Testing Whether Rate and Variance Are Positively Correlated
Summary
The document examines a claim from the Longstaff–Schwartz interest-rate model that the short rate, expressed as a weighted sum of two factors, is positively correlated with the variance factor. The answer reduces the problem to the sign of the covariance, normalizes one coefficient, and writes the covariance in terms of the factors’ variances and covariance. It then considers the most adverse factor correlation, namely perfect negative correlation, and minimizes the resulting quadratic expression over the relative standard deviation.
This calculation shows that positivity does not follow from positive coefficients alone: for a sufficiently small second coefficient and a suitable variance ratio, the covariance can be negative. The response therefore supplies a counterexample to an unconditional claim rather than establishing a general positive-correlation result. Its conclusion depends on the assumed factor definitions and parameter restrictions; any positivity assertion in the original model must rely on additional hypotheses not included in the question.
Key ideas
- The sign of correlation can be determined by the sign of covariance when both variables have nonzero variance.
- The covariance of the rate and variance expressions depends on both factor variances and their cross-covariance.
- Positive factor weights do not by themselves guarantee positive correlation between the derived quantities.
- Perfect negative correlation between the factors provides a useful worst-case test for the covariance sign.
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Full text
# Why does the correlation between r and V in Longstaff and Schwartz 1992 model is positive?
# 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?
## Answer by M. Jeunesse (score 0, accepted)
https://quant.stackexchange.com/a/26404
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.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.