Simple linear regressions among three pairs of variables Let the "ordinary-least-squares regression of $Y$ on $X$" be given by
$$\hat{y}_i = \hat{\beta}_0 + \hat{\beta}_1 x_i\text{.}$$
Suppose I run the following:

*

*The OLS regression of $Y$ on $X$

*The OLS regression of $Y$ on $Z$

*The OLS regression of $X$ on $Z$
and in all three cases, their slope coefficients $\hat{\beta}_1 = 2$.

I am interested in the slope coefficient of $Y$ on $X + Z$. How does this compare to the value $2$ (i.e., less than, equal to, greater than, or impossible to know)?

Attempt. Let $\hat{\beta}_{Y, X}$ be $\hat{\beta}_1$ in the case of the OLS regression of $Y$ on $X$, and similarly for the other three cases. Then we know that
$$\hat{\beta}_{Y, X} = \dfrac{\sum(x_i - \bar{x})(y_i - \bar{y})}{\sum(x_i - \bar{x})^2}$$
hence
$$\dfrac{\sum_{i=1}^{n}(x_i - \bar{x})(y_i - \bar{y})}{\sum_{i=1}^{n}(x_i - \bar{x})^2} = \dfrac{\sum_{i=1}^{n}(z_i - \bar{z})(y_i - \bar{y})}{\sum_{i=1}^{n}(z_i - \bar{z})^2} = \dfrac{\sum_{i=1}^{n}(x_i - \bar{x})(z_i - \bar{z})}{\sum_{i=1}^{n}(z_i - \bar{z})^2} = 2\text{.}$$
We also know that regression coefficients are unaffected by centering, so without loss of generality, I assume that all variables are centered and that $\sum_{i=1}^{n}(x_i - \bar{x})^2 = \sum_{i=1}^{n}(z_i - \bar{z})^2 = 1$, leading to
$$\sum_{i=1}^{n}x_iy_i = \sum_{i=1}^{n}x_iz_i = 2\text{.}$$
Thus, by linearity of the arithmetic mean, we have
$$\hat{\beta}_{Y, X+Z} = \dfrac{\sum_{i=1}^{n}y_i(x_i + z_i)}{\sum_{i=1}^{n}(x_i + z_i)^2} = \dfrac{4}{\sum_{i=1}^{n}(x_i^2 + 2x_iz_i + z_i^2)}\text{.}$$
Here's where I'm stuck. Can we do something clever with the above quantity?
I am told the answer is that $\hat{\beta}_{Y, X+Z}< 2$.

Note: OP states there is likely an error in the original question. My answer assumes everything OP states is correct thus the following answer may be incorrect
Let's focus on the denominator of the final expression. Let
$$A = \sum_{i = 1}^n x_i^2 + 2x_i z_i + z_i^2$$
If we can show that $A > 2$ then we know that $\hat{\beta}_{Y, X+Z} = \frac{4}{A} < 2$.
Now you've assumed that $\sum_{i=1}^n x_i z_i = 2$. We also know that each $x_i^2$, $z_i^2$ $\geq 0$.
Therefore:
\begin{align}
A &= \sum_{i = 1}^n x_i^2 + 2x_i z_i + z_i^2  \\
  &= \sum_{i = 1}^n x_i^2 + 2\sum_{i = 1}^nx_i z_i + \sum_{i = 1}^n z_i^2   \\
  &\geq \sum_{i = 1}^n 0 + 2 \times 2 + \sum_{i = 1}^n 0
\end{align}
(because $x_i^2, z_i^2 \geq 0$ and $\sum x_i z_i = 2$)
$$ \geq 4$$
Now since $A > 4$ (and specifically, $A > 2$) we can conclude that $\hat{\beta}_{X, Y+X} < 2$
