My statistics course requires me to find the $\text{MLE}$ of $A$ (denoted $A_{ML}$) where $X_1, X_2, X_3\sim \text{Unif}[A,2A]$ and $A>0$ and then find the constant $k$ such that $E(kA_{ML})=A.$ I have done the first part and found that $A_{ML}=\frac{1}{2} \max \{X_1, X_2, X_3\},$ and simplified $E(kA_{ML})=\frac{k}{2}E(\max\{X_1, X_2, X_3\}),$ but have gotten stuck trying to find $E(\max\{X_1, X_2, X_3\}).$

I have seen others go about this by using the $\text{CDF}:$ $P(\max\{X_1, X_2, X_3\} \le y)= P(X_1, X_2, X_3 \le y)= \text{ (by independence) }$ $P(X_1 \le y)P(X_2 \le y)P(X_3 \le y)$ but I am not sure if you can assume independence from the wording of the question: "Let $X_1, X_2, X_3$ be a random sample of size three from a Unif$[A, 2A]$ distribution, where $A>0$ is a model parameter.".

Does anyone have any suggestions?


1 Answer 1


It is very easy to realize that


Thus the CDF of the max is the following


With density


You HAVE to assume independence basing your assumption on the property of the random sample ( $X_1,..X_n$ are iid with the same population's distribution)... note that you assumed independence also when calculating the MLE

now I think you can conclude by yourself

EDIT: further explanation

As you should know, with independence, the CDF of the Max is the product of the CDF's, so letting $T=max(X_1,X_2,X_3)$ and using the fact that $X_i$ are independend and IDENTICALLY DISTRIBUTED,




  • $\begingroup$ Is $Z$ here the standard normal? $\endgroup$
    – Tikak
    Oct 1, 2020 at 7:02
  • $\begingroup$ @Tikak: WHAT?????? z is just a letter....write z,t,u,v....on your convenience $\endgroup$
    – tommik
    Oct 1, 2020 at 7:03
  • $\begingroup$ Lol okay sorry in my course we often use $Z$ to denote that, so I was a bit confused. Why in the CDF of the max is there no $X_1, X_2, X_3$? And if you did implement those, would it mean your function would be $\frac{(X_1 - A)(X_2 - A)(X_3 - A)}{A^3}$ which would be a bit nastier to differentiate. I'm just a bit confused as to how the $Z$ translates. $\endgroup$
    – Tikak
    Oct 1, 2020 at 7:08
  • $\begingroup$ @Tikak : look at my edit in the answer. Used T in order to avoid confusion with Standard Gaussian $\endgroup$
    – tommik
    Oct 1, 2020 at 7:14
  • 1
    $\begingroup$ @Tikak : $X_1$ is the random variable while $t$ or any letter you prefer is a value of the rv. If T is the rv "max" this rv has a CDF that is (in function of t) the product of the 3 CDF, that is one of the CDF^3 $\endgroup$
    – tommik
    Oct 1, 2020 at 7:31

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