# Tag Info

### Why are we using a biased and misleading standard deviation formula for $\sigma$ of a normal distribution?

For the more restricted question Why is a biased standard deviation formula typically used? the simple answer Because the associated variance estimator is unbiased. There is no real ...
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• 616
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### UMVUE for $g(\theta)=\theta^2$ of Poisson random variables

Yes, that's correct. To make it rigorous, you can apply Lehmann–Scheffé Theorem with $Y = T(X)$ and $\varphi(Y) = \frac{1}{n^2}(Y^2 - Y)$. After a closer look at the Wikipedia link above, it seems ...
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### Find the joint distribution of $X_1$ and $\sum_{i=1}^n X_i$

Correct me if I am wrong, but I don't think one needs to find the conditional distribution to find the conditional expectation for the UMVUE. We can find the conditional mean using well-known ...
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### Unbiased Estimator for the CDF of a Normal Distribution

As a comment suggested, an unbiased estimator is (one minus) the empirical distribution function $$\hat P(X_1 > 0) = 1-\hat F_X(0) = 1-\frac 1n \sum_{i=1}^n I\{x_i \leq 0\}$$ where $I\{\}$ is the ...
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### UMVUE of $e^{-\lambda}$ from poisson distribution

Yes, Lehman Scheffe gives you the sufficiency of $\bar{X}$ for $e^{-\lambda}$. The problem is that $\bar{Y}$ does not actually condition on the sufficient statistic. Although the $Y$ are a function of ...
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### UMVU estimator for non-linear transformation of a parameter

Your final answer is not quite right. The conclusion due to the sample mean $\bar X$ being only sufficient for $\mu$ also looks faulty. Recall that $T(X_1,X_2,\cdots,X_n)=\sum_{i=1}^n X_i$ is a ...
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### Find UMVUE of $\theta$ where $f_X(x\mid\theta) =\theta(1 +x)^{−(1+\theta)}I_{(0,\infty)}(x)$

From your previous question, you already have the complete sufficient statistic: $$T(\mathbf{X}) = \sum_{i=1}^n \ln(1+X_i).$$ The simplest way to find the UMVUE estimator for $\theta$ is to appeal ...
• 130k
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• 6,355