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How Correlation Affects the Expected Product of Increasing Functions

Article Quant Q&A · Author: DeepInTheQF
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Summary

The note asks whether the expected product of increasing functions of two correlated stochastic processes rises as their instantaneous correlation rises. The response sketches an Itô-calculus argument in a simplified setting: it assumes zero drift and unit, constant volatility, represents one Brownian motion using the other and an independent Brownian motion, and tracks the cross-variation term in the product.

Under those assumptions, the expected product contains a term proportional to correlation, with a sensitivity expressed as an integral of the product of the functions’ derivatives. This supports a positive relationship when both functions are differentiable and increasing. The argument is not a general proof for the original dynamics: it simplifies away the stated drifts and stochastic volatilities, and monotonicity alone does not guarantee differentiability. Applying the conclusion more broadly requires additional regularity and model assumptions.

Key ideas

  • The covariance contribution to the expected product depends on correlation between the processes.
  • In the simplified zero-drift, unit-volatility setup, Itô’s formula yields a correlation sensitivity involving both derivatives.
  • The sensitivity is positive when the differentiable functions are increasing.
  • The sketch does not establish the result for arbitrary drifts or stochastic volatilities.

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StatisticsDerivatives pricing
Full text
# Lemma (maybe) to imply the sign of the sensitivity to correlation


# Lemma (maybe) to imply the sign of the sensitivity to correlation












Can anybody please help me to understaind if this result is true ?

Let $\pi=\mathbb{E}\left(f(X_{T})g(Y_{T})\right)$

where $f$ and $g$ are increasing functions.

Hence, $\pi$ is increasing with respect to $\rho^{X,Y}$ : the instantanous correlation between $X_{t}$, $Y_{t}$, defined by $$dW_{t}^{X}dW_{t}^{Y}=\rho^{X,Y}dt$$ and $X$,$Y$ have the following dynamics :

$$dX_{t}=\mu^{X}dt+\sigma_{t}^{X}dW_{t}^{X}$$

$$dY_{t}=\mu^{Y}dt+\sigma_{t}^{Y}dW_{t}^{Y}$$

and $W_{t}^{X}$,$W_{t}^{Y}$ are Brownian motions.

I would like to know whether $\pi$ is increasing w.r.t $\rho^{X,Y}$?

Thanks in advance.

## Answer by Kermittfrog (score 3)

https://quant.stackexchange.com/a/68834

I'd argue as follows.

Let's simplify and assume $\mu_i=0,\sigma_i=1$ and let us set

$$ \begin{align} dW_t^{X}&=dW_t^{(1)}\\ dW_t^{Y}&=\rho dW_t^{(1)}+\sqrt{1-\rho^2}dW_t^{(2)} \end{align} $$

Using Ito's lemma,

$$ dF(x,y)=F_xdx+F_ydy+\frac{1}{2}\left(F_{xx}dx^2+F_{yy}dy^2+2F_{xy}dxdy\right) $$

In our case:

$$ d\pi = \rho f_xg_y dt+\frac{1}{2}(gf_{xx}+fg_{yy})dt+\left(gf_x+\rho fg_y\right)dW_t^{(1)}+fg_y\sqrt{1-\rho^2}dW_t^{(2)} $$

i.e.

$$ \pi(X_t,Y_t)=\pi(X_0,Y_0)+\int\limits_0^t\rho f_xg_y +\frac{1}{2}(gf_{xx}+fg_{yy})ds+\int\limits_0^t gf_x+\rho fg_ydW_s^{(1)}+\int\limits_0^t fg_y\sqrt{1-\rho^2}dW_s^{(2)} $$

With corresponding expectation

$$ E(\pi(X_t,Y_t))=\pi(X_0,Y_0)+\int\limits_0^t\rho f_xg_y +\frac{1}{2}(gf_{xx}+fg_{yy})ds, $$

which is increasing in $\rho$:

$$ \frac{\partial E(\pi(X_t,Y_t))}{\partial \rho}=\int\limits_0^t f_xg_y ds $$

which is positive since $f,g$ are increasing functions.

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.