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 :



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.

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

1 Answer 1


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:

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


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

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.

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

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