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Can we express $Cov(XY,X)$ in terms of moments of $X$ and $Y$ (instead of joint moments)?

Is there an alternative simplification than $$Cov(XY,X)=E[X^2Y]-E[XY]E[X]$$

If not, is there perhaps an approximation for $Cov(XY,X)$?

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    $\begingroup$ Any such simplification would require $X$ and $Y$ to be independent: is that what you are assuming? $\endgroup$
    – whuber
    Jun 20, 2020 at 14:22
  • $\begingroup$ @whuber Thanks for your comment. No, I can't assume this. The random variables $X$ and $Y$ are negatively correlated. $\endgroup$
    – Alex
    Jun 20, 2020 at 14:25
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    $\begingroup$ That's just one way of saying the result must depend on joint moments. For an approximation, what do you have in mind? What form would an approximation take and what information would it use? $\endgroup$
    – whuber
    Jun 20, 2020 at 16:32
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    $\begingroup$ no, you'll need a joint moment at some point $\endgroup$
    – Aksakal
    Jun 20, 2020 at 17:51
  • $\begingroup$ @whuber You're fully right. Obviously. I see your point now. Equality requires the joint distribution. I still wonder whether any combination of individual moments of $X$ and $Y$ could yield a quantity approximating $Cov(XY,X)$? $\endgroup$
    – Alex
    Jun 20, 2020 at 22:47

1 Answer 1

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If you calculate $E(Y|X)=g(X)$ (Since E(Y|X) is a function of $X$) so \begin{align} Cov(XY,X)&=E(X^2Y)-E(XY)E(X) \\ &= E\color{blue}(E(X^2Y\mid X)\color{blue})-E\color{blue}(E(XY|X) \color{blue})E(X) \\ &=E\color{blue}(X^2E(Y\mid X)\color{blue})-E\color{blue}(XE(Y|X)\color{blue})E(X) \\ &= E(X^2g(X))-E(Xg(X))E(X) \end{align}

For example consider $$(X,Y)\sim Normal(\mu_x,\mu_y,\sigma_x,\sigma_y,\rho)$$

$E(Y|X)=\mu_y+\rho \frac{\sigma_y}{\sigma_x}(X-\mu_x)$

so \begin{align} Cov(XY,X)&= E(X^2g(X))-E(Xg(X))E(X) \\ &= E(X^2\color{red}(\color{blue}{\mu_y+\rho \frac{\sigma_y}{\sigma_x}(X-\mu_x)}\color{red})) -E(X\color{red}(\color{blue}{\mu_y+\rho \frac{\sigma_y}{\sigma_x}(X-\mu_x)}\color{red}))E(X) \end{align} that can be calculated.

Consider situation that $(X,Y)$ has a complicated distribution but you can calculate $E(Y|X)=g(X)$. if you simulate $x_1,\cdots ,x_N $ from $X$ distribution, by Monte Carlo methods you can simply approximate $$Cov(XY,X)=E(X^2g(X))-E(Xg(X))E(X)$$ by $$\frac{1}{N} \sum_{i=1}^{N} x_i^2 g(x_i)-\color{blue}( \frac{1}{N} \sum_{i=1}^{N} x_i g(x_i) \color{blue}) \color{green}( \frac{1}{N} \sum_{i=1}^{N} x_i \color{green})$$

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