Consider the variables $x_i \text{~} \mathcal{N}(\mu, \sigma^2,a,b)$ iid with truncation points $a$ and $b$, i.e. $a < x_i < b$.

Suppose all 4 parameters, namely $\mu, \sigma, a, b$ are all unknown.

Are there unbiased estimators for $a$ and $b$?

In practice what is the best approach to obtain unbiased estimators for $a$ and $b$?

MLEs are biased

I can show that the maximum likelihood estimators are biased. In fact we can see that

$$\hat{a}_{MLE} = \min(x_i)$$ and $$\hat{b}_{MLE} = \max(x_i)$$

Clearly $\hat{a}_{MLE}$ is an over-estimate and $\hat{b}_{MLE}$ an under-estimate.

There is an unbiased estimator for $\Phi^{-1}(\frac{b- \mu}{ \sigma})$

It is well known that if $y_i \text{~} Unif(0,\lambda)$ then $$ \hat{\lambda} = \frac{n+1}{n}\max(y_i)$$ is an unbiased estimator of $\lambda$ i.e. $\mathbb{E}(\hat{\lambda}) = \lambda$.

Let $\Phi$ be the standard normal CDF, using the above result and the fact that $$\Phi^{-1}\left(\frac{x_i-\mu}{\sigma}\right) \text{~} Unif\left(0,\Phi^{-1}(\frac{b- \mu}{ \sigma})\right)$$ we get $$\frac{n+1}{n}\max\left(\Phi^{-1}\left(\frac{x_i-\mu}{\sigma}\right)\right)$$ is an unbiased estimator for $$\Phi^{-1}(\frac{b- \mu}{ \sigma})$$.

However clearly one can not simply solve for $b$ in the above to get an unbiased estimator for $b$. Also $\mu$ and $\sigma$ are unknown hence it complicates the estimation further in practice.

  • $\begingroup$ You should check the last part, I don't think its correct. I can see no other solution than maybe something approximate based on simulation. $\endgroup$ – kjetil b halvorsen Feb 3 '19 at 16:37

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