$f(y) = ay^{a-1}/θ^a, 0<y<θ$

$ \hat{\Theta} = max(Y_1, Y_2, . . . , Y_n).$

How do I find the $E[\hat{\Theta}]$ ?

I'm trying to show that it's a biased estimator, then I'm going to find an unbiased version and derive MSE.

  • 2
    $\begingroup$ Find the CDF of the maximum. (It's an easy calculation.) $\endgroup$ – whuber Feb 1 '19 at 20:20
  • $\begingroup$ Add the self study tag. $\endgroup$ – Michael R. Chernick Feb 1 '19 at 20:35
  • $\begingroup$ @MichaelChernick I've added the tag, thanks I'm new here. $\endgroup$ – Gire Feb 1 '19 at 21:36
  • $\begingroup$ @whuber I know I need the distribution function of theta_hat but I don't know what it is. How would I go about finding the CDF? I'm sorry if these are really basic, dumb questions, I'm new to mathematical statistics and having only done applied statistics in my field (psychology) I get easily confused. $\endgroup$ – Gire Feb 1 '19 at 21:39
  • 1
    $\begingroup$ While this is a simple algebraic problem, it's clear that it must be biased just from general reasoning. The distribution is continuous, and the observations are bounded above by $\theta$, so every observation is $<\theta$ with probability 1. Therefore the maximum is $<\theta$ with probability 1. Therefore the mean of the distribution of the maximum is $<\theta$. $\endgroup$ – Glen_b -Reinstate Monica Feb 2 '19 at 2:19

Here's the way I like to explain this to my class.

If the $Y_i$ represent the heights of a group of people, then $\hat\theta$ is the height of the tallest person. If the tallest person can walk through a door without hitting his head, then so can everybody in the group! Hopefully you agree with the logic behind this. Mathematically, this is equivalent to saying $$P(\hat\theta \leq y) = P(\text{all of the $Y_i \leq y$}) = P\left(\bigcap_{i=1}^n[Y_i \leq y]\right)$$ From here the calculation is easy! (Use independence)

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