<p>For multivariate regressions, we can utilize vector notations and write the regression equation for observation <span class="math-container">$i$</span> as <span class="math-container">$y_i = x_i' \beta + u_i$</span> where <span class="math-container">$y_i$</span> is of dimension <span class="math-container">$1 \times 1$</span>, <span class="math-container">$x_i$</span> is of dimension <span class="math-container">$K \times 1$</span>, <span class="math-container">$\beta$</span> is of dimension <span class="math-container">$K \times 1$</span>, and <span class="math-container">$e_i$</span> is of dimension <span class="math-container">$1 \times 1$</span>.</p>
<p>We can then stack the <span class="math-container">$n$</span> equations (<span class="math-container">$n$</span> is the number of observations) to obtain the matrix form <span class="math-container">$y = X \beta + e$</span>, where <span class="math-container">$Y = (y_1, y_2, ..., y_n)'$</span>, <span class="math-container">$X = (x_1', x_2', ..., x_n')'$</span>, and <span class="math-container">$U = (u_1, u_2, ..., u_n)'$</span>.</p>
<p>Now the OLS estimator expressed in matrix notation is just <span class="math-container">$\hat{\beta} = (X'X)^{-1}  X'Y$</span>. Using the fact that <span class="math-container">$Y = X \beta + U$</span>, we can rewrite that as:</p>
<p><span class="math-container">$$
(X'X)^{-1} X'(X \beta + U) =
\beta + (X'X)^{-1} X'U
$$</span></p>
<p>For consistency, by Law of Large Numbers, we have <span class="math-container">$n^{-1} X'X = n^{-1}\sum{x_i x_i'} \rightarrow E(x x')$</span> in probability, and <span class="math-container">$n^{-1} X'U = n^{-1}\sum{x_i' y_i} \rightarrow E(x u)$</span> in probability.</p>
<p>Now by Continuous Mapping Theorem, we have <span class="math-container">$\hat{\beta} = \beta + (X'X)^{-1}  X'U = \beta + (n^{-1} X'X)^{-1} n^{-1} X'U \rightarrow (E(x x'))^{-1} E(x u)$</span> in probability. As <span class="math-container">$E(xu) = 0$</span> (this is generally assumed for OLS, as otherwise you have inconsistency, as we just showed) we have that <span class="math-container">$\hat{\beta} \to \beta$</span> in probability</p>
<p>For bias, look at:
<span class="math-container">$$
E(\hat{\beta} | X)
$$</span>
and try to prove that it is equal to <span class="math-container">$\beta$</span> no matter your choice of <span class="math-container">$X$</span></p>

CC BY-SA 4.0
2022-12-04T11:48:29.770Z
Substitute $y=X\beta+e$ so that $\hat\beta = \beta + (X'X)^{-1} X'e$ and then follow @easyliving
chan1142
CC BY-SA 4.0
2022-12-04T07:00:10.210Z
2022-12-06T09:34:25.730Z
2022-12-06T09:34:25.730Z
easyliving
<p>If we have a normal OLS estimator, then the probability limit of Beta_1 hat as n -&gt; infinity equals beta_1 plus (cov(x, error) / var(x)).</p>
<p>How is this calculated for a vector of betas?</p>

<econometrics>
How is bias and consistency calculated for multiple regression?
user42504
# How is bias and consistency calculated for multiple regression?

(Asked by: username_0 on 2022-12-04)

If we have a normal OLS estimator, then the probability limit of Beta\_1 hat as n -> infinity equals beta\_1 plus (cov(x, error) / var(x)).

How is this calculated for a vector of betas?

**Tags:** econometrics

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## 1 answers

### Answer by username_1 on 2022-12-04. Score: 2

For multivariate regressions, we can utilize vector notations and write the regression equation for observation $i$ as $y\_i = x\_i' \beta + u\_i$ where $y\_i$ is of dimension $1 \times 1$, $x\_i$ is of dimension $K \times 1$, $\beta$ is of dimension $K \times 1$, and $e\_i$ is of dimension $1 \times 1$.

We can then stack the $n$ equations ($n$ is the number of observations) to obtain the matrix form $y = X \beta + e$, where $Y = (y\_1, y\_2, ..., y\_n)'$, $X = (x\_1', x\_2', ..., x\_n')'$, and $U = (u\_1, u\_2, ..., u\_n)'$.

Now the OLS estimator expressed in matrix notation is just $\hat{\beta} = (X'X)^{-1} X'Y$. Using the fact that $Y = X \beta + U$, we can rewrite that as:

$$
(X'X)^{-1} X'(X \beta + U) =
\beta + (X'X)^{-1} X'U
$$

For consistency, by Law of Large Numbers, we have $n^{-1} X'X = n^{-1}\sum{x\_i x\_i'} \rightarrow E(x x')$ in probability, and $n^{-1} X'U = n^{-1}\sum{x\_i' y\_i} \rightarrow E(x u)$ in probability.

Now by Continuous Mapping Theorem, we have $\hat{\beta} = \beta + (X'X)^{-1} X'U = \beta + (n^{-1} X'X)^{-1} n^{-1} X'U \rightarrow (E(x x'))^{-1} E(x u)$ in probability. As $E(xu) = 0$ (this is generally assumed for OLS, as otherwise you have inconsistency, as we just showed) we have that $\hat{\beta} \to \beta$ in probability

For bias, look at:
$$
E(\hat{\beta} | X)
$$
and try to prove that it is equal to $\beta$ no matter your choice of $X$

> **Comments:**
> **username_2**: Substitute $y=X\beta+e$ so that $\hat\beta = \beta + (X'X)^{-1} X'e$ and then follow @easyliving  

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