0
$\begingroup$

Suppose that for $1\le i \le N$ $$\begin{align} Y_i^j &= f(X_i) + \epsilon_y \qquad &1 \le j \le R_y^i \\ Z_i^j &= af(X_i)+b + \epsilon_z \qquad &1 \le j \le R_z^i \end{align}$$ where $X_i$ is an unordered categorical variable. Observations $Y_i^j$ and $Z_i^j$ are not paired, and $R_y^i \ne R_z^i$ are the number of repetitions for $i$. We don't know anything about $f$, nor do we care. $\epsilon_y$ and $\epsilon_z$ can be assumed to be normal variates with different standard deviations.

We can compute the means and standard deviations at fixed $X_i$ $$\begin{equation} \bar{Y}_i \equiv \frac{1}{R_y^i} \sum_{j=1}^{R_y^i} Y_i^j \qquad \sigma^2_{y,i} \equiv \frac{1}{R_y^i-1} \sum_{j=1}^{R_y^i} (Y_i^j-\bar{Y}_i)^2 \end{equation}$$ (similarly for $Z$) and the results can be visualized by plotting $\bar{Z}_i$ as a function of $\bar{Y}_i$, with accompanying error bars on both axes.

What is a clean way to compute the expected values for $a$ and $b$, and a measure of association between $Z$ and $Y$ ?

In a brute force approach, I would randomly and repeatedly pair up $Y_i$s and $Z_i$s, do a linear regression, and report mean and standard deviation of the estimates of $a$, $b$ and $R^2$. But I would like to know the correct approach for this.

$\endgroup$
0
$\begingroup$

I would average observations over $j$: $$\begin{align} \bar{Y}_i &= f(X_i) + \bar{\epsilon}_y \\ \bar{Z}_i &= af(X_i)+b + \bar{\epsilon}_z \end{align}$$

And solve linear regression: $$\bar{Z}_i = a\bar{Y}_i +b + (\bar{\epsilon}_z - a\bar{\epsilon}_y) $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.