Let $X_1, \ldots, X_n$ be a random sample from the exponential distribution $\exp(\lambda)$. Let $$M_n=\max\{X_1, \ldots, X_n\}$$ with probability density function $$g_{M_n}(x)=n\lambda e^{-\lambda x}(1-e^{-\lambda x})^{(n-1)}, \qquad x>0$$ Q1. If $M_n$ is the only information that you have from the sample, find a maximum likelihood estimator (mle) $\hat{\lambda}_n$ of $\lambda$.
Q2. Using $(1+x)^n>1+nx$ prove that $\hat{\lambda}_n$ is consistent, i.e. that $P(| \hat{\lambda}_n-\lambda|>\epsilon)\longrightarrow0$, for $n\rightarrow \infty$


  • $\begingroup$ For your first question, perhaps you can use the invariance property of the MLE? $\endgroup$
    – JohnK
    Mar 1, 2014 at 11:06
  • $\begingroup$ @JohnK. You mean functional invariance, don't you? So, to find as usual the MLE and then replace in the Maximum? But my problem is that according to the exercise I may only use the maximum of the $X_i$ and not all the $X_i$ explicitly. So, if the MLE of $\lambda$ depends on the individual values of the $X_i$ then I cannot use it. And it is an exam question, so it should be solvable without monster calculations. Thanks a lot, I will think more over it $\endgroup$
    – Jimmy R.
    Mar 1, 2014 at 11:48
  • $\begingroup$ @JohnK Wow, I did not know about the other site, many thanks John, I will post it there! $\endgroup$
    – Jimmy R.
    Mar 1, 2014 at 12:44

1 Answer 1


As you already know, considering the function $$G_n(x)=\log\left(n+\frac{n-1}{x-1}\right),$$ defined on $(1,+\infty)$, $\hat\lambda_n$ solves the identity $$\hat\lambda_n M_n=G_n(\hat\lambda_n M_n).$$ As you noted, this identity has no analytical solution. However, the function $G_n$ decreases on $(1,+\infty)$ from $G_n(1)=+\infty$ to $G_n(+\infty)=\log(n)$ and, perhaps amazingly, this is all one needs to solve the exercise.

To wit, first note that $G_n\gt\log(n)$ on $(1,+\infty)$ hence the unique solution of the identity above is such that $\hat\lambda_n M_n\gt\log(n)$. In particular, $\hat\lambda_n M_n\to\infty$ hence $$G_n(\hat\lambda_n M_n)=\log(n+o(n))=\log(n)+o(1),$$ which implies $$\hat\lambda_n M_n/\log n\to1\ \text{almost surely.}\tag{$\ast$}$$ On the other hand, classic estimates of $M_n$ are as follows. For every nonnegative $x$, $$P(M_n\leqslant x)=(1-\mathrm e^{-\lambda x})^n,$$ hence, for every $\varepsilon$ in $(0,1)$, $$P(M_n\leqslant(1+\varepsilon)\log n/\lambda)=(1-n^{-1-\varepsilon})^n\to1,$$ and$$P(M_n\leqslant(1-\varepsilon)\log n/\lambda)=(1-n^{-1+\varepsilon})^n\to0,$$ that is, $$M_n/\log n\to1/\lambda\ \text{in probability.}\tag{$\dagger$}$$ Putting $(\ast)$ and $(\dagger)$ together, one sees that $\hat\lambda_n\to\lambda$ in probability, as desired.

  • 2
    $\begingroup$ Two remarks: 1. In view of the above, one can find rather questionable the formulation of Q1 and Q2. 2. All in all, math.SE might be the best site to get math answers... :-) $\endgroup$
    – Did
    Nov 24, 2014 at 10:07
  • 1
    $\begingroup$ Thanks for the great answer! Except that it answers (finally) my question, it gives also a motivation to keep studying :)! Thanks again! $\endgroup$
    – Jimmy R.
    Dec 1, 2014 at 13:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .