I have a question.

Suppose that $X_1,\ldots,X_n$ are iid $U(0,\lambda)$ and let $X(n)$ denote the nth order statistic.

Suppose $\lambda$ is unknown and should be estimated from the sample.

Take $\lambda^* = \frac{(n+1)X(n)}n$ as the estimator of lambda. Find the distribution of $\lambda^*$.

I started with taking the CDF of a uniform distribution as $\frac{x}{\lambda}$. The CDF of the nth order statistic is $(\frac{x}{\lambda})^n$ after some operations I came up to the cdf of $\lambda^*$ is $(\frac{(n+1)x}{n\lambda})^n$

Now I am trying to find the expectation of $\lambda^*$.

I found the pdf $n(\frac{(n+1)x}{n\lambda})^{n-1}$ and am struggling to find the expectation. I suspect that it is $\lambda$ but am having trouble evaluating $$\int_0^\lambda xn(\frac{(n+1)x}{n\lambda})^{n-1}$$

Can anyone confirm that I did this correctly and help finding the expectation?

  • $\begingroup$ You're already using a great Wysiwyg MathJax editor. Just enclose $\TeX$ in dollar signs and look at the preview in the rendering window beneath your input window. This will help you clarify your question, which in its current form appears to have little connection with the image (and proposes a truly terrible estimator of $\lambda$: as $n$ grows larger the expectation of this estimator diverges to infinity!). I converted your text to $\TeX$ to serve as an illustration. $\endgroup$
    – whuber
    Nov 24, 2014 at 19:34
  • $\begingroup$ You don't want to take the $n^\text{th}$ power of $\frac{n+1}{n}$ in the CDF! A helpful formula: when $F$ is the CDF of a continuous distribution on the finite interval $[0,\lambda]$ then the expectation of that distribution is $\int_0^\lambda(1-F(x))dx$. $\endgroup$
    – whuber
    Nov 24, 2014 at 20:52
  • $\begingroup$ thanks for the info! are you saying my cdf is incorrect or just use that formula to calculate the expectation? $\endgroup$ Nov 24, 2014 at 20:57
  • $\begingroup$ Using the formula will also show you that the CDF is incorrect :-). The expectation of $\lambda^*$ obviously is just $(n+1)/n$ times the expectation of $X(n)$, so apply the formula to the CDF of $X(n)$ to compute the latter. $\endgroup$
    – whuber
    Nov 24, 2014 at 21:00
  • 1
    $\begingroup$ Yes: I gave a strong one in my second comment. $\endgroup$
    – whuber
    Nov 24, 2014 at 22:37

1 Answer 1


Taking your CDF and using whuber's hint, we have the expectation of the maximum as $$E[X_{(n)}] = \int_0^\lambda \left[ 1 - \left( {x \over \lambda} \right)^n \right]dx=\lambda - {\lambda \over {n+1}} $$ This simplifies to $$E[X_{(n)}]= \lambda \left[ {n \over {n+1}} \right]. $$ Then for your altered estimator we get $$E[\lambda^*]=\left[ {{n+1} \over n} \right] E[X_{(n)}]= \lambda.$$


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