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

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 $$\rho^{X,Y}$$?

• What is the question? Whether $\pi$ is increasing w.r.t $\rho$? Commented Nov 18, 2021 at 17:03
• Yes, I will add that in the question. Thanks Commented Nov 18, 2021 at 17:15

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.

• How do we conclude from $f$ and $g$ are increasing that the last integral cannot be negative? Commented Nov 20, 2021 at 2:30
• I‘d argue that the first derivative of an increasing function cannot become smaller than zero. Would that make sense? Commented Nov 20, 2021 at 4:05