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enter image description hereHow do you determine the derivative of an average to use the delta method to approximate the variance of an estimator?

For example for the problem above, I came across the estimators $\hat{\theta}=\frac{1}{\bar{X}}-1$ and $\hat{\theta}=-\frac{1}{\overline{\ln(1-X)}}$ using method of moments and maximum likelihood and I am supposed to determine the variance using the delta method. I know the formula is $\sigma_{\bar{\theta}}\approx |\hat{\theta}'(\mu_\theta)|\frac{\sigma_\theta}{\sqrt{n}}$ how would I do this, replace $x$ with $\mu_\theta$?

ernesto
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    The estimators are incorrect. I got $$\hat{\theta}{\text{MLE}} = \dfrac{n}{\sum{i=1}^{n}\ln(1-X_i)}$$ and $$\hat\theta_{\text{MOM}} = \dfrac{1}{\overline{X}}-1\text{.}$$ – Clarinetist Jun 25 '21 at 23:38
  • @Clarinetist After that how would you use the delta method? – ernesto Jun 25 '21 at 23:39
  • I think I might have messed up on $\hat{\theta}{\text{MLE}}$ - check my work there. Anyway, I don't have time to sketch this out completely, but the idea is to use the CLT to gather that $$\sqrt{n}\left(\overline{X}_n - \dfrac{1}{1+\theta}\right) \to \mathcal{N}\left(0, \dfrac{\theta}{\theta^2(\theta + 2)}\right)$$ in distribution. Then, by the Delta method, set $g(\theta) = \dfrac{1}{\theta} - 1$ and that's what you have for $\hat\theta{\text{MOM}}$. Recall that the Delta Method then tells you that $$\sqrt{n}[g(\overline{X}_n) - g(\mu)] \to N(0, \sigma^2[g^{\prime}(\theta)]^2)$$ – Clarinetist Jun 26 '21 at 00:02
  • ... where $\mu$ is the mean of the Beta distribution and $\sigma^2$ is the variance of the Beta distribution. You then are looking for $\sigma^2[g^{\prime}(\theta)]^2$ square rooted to yield the standard deviation. I don't have time to figure out where I messed up with the MLE one, but hopefully that gives you a start. – Clarinetist Jun 26 '21 at 00:02

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The MLE of $\theta$ is actually $\hat{\theta}_{MLE}=-\frac{n}{\ln(1-X_1)+\dots +\ln(1-X_n)}$. It's easy to show that $-\ln(1-X_j)\sim \text{Exponential}(\theta)$ for each $1\leq j \leq n$. So if we take $$\overline{Y}=-\frac{\ln(1-X_1)+\dots +\ln(1-X_n)}{n}$$ Then $\overline{Y}$ is asymptotically $\mathcal{N}\Big(\frac{1}{\theta},\frac{1}{n\theta^2}\Big)$ by CLT i.e. $$\sqrt{n}\Bigg(\overline{Y}-\frac{1}{\theta}\Bigg)\longrightarrow \mathcal{N}\Bigg(0,\frac{1}{\theta^2}\Bigg)$$ Taking $g(x)=1/x$ in the statement of the delta method yields $$\sqrt{n}\Bigg(\hat{\theta}_{MLE}-g(1/\theta)\Bigg)\longrightarrow \mathcal{N}\Big(0,\theta^2\Big)$$ So the estimated standard deviation of $\hat{\theta}_{MLE}$ is $\frac{\theta}{\sqrt{n}}$.

Matthew H.
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