Come si calcolano gli errori standard per una trasformazione dell'MLE?

Devo dedurre un parametro positivo . Per acomodare la positività ho ri-parametrizzato . Usando la routine MLE ho calcolato la stima puntuale e se per . La proprietà di invarianza dell'MLE mi dà direttamente una stima puntuale per , ma non sono sicuro di come calcolare se per . Grazie in anticipo per qualsiasi suggerimento o riferimento. $p$ $p=\exp(q)$ $q$ $p$ $p$

maximum-likelihood standard-error delta-method

— Marcel
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Non puoi usare la stessa routine MLE per calcolare una stima puntuale e cercare direttamente ?

p

$p$

— whuber

Risposte:

Il metodo Delta viene utilizzato per questo scopo. Secondo alcune ipotesi di regolarità standard , sappiamo che l'MLE, per è approssimativamente (cioè asintoticamente) distribuito come $\hat{\theta}$ $\theta$

\hat{θ} \sim N (θ, I^{- 1} (θ))

$\hat{\theta} \sim N(\theta, \mathcal{I}^{-1}(\theta))$

where $\mathcal{I}^{-1}(\theta)$ is the inverse of the Fisher information for the entire sample, evaluated at $\theta$ and $N(\mu,\sigma^{2})$ denotes the normal distribution with mean $\mu$ and variance $\sigma^{2}$ . The functional invariance of the MLE says that the MLE of $g(\theta)$ , where $g$ is some known function, is $g(\hat{\theta})$ (as you pointed out) and has approximate distribution

g (\hat{θ}) \sim N (g (θ), I^{- 1} (θ) [g^{'} (θ)]^{2})

$g(\hat{\theta}) \sim N( g(\theta), \mathcal{I}^{-1}(\theta) [g'(\theta)]^{2} )$

where you can plug in consistent estimators for the unknown quantities (i.e. plug in $\hat{\theta}$ where $\theta$ appears in the variance). I would assume the standard errors you have are based on the Fisher information (since you have MLEs). Denote that standard error by $s$ . Then the standard error of $e^{\hat{\theta} }$ , as in your example, is

\sqrt{s^{2} e^{2 \hat{θ}}}

$\sqrt{s^{2}e^{2 \hat{\theta}}}$

I may be interpreting you backwards and in reality you have the variance of the MLE of $\theta$ and want the variance of the MLE of $\log(\theta)$ in which case the standard would be

\sqrt{s^{2} / {\hat{θ}}^{2}}

$\sqrt{ s^{2}/\hat{\theta}^{2} }$

— Macro
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Just a side note: there are also appropriate multivariate extensions whereby the derivatives are replaced by gradients, and the multiplications have to be matrix multiplications, so there's a bit more headache in figuring out where the transpose goes.

— StasK

Thanks for pointing that out StasK. I believe in the multivariate case the asymptotic covariance of

g (\hat{θ})

$g(\hat{\theta})$ is

\nabla g (θ)^{'} I (θ)^{- 1} \nabla g (θ)

$\nabla g(\theta)' \mathcal{I}(\theta)^{-1} \nabla g(\theta)$

— Macro

(+1) I added a link to the regularity assumptions (and some other things) since it isn't clear whether these are satisfied in the OP's problem. I might have said that

\hat{θ}

$\hat{\theta}$ is asymptotically normal and not approximately normal, since the convergence rates can be slow at times.

— MånsT

Thank you @MånsT, I also did clarify that I meant asymptotically when I said approximately :)

— Macro

Macro gave the correct answer on how to transform standard errors via the delta method. Though the OP specifically asked for the standard errors, I suspect that the objective is to produce confidence intervals for $p$ . Besides computing estimated standard errors of $\hat{p}$ you can directly transform a confidence interval, $[q_1, q_2]$ , in the $q$ -parametrization to a confidence interval $[\exp(q_1), \exp(q_2)]$ in the $p$ -parametrization. This is perfectly valid, and it may even be a better idea depending on how well the normal approximation used to justify a confidence interval based on standard errors works in the $q$ -parametrization versus the $p$ -parametrization. Moreover, the directly transformed confidence interval will fulfill the positivity constraint.

— NRH
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