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Let $D, X_1, X_2, Z$ be some arbitrary random variables. If it is easier to think, we can assume they are mean zero.

I'm curious if there is any relationship between the correlation $E(D Z)$ and the nonparametric partial correlation $\eta^2_{D\sim Z\mid X_1, X_2}$ between random variables $D$ and $Z$? For example, if I know $E(D Z)$, how can I use it to write $\eta^2_{D\sim Z\mid X_1, X_2}$? or vice versa. It is also clear to me that extra quantities must be identified in order to fully bridge those two types of correlations.

Let me explain the notation $\eta^2_{D\sim Z\mid X_1, X_2}$ of the nonparametric partial correlation between the variables $D$ and $Z$ given $X_1, X_2$ a little bit more here.

\begin{align*} \eta^2_{D\sim Z\mid X_1, X_2} &= Corr^2\big(\{D - E(D\mid X_1, X_2)\} ,E\left[\{D- E(D\mid X_1, X_2)\}\mid X_1, X_2, Z\right]\big) \end{align*}

Since I didn't assume $E[D\mid X_1, X_2, Z]$ to be a linear regression function, notice $\eta^2_{D\sim Z\mid X_1, X_2}$ is not the partial correlation between the variables $D$ and $Z$ given controlling variables $X_1, X_2$.

Omega
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1 Answers1

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Since $X_1$ and $X_2$ always show up together in the OP, I will just use $X$ to denote them. Also, the terms "correlation" and "covariance" are used casually in the OP. I'll specifically study "covariance" here.

The covariance part in $\eta^2$ is

$$\newcommand{\C}{\mathbb{C}\mathrm{ov}} \newcommand{\V}{\mathbb{V}\mathrm{ar}} \newcommand{\E}{\mathbb{E}} \begin{align} &\phantom{{}={}} \C\left\lbrace D-\E(D\mid X),\quad \E\left[D-\E(D\mid X) \mid X,Z \right] \right\rbrace \\ &=\C\left\lbrace D-\E(D\mid X), \quad \E(D\mid X,Z)-\E(D\mid X) \right\rbrace \\ &=\C\left\lbrace D-\E(D\mid X), \quad \E(D\mid X,Z)\right\rbrace - \\ &\qquad\qquad \underset{0}{\underbrace{ \C\left\lbrace D-\E(D\mid X), \quad \E(D\mid X,)\right\rbrace }} \\ &=\E\left \lbrace \C\left[D-\E(D\mid X),\quad \E(D\mid X,Z)\mid X\right] \right\rbrace + \\ &\qquad\qquad \C\left( \underset{0}{\underbrace{ \E\left[D-\E(D\mid X) \mid X\right] }},\quad \E\left[\E(D\mid X,Z)\mid X\right] \right) \\ &=\E\left\lbrace \C\left[D,\quad \E(D\mid X,Z)\mid X\right] - \right. \\ &\qquad\qquad \underset{0}{\underbrace{ \C\left[\E(D\mid X), \quad\E(D\mid X,Z)\mid X\right] }} \rbrace\\ &=\E\left\lbrace \C\left[D,\quad \E(D\mid X,Z)\mid X\right] \right\rbrace \tag{1}\label{1} \\ &=\E\left\lbrace \underset{0}{\underbrace{\C\left[D-\E(D\mid X,Z),\quad \E(D\mid X,Z)\mid X\right]}} + \V\left[ \E(D\mid X,Z) \mid X \right] \right\rbrace \\ &=\E\left\lbrace \V\left[ \E(D\mid X,Z) \mid X \right] \right\rbrace \\ &\geq 0. \end{align}$$

So, even without squaring, this covariance is already non-negative. Compare \eqref{1} with the (not necessarily non-negative) $$ \C(D,\ Z)=\E\left\lbrace \C(D,\quad Z\mid X)\right\rbrace + \C\left[\E(D\mid X), \quad\E(Z\mid X)\right]. $$

Apart from the extra second term in $\C(D,\ Z)$, it seems more interesting to me to compare the square of terms inside their first braces:
$$ \left\lbrace\begin{array}{ccc} \C^2 [\,D, & \E(D\mid X,Z) & {}\mid X\,] \\ \C^2 [\,D, & Z & {}\mid X\,]\\ \end{array}\right.$$ I intuitively think that the upper line will have a larger value, because $\E(D\mid X,Z)$ (as a function of $Z$) is made specifically to approximate $D$, compared with $Z$ itself.

We can also rewrite $\C(D,\ Z \mid X)$ as $$\begin{align} \C(D,\ Z \mid X) &=\E\left[\underset{0}{\underbrace{\C(D,\ Z\mid X,Z)}} \mid X\right]+\C\left[\E(D|X,Z),\ \E(Z\mid X,Z) \mid X\right] \\ &=\C\left[Z,\ \E(D|X,Z) \mid X\right]. \end{align}$$ This provides another comparison with \eqref{1}: $$\left\lbrace\begin{array}{ccc} \C [\,D, & \E(D\mid X,Z) & {}\mid X\,] \\ \C [\,Z, & \E(D\mid X,Z) & {}\mid X\,]\\ \end{array}\right.$$

Zack Fisher
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