All random variables below are square-integrable and centered: 𝔼[x]=𝔼[z]=𝔼[y]=0

All random variables below are square-integrable and centered: $\mathbb{E}[x]=\mathbb{E}[z]=\mathbb{E}[y]=0$. The structural equation is \[ y=\beta_1 x+\beta_2 z+u, \qquad \mathbb{E}[u\mid x,z]=0. \] A researcher regresses $y$ on $x$ alone and forms the population (and, by the WLLN, probability-limit) slope \[ \mathrm{plim}\,\hat\beta_1 =\frac{\mathrm{Cov}(x,y)}{\mathrm{Var}(x)}. \] 1. Prove the omitted-variable identity \[ \mathrm{plim}\,\hat\beta_1 =\beta_1+\beta_2\frac{\mathrm{Cov}(x,z)}{\mathrm{Var}(x)}. \] 2. Specialize to $\beta_1=2$, $\beta_2=3$, $\mathrm{Var}(x)=4$, and $\mathrm{Cov}(x,z)=2$. Prove \[ \mathrm{plim}\,\hat\beta_1 =\frac72. \] 3. The same covariance is realized by the linear projection $z=\tfrac12 x+v$ with $\mathrm{Cov}(x,v)=0$. Prove that this representation reproduces $\mathrm{Cov}(x,z)=2$ and the same probability limit $7/2$. 4. Prove that the bias term vanishes if $\beta_2=0$ or $\mathrm{Cov}(x,z)=0$, and that the sign of the bias equals the sign of the product $\beta_2\,\mathrm{Cov}(x,z)$ (for $\mathrm{Var}(x)>0$). 5. Audit both sentences: (i) “the probability limit is $\beta_1+\beta_2\mathrm{Var}(x)/\mathrm{Cov}(x,z)$, or $\beta_1+\beta_2\mathrm{Cov}(x,z)$, giving $14$ or $8$”; (ii) “bias remains when $\beta_2=0$, or the sign follows $\beta_1$ alone, or $z=\tfrac12 x+v$ changes the limit, or this is a WLS / ridge exercise.” Stay inside the centered linear structural equation with $\mathbb{E}[u\mid x,z]=0$. Do not introduce an intercept estimation problem, and do not replace the omitted-variable identity by a ridge shrinkage.

$\mathrm{Cov}(x,y)=\beta_1\mathrm{Var}(x)+\beta_2\mathrm{Cov}(x,z)$ because $\mathrm{Cov}(x,u)=0$. Divide by $\mathrm{Var}(x)$.