# Why does the correlation between r and V in Longstaff and Schwartz 1992 model is positive?

I am reading the Longstaff and Schwartz's 1992 and 1993.

From $r = \alpha x + \beta y$ and $V = \alpha^2 x + \beta^2 y$. It was mentioned in the paper that the $r$ is positive correlated with $V$.

But I could not show that the $\operatorname{corr}(r, V)$ is positive. Would you please give me the proof or hints?

Humm....you must miss some hypothesis.

Assuming $\alpha>0,\beta>0$, you can without loss of generality set $\alpha=1$. (since $\frac{r}{\alpha}$ and $\frac{V}{\alpha^2}$ being positively correlated is the same as $r$ and $V$ positively correlated)

now, positive correlation is equivalent to positive covariance.

Working with covariance and above hypothesis, I have :

$$\text{Cov}(r,V)=\text{Var}(x)+\beta^3\text{Var}(y)+(\beta+\beta^2)\text{Cov}(x,y)$$

Since $\text{Cov}(x,y)=\rho_{xy}\sqrt{\text{Var}(x)\text{Var}(y)}$ with $\rho_{xy}$ being the correlation between $x$ and $y$.

we finally get by dividing by $\text{Var}(x)$ and by setting $z=\sqrt{\frac{\text{Var}(y)}{\text{Var}(x)}}$

$$\frac{\text{Cov}(r,V)}{\text{Var}(x)}\geq 1+\beta^3z^2-(\beta+\beta^2)z$$

which is an equality if $\rho_{xy}=-1$

now $$1+\beta^3z^2-(\beta+\beta^2)z\geq 1-\frac{(\beta+\beta^2)^2}{4\beta^3}$$

which is hit when $z=\frac{\beta+\beta^2}{2\beta^3}$

With $\beta$ small enough, rhs is negative.

So by setting $x$ and $y$ with $\rho_{xy}=-1$ and $\text{Var}(y)=\left(\frac{\beta+\beta^2}{2\beta^3}\right)^2\text{Var}(x)$, and $\beta$ small enough, you get a counter example.

• I got your points. Does it relate to the $dx = (\gamma - \delta x)dt + \sqrt{x}dW$ and $dy = (\eta - \xi y)dt + \sqrt{y}dZ$, where $W$ and $Z$ are uncorrelated increments to Gauss-Wiener processes? So, can we always choose $\rho_{xy} = -1$? – imp Jun 2 '16 at 15:49
• If $W\perp Z$, then $X\perp Y$. – MJ73550 Jun 2 '16 at 15:59
• Is it true in general or just these specific $x$ and $y$ processes? Does $X \perp Y$ imply $\operatorname{cov}(X, Y) = 0$? Is $W \perp Z$ the same as $[W, Z] = 0$? I am a bit confused. – imp Jun 2 '16 at 16:11
• ''Point 1 :'' It is true if $X$ satisfy SDE with no $Y$ dependence and resp. for $Y$. "Point 2 :" $X\perp Y\Rightarrow \text{cov}(X,Y)=0$. $\text{cov}(X,Y)=E(XY)-E(X)E(Y)$ now if $X\perp Y$, then $E(XY)=E(X)E(Y)$. "Point 3 :" Two independents brownian motion will give you $[W,Z]\equiv 0$. If $(W_{t+h}-W_t,Z_{t+h}-Z_t)$ is a gaussian vector for all $t,h\geq 0$, then $[W,Z]=0\Rightarrow W\perp Z$. – MJ73550 Jun 2 '16 at 16:39
• Got it, very helpful! So, am I right that we need to be careful when writing the specification between "$W$ and $Z$ are uncorrelated increments to Gauss-Wiener processes" and "$W$ and $Z$ are two independents Brownian motion"? – imp Jun 2 '16 at 20:38