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We have a sample $X_1, ..., X_n$ from $U[M,2M]$, $M>0$. Define $U=\frac{(n+1)}{(2n+1)}\max(X_i)$. I wish to show that $U$ is an unbiased estimator for $M$, but I am getting a wrong result. This is how I am trying to do it.

$f(x)=\frac{1}{2M-M}=\frac{1}{M}$

$P(X_i \leq x)=\int_M^x \frac{1}{M} dy = \frac{x-M}{M}$

$P(\max(X_i) \leq x) = P(X_1 \leq x, ... , X_n \leq x)=(\frac{x-M}{M})^n$

$E(\max(X_i))=\int_M^{2M} {1- (\frac{y-M}{M})^n } dy = \frac{n}{n+1}M$

$E(U)=E(\frac{(n+1)}{(2n+1)}\max(X_i))=\frac{(n+1)}{(2n+1)}E(\max(X_i))=\frac{(n+1)}{(2n+1)}*\frac{n}{n+1}M$

but I have to show that $ E(U)=M$ so it is an unbiased estimator, I really don't see where I am going wrong!

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  • $\begingroup$ When computing E(max(X_i)) you forgot to integrate P(max(X_i)>x)=1 from x=0 to x=M. $\endgroup$ – Did Jun 8 '14 at 20:39
  • $\begingroup$ Thank you, that was indeed the mistake. $\endgroup$ – Shawn Jun 8 '14 at 20:54
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It looks like your problems start occurring when you try to calculate the Expected Value of X.

Go back one step: When you calculate the probability that all X's are less than "x", you're in effect calculating the CDF, right? As a result, you want to take the derivative of this CDF, giving you your new PDF to work with [i.e. n/M(X/M -1)^(n-1)]. [See: Expected Value of Max of IID Variables ]

The next step would be to multiply by X and integrate from M to 2M (performing the necessary u-substitution, etc.), leaving you with the Expected Value of X.

From there it's easy since the "2n+1" and "n+1" cancel out the "2n+1" and "n+1" earlier...

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