E[ϵi∣xi]=0  ⟹  Cov(x,ϵ)=0E[\epsilon_i | \pmb x_i] = 0 \implies Cov(\pmb x, \epsilon) = \pmb 0

Claim

E[ϵi∣xi]=0  ⟹  Cov(x,ϵ)=0E[\epsilon_i | \pmb x_i] = 0 \implies Cov(\pmb x, \epsilon) = \pmb 0

Proof

Recall,

Cov(xi,ϵi)=E[(xi−E[xi])(ϵi−E[ϵi])]=E[xiϵi]−E[xi]E[ϵi]\begin{align} Cov(\pmb x_i, \epsilon_i) &= E[(\pmb x_i - E[\pmb x_i])(\epsilon_i - E[\epsilon_i])] \\&= E[\pmb x_i \epsilon_i] - E[\pmb x_i]E[\epsilon_i] \end{align}

Furthermore, by the law of iterated expectation we have,

Cov(xi,ϵi)=E[xiϵi]−E[xi]E[ϵi]=E(E[xiϵi∣xi])−E[xi]E(E[ϵi∣xi])=E(xiE[ϵi∣xi])−E[xi]E(E[ϵi∣xi])=E[xi⋅0]−E[xi]⋅0=0\begin{align} Cov(\pmb x_i, \epsilon_i) &= E[\pmb x_i \epsilon_i] - E[\pmb x_i]E[\epsilon_i] \\ &= E\big(E[\pmb x_i \epsilon_i| \pmb x_i]\big) - E[\pmb x_i]E\big(E[\epsilon_i| \pmb x_i] \big)\\ &= E\big(\pmb x_i E[ \epsilon_i| \pmb x_i]\big) - E[\pmb x_i]E\big(E[\epsilon_i| \pmb x_i] \big) \\ &= E[\pmb x_i \cdot 0] - E[\pmb x_i]\cdot 0 \\ &= 0 \end{align}