Consider the following estimator $$\hat\beta = \left(\sum_{i=1}^{N}x_ix_i' + \lambda I_k\right)^{-1}\left(\sum_{i=1}^Nx_iy_i\right)$$ where $x_i$ is a column vector $k\times1$ from $X$ and $\lambda > 0$ is a scalar and $\mathbb{E}(x_ie_i) = 0$ .

  1. Define bias and show that $\hat\beta$ is biased
  2. Define consistency and show that $\hat \beta$ is consistent
  3. Define conditional variance of $\hat\beta$.

For number 1 I have \begin{equation} \begin{aligned} \hat\beta &= \left(\sum_{i=1}^{N}x_ix_i' + \lambda I_k\right)^{-1}\left(\sum_{i=1}^Nx_iy_i\right) \\ & = (X'X + \lambda I_k)^{-1}X'y \\ & = (X'X + \lambda I_k)^{-1}X'(X\beta + e)\\ & = (X'X + \lambda I_k)^{-1}X'X\beta \end{aligned} \end{equation} and hence it is unbiased only if $\lambda = 0$.

I am stuck on number 2. For $\hat\beta$ to be consistent I need $$(X'X + \lambda I_k)^{-1}X'X \xrightarrow{p} I$$ but how it could be the case for $\lambda > 0$?

  • $\begingroup$ Looks like ridge regression (just to note). $\endgroup$ Commented Nov 13, 2016 at 20:01
  • $\begingroup$ @tosik, the last line of the derivation is true in expectation only. $\endgroup$ Commented Dec 2, 2021 at 14:51

1 Answer 1


Note that $$ \text{plim} \Big[(X'X + \lambda I_k)^{-1} X'X\Big] =\text{plim}(n^{-1}X'X + n^{-1}\lambda I_k)^{-1}\text{plim}(n^{-1}X'X)$$

The second plim converges by asumption. For the first we have $$\text{plim}(n^{-1}X'X + n^{-1}\lambda I_k)^{-1}=\left(\text{plim}n^{-1}X'X + \text{plim}n^{-1}\lambda I_k\right)^{-1} $$

and that

$$\text{plim}n^{-1}\lambda I_k = \text{lim}n^{-1}\lambda I_k = 0$$

leading to the desired consistency result. Intuitively the purpose of adding a term like $\lambda I_k$ is to handle a "bad sample", i.e. it is a finite-sample "tactic" to get results, but whose effect is eliminated asymptotically.

  • $\begingroup$ I accepted because I understood what Alecos meant $\endgroup$
    – tosik
    Commented Nov 14, 2016 at 12:43
  • $\begingroup$ @MrFrog It is not. Notice that in the first plim expression there is a ${-1}$ outside, so essentially the two $n^{-1}$ cancel out. I just added $n^{-1}/n^{-1} =1$ and then split them. $\endgroup$ Commented Mar 25, 2022 at 20:15

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