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Let $X_i$ be i.i.d. with density $p(x) = ce^{-x^4/12}$, $c$ is a normalizing constant.
Consider the location model $p(x-\theta)$. Aim is to compute the limiting distribution of the MLE.

If I have done it right MLE of $\theta$, say $\hat{\theta}$, is the solution of $$m_3 - 3\theta m_2 + 3\theta^2 m_1 - \theta^3 = 0$$ where $m_j = \frac{1}{n}\sum x_i^j$.
I am not sure if it is possible to write the solution explicitly (I mean, in a reasonably simple form).
Is it possible to determine the limiting distribution of $\hat{\theta}$?

To clarify notation, above equation is $\partial_\theta \log \left[c^n\exp\left(\frac{-1}{12}\sum_i^n (x_i - \theta)^4\right)\right] = 0$.

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  • $\begingroup$ It's not clear what facts we can assume you will be able to use; there's certainly results relating to asymptotic behavior of MLEs that should apply. Without invoking those results, I suppose one possibility is to write $x_i-\theta$ as $(x_i-\bar{x})+(\bar{x}-\theta)$, and expand in central rather than raw moments, where I believe you will obtain a depressed cubic from which to solve for $\bar{x}-\theta$ and can immediately apply the Cardano formula, resulting in a somewhat complicated function of the central sample moments. $\endgroup$
    – Glen_b
    Commented Jun 17, 2019 at 7:43
  • $\begingroup$ Getting from there to something useful would involve multiple stages of argument, though so I doubt that this is a productive line of approach. $\endgroup$
    – Glen_b
    Commented Jun 17, 2019 at 7:45
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    $\begingroup$ Expanding the MLE as you have done unfortunately obscures the essence of the problem. Instead, leave it unexpanded and consider the second derivative of the log likelihood: that should suffice to demonstrate the MLE is unique, which is the principal obstacle to finding the limiting distribution of a suitably standardized version of the estimator. $\endgroup$
    – whuber ♦
    Commented Jun 17, 2019 at 17:11
  • $\begingroup$ This does not address the main question as to the limiting distribution but the exact form of the maximum likelihood estimator for $\theta$ is $\frac{\sqrt[3]{2} \left(m_2-m_1^2\right)}{\sqrt[3]{a}}-\frac{\sqrt[3]{a}}{\sqrt[3]{2}}+m_1$ where $a=-2 m_1^3+3 m_2 m_1-m_3+\sqrt{4 m_3 m_1^3-3 m_2^2 m_1^2-6 m_2 m_3 m_1+4 m_2^3+m_3^2}$ (using Mathematica). $\endgroup$
    – JimB
    Commented Jul 1, 2019 at 13:04

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