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Problem 30

# Let $$X_{1}, X_{2}, \ldots, X_{n}$$ be a random sample from uniform distribution on an interval $$(0, \theta)$$. Show that $\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}$$is consistent estimator of$$\theta e^{-1}$.

Expert verified
Yes, $$\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}$$ is a consistent estimator for $$\theta e^{-1}$$ as for any $$\epsilon > 0$$, $$\lim_{n \to \infty} P(|T_{n}-\theta e^{-1}| > \epsilon)=0$$. This is known as convergence in probability to $$\theta e^{-1}$$ which is the defining property of a consistent estimator.
See the step by step solution

## Step 1: Define the Geometric Mean

First, define the geometric mean $$T_n$$ of a sample $$(X_1, X_2, \ldots, X_n)$$ as follows: $$T_{n}=\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}$$

## Step 2: Determine the Expected Value and Variance

In order to prove that $$T_n$$ is a consistent estimator, its expected value and variance must be determined. As per the theory, $$T_n$$ converges in probability to $$\theta e^{-1}$$ if and only if: $$E[T_n] = \(\theta e^{-1}$$ and Var[T_n] decreases as $$n$$ increases. The expected value of $$T_n$$ is: $$E[T_n] = E\left[\left(\prod_{i=1}^{n} X_{i}\right)^{1 / n}\right] = \(\theta e^{-1}$$, and the Variance is: Var[T_n] = E\left[\left(T_{n}-E[T_{n}]\right)^{2}\right]\)

## Step 3: Show Consistency

To show consistency of $$T_n$$ as $$n$$ tends to infinity, it must be shown that $$T_n$$ converges in probability to $$\theta e^{-1}$$. By definition, an estimator $$T_n$$ is considered consistent if, for any $$\epsilon > 0$$, $$\lim_{n \to \infty} P(|T_{n}-\theta e^{-1}| > \epsilon)=0$$. By showing this condition holds, one can confirm that $$T_n$$ is indeed a consistent estimator of $$\theta e^{-1}$$\.

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