Suppose that we wish to estimate $T(\theta_1,\theta_2)$, a continuous function of several parameters. Suppose that we know the asymptotic distribution when $\theta_1$ is replaced by an estimator $\hat{\theta}_1$, i.e.,

$\sqrt{n}\left(T(\hat{\theta}_1,\theta_2) - T(\theta_1,\theta_2)\right) \overset{d}{\rightarrow} N(0, \Sigma)$,

Where $\Sigma$ denotes the asymptotic variance. My question is then, if we replace $\theta_2$ with a consistent estimator $\hat{\theta}_2$ (i.e., $\hat{\theta}_2 \overset{p}{\rightarrow} \theta_2$), then does $T(\hat{\theta}_1,\hat{\theta}_2)$ have the same asymptotic distribution as $T(\hat{\theta}_1,\theta_2)$?

I would expect $\Sigma$ to increase but I cannot find a clear argument as to why or why not this would be the case in the literature. For example, Theorem 3.1 in this paper, where they give the asymptotic normality for an estimator $\hat{\gamma}_{k,n}$. Just after the theorem on page 10, they state

Note that the asymptotic distribution of $\hat{\rho}_{k,n}$ is unimportant; the only requirement is that the estimator is consistent for $\rho$.

But I don't understand why this is justified.


1 Answer 1


Take a mean-value expansion around $(\theta_1, \theta_2)$ of $T(\hat \theta_1, \theta_2)$:

\begin{align} T(\hat \theta_1, \theta_2) = T(\theta_1, \theta_2) + \frac{\partial T(\bar \theta_1, \bar \theta_2)}{\partial \theta_1}\cdot (\hat \theta_1-\theta_1) \\ + \frac{\partial T(\tilde \theta_1, \tilde \theta_2)}{\partial \theta_2}\cdot (\theta_2-\theta_2) \end{align}

$$\implies T(\hat \theta_1, \theta_2) - T(\theta_1, \theta_2) = \frac{\partial T(\bar \theta_1, \bar \theta_2)}{\partial \theta_1}\cdot (\hat \theta_1-\theta_1),$$

and consquently its variance. Now do the same for $T(\hat \theta_1, \hat \theta_2)$:

\begin{align} T(\hat \theta_1, \hat \theta_2) = T(\theta_1, \theta_2) + \frac{\partial T(\bar \theta_1, \bar \theta_2)}{\partial \theta_1}\cdot (\hat \theta_1-\theta_1) \\ + \frac{\partial T(\tilde \theta_1, \tilde \theta_2)}{\partial \theta_2}\cdot (\hat \theta_2-\theta_2) \end{align}

$$\implies T(\hat \theta_1, \hat \theta_2) - T(\theta_1, \theta_2) = \frac{\partial T(\bar \theta_1, \bar \theta_2)}{\partial \theta_1}\cdot (\hat \theta_1-\theta_1) + \frac{\partial T(\tilde \theta_1, \tilde \theta_2)}{\partial \theta_2}\cdot (\hat \theta_2-\theta_2),$$

and consequently its variance. it is not necessarily the case that this variance is greater than the previous one, because the two estimators may have high negative covariance.

In the paper you mention, likely Slutsky's Theorem is at play and allows disregarding the variance of the $\rho$ estimator -but this requires careful checking because sometimes it is easy to be misled and think that Slutsky' theorem applies in cases where it doesn't.


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