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Let $X,Y,Z$ be three standard normal distribution. Let $\rho_{XY},\rho_{YZ},\rho_{XZ}$ be the correlation between those random variables.

Let $f()$ be a monotone, odd, bounded, and differentiable function. More precisely, $f(x)\geq f(y)$ iff $x\geq y$, $f(x)=-f(-x)$, $|f(x)| \leq c$. For example, tanh could be one such function.

If $\rho_{XY}=\rho_{YZ}=\rho_{XZ}=0$, i.e., $X,Y,Z$ are independent, then we have

$E(f(X)f'(Y)Z) = 0$.

Can we say anything about $E(f(X)f'(Y)Z)$ when $\rho_{XY},\rho_{YZ},\rho_{XZ}$ are not 0 but very small?

In other words, I am looking for something like

$E(f(X)f'(Y)Z) \rightarrow 0 $ as $\rho_{XY},\rho_{YZ},\rho_{XZ}\rightarrow 0$,

or perhaps

$|E(f(X)f'(Y)Z)| \leq g(\rho_{XY},\rho_{YZ},\rho_{XZ}) \sqrt{E(f^2(X)f'^2(Y)Z^2)} $, where

$g(\rho_{XY},\rho_{YZ},\rho_{XZ}) \rightarrow 0$ as $\rho_{XY},\rho_{YZ},\rho_{XZ}\rightarrow 0$

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  • $\begingroup$ I think so. You can write $(X,Y,Z)^T$ as $A(N_1,N_2,N_3)^T$ for a matrix that is upper triangular and close to the identity. If you then expand out, don't you get what you need? $\endgroup$
    – Anthony Quas
    Commented May 10, 2017 at 4:49
  • $\begingroup$ I did not quite get it. Could you explain in more detail? Thanks! $\endgroup$
    – clj
    Commented May 10, 2017 at 4:53
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    $\begingroup$ If the correlations are small, then $(X,Y,Z)^T$ has the same distribution as $A(N_1,N_2,N_3)^T$ (where $(N_1,N_2,N_3)$ are standard independent normal, and where $AA^T$ is equal to the covariance matrix). There are many possible choices of $A$, and you can choose any one that is convenient. It does not affect the distribution; only the covariance does. It's convenient to choose $A$ to be a triangular matrix. Since the covariance matrix is close to the identity, you can choose $A$ to be close to the identity matrix. $\endgroup$
    – Anthony Quas
    Commented May 10, 2017 at 5:20
  • $\begingroup$ If I understand correctly, you were suggesting changing the representation of the random variables. But how does that relate to the function f and the expectation? How can "A near identity" be translated to "expectation of function of N1 N2 N3 is near 0"? $\endgroup$
    – clj
    Commented May 10, 2017 at 18:30
  • $\begingroup$ You should be able to write $X=N_1+\text{small perturbation}$ etc. Now use your favourite expansion of $f$ etc. to compare the expression with $X$,$Y$ and $Z$ in it to the expression with $N_1$, $N_2$ and $N_3$ in it. $\endgroup$
    – Anthony Quas
    Commented May 10, 2017 at 19:18

1 Answer 1

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By properties of multivariate Gaussian distribution, Z given X,Y is normal with mean

$E(Z|X,Y)=(1-\rho_{XY}^2)^{-1}(\rho_{XZ}-\rho_{YZ}\rho_{XY})X+(1-\rho_{XY}^2)^{-1}(-\rho_{XZ}\rho_{XY}+\rho_{YZ})Y$

So for any measurable function $h(x,y)$, it holds that

$|E(h(X,Y)Z)|=|E(E(Z|X,Y)h(X,Y))|=(1-\rho_{XY}^2)^{-1}|E({(\rho_{XZ}-\rho_{YZ}\rho_{XY})Xh(X,Y)+(-\rho_{XZ}\rho_{XY}+\rho_{YZ})Y}h(X,Y))|\le (1-\rho_{XY}^2)^{-1}|\rho_{XZ}-\rho_{YZ}\rho_{XY}|\times E|Xh(XY)|+(1-\rho_{XY}^2)^{-1}|-\rho_{XZ}\rho_{XY}+\rho_{YZ}|\times E|Yh(XY)|$.

The result follows if $E|Xh(XY)|$ and $E|Yh(X,Y)|$ are bounded.

Take $h(x,y)=f(x)f'(y)$ in the above equations to get your result.

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