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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?

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1 Answer 1

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It is very easy to realize that

$$F_X(x)=\frac{x-A}{A}$$

Thus the CDF of the max is the following

$$F_Z(z)=\frac{(z-A)^3}{A^3}$$

With density

$$f_Z(z)=\frac{3(z-A)^2}{A^3}\mathbb{1}_{[A;2A]}(z)$$

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,

$$F_T(t)=[F_{X_1}(t)]^3$$

Concluding...

$$\mathbb{E}[T]=\int_A^{2A}\frac{3}{A^3}t(t-A)^2dt=...=\frac{7}{4}A$$

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  • $\begingroup$ 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
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    $\begingroup$ @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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