How to Compute the Mean of a Non-linear Estimator?

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The discussion focuses on computing the expected value E{θ̂(N)} for a non-linear estimator defined as the minimum of N independent samples from a distribution f_X(x;θ). The initial approach involved attempting to apply the mean-value definition through integration. A suggested solution was to derive the probability density function (pdf) of the minimum and then compute the expected value based on that pdf. The original poster successfully implemented this suggestion to solve the problem. This highlights the importance of understanding the distribution of the estimator for accurate calculations.
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Hi,
let's suppose we are given N statistically independent samples x_1,\ldots,x_n from a certain distribution f_X(x;\theta) depending on a parameter \theta.
We are also given an estimator for \theta defined as follows:

\hat{\theta}}(N) = \min\{ x_i \\ : \\ i=1..N \}

How am I supposed to compute E\{ \hat{\theta}(N) \}?

I tried to apply the definition of mean-value as follows, but I can't go any further:

\int_{\mathbb{R}}\ldots\int_{\mathbb{R}} \min\{ x_1,\ldots,x_N \} \\ f_X(x_1)\ldots f_X(x_N)dx_1\ldots dx_N

Any idea?
 
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Get the pdf of the minimum, and then compute the expected value with respect to that.
 
Thanks.
I did as you suggested and succesfully solved the problem.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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