Q}(t)dB^\mathcal{Q}(t)=\rho dt$ and with the money-market account evolving as: $$dM(t) = rM(t)dt.$$ The change of measure from $\mathcal{Q}$ to $\mathcal{N}$ is given by the following Radon-Nikodym derivative (see again Theorem 1 in the paper): $$\frac{d\mathcal{Q}}{d\mathcal{N}}=\frac{M(t)N_0}{M_0N(t)}=e^{\frac{1}{2}\varsigma^2 t-\varsigma B^\mathcal{Q}(t)}$$ According to Girsanov theorem, we can then define a new measure which we will name $\mathcal{N}$ such that the Brownian Motion there is given by: $$\begin{align} B^\mathcal{N}(t)&:=B^\mathcal{Q}(t)-\varsigma t \end{align}$$ Using the Cholesky decomposition of two correlated Brownian Motions to represent $W$, we get that under the new measure: $$W^\mathcal{N}(t)=\rho B^\mathcal{N}(t)+\sqrt{1-\rho^2}Z^\mathcal{N}(t)=W^\mathcal{Q}(t)-\rho\varsigma t$$ where $Z$ is a third Brownian Motion independent of $B$. Hence the dynamics under the new measure are: $$\begin{align} dX(t) &= (r+\rho\sigma\varsigma)X(t)dt+\sigma X(t)dW^\mathcal{N}(t) \\ dN(t) &= (r+\varsigma^2)N(t)dt+\varsigma N(t)dB^\mathcal{N}(t) \end{align}$$ That is: $$\begin{align} X(t) &= X_0e^{(r+\rho\sigma\varsigma-\frac{1}{2}\sigma^2)t+\sigma W^\mathcal{N}(t)} \\ N(t) &= N_0e^{(r+\frac{1}{2}\varsigma^2)t+\varsigma B^\mathcal{N}(t)} \end{align}$$ Hence the asset $X(t)$ divided by the new numéraire $N(t)$ is equal to: $$\frac{X(t)}{N(t)}=\frac{X_0}{N_0}e^{(\rho\sigma\varsigma-\frac{1}{2}(\sigma^2+\varsigma^2))t+\sigma W^\mathcal{N}(t)-\varsigma B^\mathcal{N}(t)}$$ Using again the Cholesky representation of $W$: $$\frac{X(t)}{N(t)}=\frac{X_0}{N_0}e^{(\rho\sigma\varsigma-\frac{1}{2}(\sigma^2+\varsigma^2))t+(\sigma\rho-\varsigma)B^\mathcal{N}(t)+\sigma\sqrt{1-\rho^2} Z^\mathcal{N}(t)}$$ The random variable $(\rho\sigma-\varsigma)B^\mathcal{N}(t)+\sigma\sqrt{1-\rho^2} Z^\mathcal{N}(t)$ is normally-distributed with zero expectation and variance: $$(\rho\sigma-\varsigma)^2t+\sigma^2(1-\rho^2)t=\varsigma^2t-2\rho\sigma\varsigma t+\sigma^2t$$ Thus by properties of log-normal variables: $$\mathbb{E}^\mathcal{N}\left(e^{(\sigma\rho-\varsigma)B^\mathcal{N}(t)+\sigma\sqrt{1-\rho^2} Z^\mathcal{N}(t)}\right)=e^{\frac{1}{2}(\sigma^2+\varsigma^2)t-\rho\sigma\varsigma t}$$ Terms cancel and we would get: $$\mathbb{E}^\mathcal{N}\left(\frac{X(t)}{N(t)}\right)=\frac{X_0}{N_0}$$ Hence the process is a proper martingale under the new measure $\mathcal{N}$.

In my change-of-measure Equations, you notice that the "shift" applied to the second Brownian Motion takes into account correlation, i.e. $W^\mathcal{N}(t)=W^\mathcal{Q}(t)-\color{blue}{\rho}\varsigma t$. This term then is injected into the drift of $X$ under the new measure: $dX(t)=(\dots+\color{blue}{\rho}\sigma\varsigma)dt+\dots$, which gets cancelled when computing the expectation of the log-normal variable.

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_{A technical point on change of measure under a Brownian setting, for completeness purposes (measure superscripts skipped unless necessary). Properly speaking, our model is actually driven by a 2-dimensional Brownian Motion: $$\textbf{W}(t)= \begin{bmatrix} B(t) \\ Z(t) \end{bmatrix}$$ where $B$ and $Z$ are independent. We then have both a volatility matrix $\Sigma$ and a Cholesky matrix $\textbf{C}$ (which is the decomposition of the correlation matrix between the Brownian Motions), which give us a weight matrix $\Phi$ for the two Brownian Motions: $$\Sigma := \begin{bmatrix} \varsigma & 0 \\ 0 & \sigma \end{bmatrix}, \qquad \textbf{C} := \begin{bmatrix} 1 & 0 \\ \rho & \sqrt{1-\rho^2} \end{bmatrix}, \qquad \Phi:=\Sigma\cdot\textbf{C}=\begin{bmatrix} \varsigma & 0 \\ \sigma\rho & \sigma\sqrt{1-\rho^2} \end{bmatrix}$$ Note that $\Phi\cdot\Phi^T$ gives us the instantaneous covariance matrix. The diffusion part of $N$ and $X$ is represented by the following vector: $$\Phi\cdot d\textbf{W}(t)=\begin{bmatrix} \varsigma dB(t) \\\sigma (\rho dB(t)+\sqrt{1-\rho^2}dZ(t)) \end{bmatrix} =\begin{bmatrix} \varsigma dB(t) \\ \sigma dW(t) \end{bmatrix}$$ where $W$ is the original Brownian Motion of $X$ introduced in the body of the text. When we change measures, we are actually applying 2-dimensional Girsanov theorem and "shifting" the whole vector $\textbf{W}$. However as you can see in the Radon-Nikodym derivative Equation, it's only the Brownian $B$ that is shifted by $\varsigma t$, while the Brownian $Z$ is shifted by $0$. Indeed we can write: $$\frac{d\mathcal{Q}}{d\mathcal{N}} =e^{\frac{1}{2}\varsigma^2 t-\varsigma B^\mathcal{Q}(t)} =e^{\frac{t}{2}(\Theta^T\cdot\Theta)-\Theta^T\cdot\textbf{W}(t)}$$ where $\Theta$ is the vector specifying the change of measure from $\mathcal{Q}$ to $\mathcal{N}$: $$\Theta := \begin{bmatrix} \varsigma \\ 0 \end{bmatrix}$$ So the Brownian Motion under the new measure becomes: $$\textbf{W}^\mathcal{N}(t) =\textbf{W}^\mathcal{Q}(t)-\Theta\times t =\begin{bmatrix} B(t)-\varsigma t \\ Z(t) \end{bmatrix}$$}