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Loss Function Asymptotic Distribution

Discussion in 'Education' started by user191919, Oct 8, 2018.

  1. user191919

    user191919 Guest

    I am trying to find the asymptotic distribution of a maximum estimator loss function of the type:

    $$ \hat{\theta} = \arg \max_\tilde{\theta} M_n(\tilde{\theta},x) $$

    $$ \theta = \arg \max_\tilde{\theta} M(\tilde{\theta},x) $$

    Having settled that $\sqrt{n} (\hat{\theta} - \theta) \to N(0,V)$ where $\theta$ is the true value, my approach to determine the asymptotic distribution of the loss function is a simple taylor expansion of $M$ around $\theta$ evaluated at $\hat{\theta}$:

    $$ n ( M(\hat{\theta},x) - M(\theta,x) ) = \sqrt{n} \nabla_\theta M(\theta,x) \sqrt{n} (\hat{\theta} - \theta) + n R(\theta,\hat{\theta},x) $$

    Here I have trouble since $\nabla_\theta M(\theta,x) = 0$ since its an interior maximizer.

    I suppose an alternative would be using a taylor expansion around $\hat{\theta}$ instead of $\theta$, which would yield:

    $$ n ( M(\hat{\theta},x) - M(\theta,x) ) = \sqrt{n} \nabla_\theta M(\hat{\theta},x) \sqrt{n} (\hat{\theta} - \theta) - n R(\theta,\hat{\theta},x) $$

    which I suppose under suitable assumptions would provide that $\sqrt{n} \nabla_\theta M(\hat{\theta},x) \to N(0,V_M)$. But then again, I now have a product of normals. Also, what would be an appropriate remainder formulation to show $R \to 0$?

    Is this a productive path? Any hints or references are greatly appreciated. I am having difficulty in finding the latter.

    Many thanks.

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