When are the asymptotic variance of OLS and 2SLS equal? - Cross Validated most recent 30 from stats.stackexchange.com 2019-10-15T04:25:42Z https://stats.stackexchange.com/feeds/question/45520 https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/45520 4 When are the asymptotic variance of OLS and 2SLS equal? Druss2k https://stats.stackexchange.com/users/13636 2012-12-09T20:17:03Z 2019-03-20T10:55:53Z <p>Assume the model <span class="math-container">$\ y = X\beta + u \$</span> with <span class="math-container">$\ W \$</span> is a <span class="math-container">$\ n\times l \$</span> so called matrix of instruments. </p> <p>The following assumptions hold. There is a law of large numbers (LLN) for 1.,2.,3. and 4. such that</p> <ol> <li><p><span class="math-container">$\text{plim}_{n\to\infty} \ \left(\frac{X^TX}{n}\right) = m_{X^TX}$</span></p> <p>holds where <span class="math-container">$m_{X^TX}$</span>, a finite, non stochastic matrix with full column rank, exsists.</p></li> <li><p><span class="math-container">$\text{plim}_{n\to\infty} \ \left(\frac{W^TW}{n}\right) = \text{lim}_{n\to\infty} \left(\frac{{\mathbb E}\left(W^TW\right)}{n}\right)$</span></p> <p>where we assume, that the RHS exists and is finite as well as positive definite.</p></li> <li><p><span class="math-container">$\text{plim}_{n\to\infty} \ \left(\frac{W^TX}{n}\right)$</span></p> <p>where we assume, that the limit <span class="math-container">$W^TX$</span> exists, is finite and <span class="math-container">$W^TX$</span> has full column rank i.e. <span class="math-container">$\text{rk}\left(W^TX\right)=k$</span>.</p></li> <li><p><span class="math-container">$\text{plim}_{n\to\infty} \ \left(\frac{W^Tu}{n}\right) = \text{lim}_{n\to\infty} \left(\frac{{\mathbb E}\left(W^Tu\right)}{n}\right) = 0$</span></p> <p>where we assume that <span class="math-container">$\text{lim}_{n\to\infty} \left(\frac{{\mathbb E}\left(W^Tu\right)}{n}\right)$</span> is equal to <span class="math-container">$0$</span>.</p></li> </ol> <p>The asymptotic variance for OLS will, under the assumption of homoscedastic errors <span class="math-container">$u$</span> i.e. that <span class="math-container">${\mathbb E}\left(uu^T|X\right) = \sigma^2_0 I_n$</span> and that the model is actually correctly specified , be</p> <p><span class="math-container">$\text{plim}_{n\to\infty} \text{Var}\left[ n^{\frac{1}{2}}\left(\boldsymbol{\widehat{\beta}}_{\text{KQ}} - \boldsymbol{\beta_0}\right)\big| X\right] = \sigma_0^2 \ m_{X^TX}^{-1}$</span></p> <p>The corresponding asymptotic variance for the 2SLS case will look like</p> <p><span class="math-container">$\text{plim}_{n\to\infty} \text{Var}\left[ n^{\frac{1}{2}}\left(\boldsymbol{\widetilde{\beta}_{\text{2SLS}}} - \boldsymbol{\beta_0}\right)\big| X\right] = \sigma_0^2 \ \text{plim}_{n\to\infty} \left(\frac{X^TP_W X}{n}\right)^{-1}$</span></p> <p>If I now consider the precision matrix rather then the variance matrix and take their difference I'll get </p> <p><span class="math-container">$\text{plim}_{n\to\infty} \ \left(\frac{X^TM_WX}{n}\right) = \left(\text{plim}_{n\to\infty} \ \frac{X^TX}{n}\right) - \left(\text{plim}_{n\to\infty} \ \frac{X^TP_WX}{n}\right)$</span></p> <p>which is a positive semidefinite matrix.</p> <p><strong>My question: When are both asymptotic variances of OLS and 2SLS equal</strong></p> <p>My guess is that we need to set <span class="math-container">$X^TX = X^T P_W X$</span>. The only possibility for such a thing is when <span class="math-container">$P_WX=X$</span>. So all the columns of <span class="math-container">$X$</span> must be in the image of <span class="math-container">$P_W$</span>. This means that all the columns of <span class="math-container">$X$</span> are actually viable instruments. By that I conclude that there are (at least asymptotically) no endogenous regressor within <span class="math-container">$X$</span> if their asymptotic variances are equal. But this seems to good to be true. There are downfalls of using a 2SLS-approach if theres no endogeneity involved. Again, this would mean that if the variances of the 2SLS and OLS estimator are equal asymptotically it does not matter if we choose 2SLS or OLS since both estimators are consistent for <span class="math-container">$\beta$</span>.</p> https://stats.stackexchange.com/questions/45520/when-are-the-asymptotic-variance-of-ols-and-2sls-equal/172537#172537 2 Answer by Alecos Papadopoulos for When are the asymptotic variance of OLS and 2SLS equal? Alecos Papadopoulos https://stats.stackexchange.com/users/28746 2015-09-15T02:29:47Z 2015-09-15T02:29:47Z <p>To close this one:</p> <blockquote> <p>When are the asymptotic variances of OLS and 2SLS equal?</p> </blockquote> <p><strong>A:</strong> Only when the "matrix of instruments" essentially contains exactly the original regressors, (or when the instruments predict <em>perfectly</em> the original regressors, which amounts to the same thing), as the OP himself concluded. </p> <p>The variance of the IV estimator (generalized or not), when the OLS is also consistent, is never smaller then the variance of the OLS.</p> <p>This is one of the main reasons why the "Hausman specification test" (or the Durbin-Wu-Hausman) is performed: because if the data show that the two estimators have the same probability limits (i.e if the null of the test is not rejected), then we should stick to OLS, which has a lower variance.</p> <p>If the null is rejected (and we have other arguments that support the consistency of the IV estimator), then the OLS is deemed inconsistent, and then we go with the IV estimator, to buy consistency at the price of higher variance. </p>