Calculate the bias of the estimator

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Main Question or Discussion Point

Suppose that Y_1,...,Y_n is a random sample where the density of each random variable Y_i is f(y) = 2*x^2*y^(-3), y >= x for some parameter x > 1. Let x^hat := min{Y_1,...,Y_n}.

I figured out that the pdf for the minimum order statistic is n*[f(y)]*[1-F(y)]^(n-1).

Also I think that 1-F(y) = Integrate[2*x^2*t^(-3), t, y, Infinity] = x^2*y^(-2)

Plugging this into the pdf for the first order statistic, we have n*[2*x^2*y^(-3)]*[x^2*y^(-2)]^(n-1).

Now to find the bias we have that B(x^hat) = E(x^hat) - x

So I think E(x^hat) = Integrate[y*n*[2*x^2*y^(-3)]*[x^2*y^(-2)]^(n-1), y, x, Infinity].

This is where I am running into problems because I am finding it very difficult to get a "nice" integral here, in fact, using the limits of integration I have listed, I'm getting an indeterminant form so I'm guessing that this might be the problem. Can anyone point me in the right direction as to what I'm doing wrong? Thanks.
 

Answers and Replies

  • #2
Stephen Tashi
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Are you asking about the integral

[tex] \int_x^\infty {y n 2 x^2 y^{-3} ( x^2 y^{-2})^{n-1} dy [/tex]

[tex] = 2 n x^{2n} \int_x^\infty y^{-2n} dy [/tex]
 
  • #3
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Are you asking about the integral

[tex] \int_x^\infty {y n 2 x^2 y^{-3} ( x^2 y^{-2})^{n-1} dy [/tex]

[tex] = 2 n x^{2n} \int_x^\infty y^{-2n} dy [/tex]
I was, but I actually was able to get the answer. Thank you.
 

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