[proofplan] The proof is an algebraic expansion of the definition of the debiased estimator. We substitute the fixed-design model $Y=X\beta^*+\varepsilon$ into the residual $Y-X\hat\beta$, isolate the stochastic term involving $X^\top\varepsilon/n$, and collect the remaining deterministic estimation-error terms. The coordinate identity follows by multiplying the vector identity on the left by the standard basis row vector $e_j^\top$. [/proofplan]

[step:Decompose the residual using the fixed-design model]From the model equation $Y=X\beta^*+\varepsilon$, the residual vector $Y-X\hat\beta \in \mathbb{R}^n$ satisfies \begin{align*} Y-X\hat\beta = X\beta^*+\varepsilon-X\hat\beta. \end{align*} Combining the two terms involving $X$ gives \begin{align*} Y-X\hat\beta = \varepsilon-X(\hat\beta-\beta^*). \end{align*}[/step]

[guided]The residual in the debiased estimator is $Y-X\hat\beta$, and the goal is to express it in terms of the true noise vector $\varepsilon$ and the estimation error $\hat\beta-\beta^*$. Substituting the model identity $Y=X\beta^*+\varepsilon$ gives \begin{align*} Y-X\hat\beta = X\beta^*+\varepsilon-X\hat\beta. \end{align*} The two terms involving $X$ can be grouped because $X$ is a [linear map](/page/Linear%20Map) from $\mathbb{R}^p$ to $\mathbb{R}^n$ represented by the matrix $X \in \mathbb{R}^{n \times p}$. Thus \begin{align*} X\beta^*-X\hat\beta = -X(\hat\beta-\beta^*), \end{align*} and therefore \begin{align*} Y-X\hat\beta = \varepsilon-X(\hat\beta-\beta^*). \end{align*} This is the only place where the model equation is used.[/guided]

[step:Expand the debiased estimator and collect the estimation error] Using the definition of $\hat b$ and the residual decomposition above, \begin{align*} \hat b-\beta^* = \hat\beta-\beta^*+\hat\Theta\frac{X^\top(Y-X\hat\beta)}{n}. \end{align*} Substituting $Y-X\hat\beta=\varepsilon-X(\hat\beta-\beta^*)$ into this expression gives \begin{align*} \hat b-\beta^* = \hat\beta-\beta^*+\hat\Theta\frac{X^\top\varepsilon}{n} -\hat\Theta\frac{X^\top X(\hat\beta-\beta^*)}{n}. \end{align*} Since $\hat\Sigma=X^\top X/n$, this becomes \begin{align*} \hat b-\beta^* = \hat\beta-\beta^*+\hat\Theta\frac{X^\top\varepsilon}{n} -\hat\Theta\hat\Sigma(\hat\beta-\beta^*). \end{align*} Factoring the common vector $\hat\beta-\beta^*$ from the first and third terms gives \begin{align*} \hat b-\beta^* = \hat\Theta\frac{X^\top\varepsilon}{n} +\bigl(I_p-\hat\Theta\hat\Sigma\bigr)(\hat\beta-\beta^*). \end{align*} This proves the vector identity. [/step]

[step:Project the vector identity onto the $j$-th coordinate] Fix $j \in \{1,\dots,p\}$. Multiplying the vector identity on the left by $e_j^\top$ gives \begin{align*} e_j^\top(\hat b-\beta^*) = e_j^\top\hat\Theta\frac{X^\top\varepsilon}{n}+e_j^\top(I_p-\hat\Theta\hat\Sigma)(\hat\beta-\beta^*). \end{align*} By the definitions of coordinates and of $\hat\theta_j$, we have $e_j^\top(\hat b-\beta^*)=\hat b_j-\beta_j^*$ and $e_j^\top\hat\Theta=\hat\theta_j^\top$. Hence \begin{align*} \hat b_j-\beta_j^* = \hat\theta_j^\top\frac{X^\top\varepsilon}{n}+e_j^\top(I_p-\hat\Theta\hat\Sigma)(\hat\beta-\beta^*). \end{align*} This is the claimed coordinate decomposition. [/step]