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Given a set of n random variables $X_1,\ldots,X_n\sim Uniform(0,\alpha)$

Let $Y=\max[X_1,\ldots,X_n]$, What is the expectation $E(Y)$?

My question came when I saw an old question but I'm interested in the case where it is $\alpha=10$ for example so: $uniform(0,10)$, or more generally $uniform(0,\alpha)$ for any $\alpha$

What does it become in that case?

Edit: the questions is closed: "Please provide additional context, which ideally explains why the question is relevant to you and our community."

It is clearly mentioned in the first lines when I saw another answer by coincidence, I was curious to see how to generalize it. The part about "our community",well, if the mod who closed the question does not see it useful and can speak for the entire community, that's his own problem. Not mine to solve. Math SE was way nicer years ago. I hate the way this site is heading. It's trying to invite people to stop thinking, being curious or asking questions. good job mods!

Thanks

user206904
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    You can use exactly the same reasoning in the other answer, but with the different pdf. Or, since different uniform RVs are all related by linear transformations, you can take the answer from $Unif(0,1)$ and multiply it by $\alpha$. Both perspectives are useful. – Aaron May 28 '22 at 23:19
  • @BrianMoehring, thanks, I misused the notation, I meant for any alpha $\in (0,+\infty)$ – user206904 May 29 '22 at 08:16
  • @Aaron Thanks for the hint! I thought about multiplying at the end, but I was not sure. Feel free to write an answer! Thanks – user206904 May 29 '22 at 08:16
  • @user206904 . What, given Aaron's succint *comment* above, prevents you to solve the problem ? – Kurt G. Jun 02 '22 at 06:37
  • @KurtG.,Huh? no it does not prevent me. My question has been answered. which is why the question is marked as answered. (thus my complaint that closing an answered question makes no sense, but whatever) – user206904 Jun 03 '22 at 00:02
  • Answering a question like that makes no sense. A göod hint will start a better learning process. – Kurt G. Jun 03 '22 at 01:55

2 Answers2

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If $X_1, \ldots, X_n$ are i.i.d. $U(0,1)$, then $aX_1, \ldots, aX_n$ are i.i.d. $U(0,a)$ and $E(\max aX_i)=aE(\max X_i)$, so we have reduced the problem to that of the cited answer.

Aaron
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Given a finite collection of iid rv's $X_{1},...,X_{n}$ with cdf $F_{X}$. You have

You have for any random variable $Y$

$\displaystyle E(Y)=\int_{(0,\infty)}(1-F_{Y}(y))\,d\lambda(y) -\int_{(-\infty,0)}F_{Y}(y)\,d\lambda(y)$ . Where $\lambda$ is the Lebesgue Measure on $\Bbb{R}$. For simplicity you can ignore Lebesgue Integrability and treat everything as Riemann Integrals. See my answer here for a proof.

Now $F_{Y}(y)=\Bbb{P}(\max(X_{1},...,X_{n})\leq y)=\\\Bbb{P}\bigg((X_{1}\leq y)\cap (X_{2}\leq y)\cap...\cap(X_{n}\leq y)\bigg)=\bigg(\Bbb{P}(X_{1}\leq y)\bigg)^{n}=(F_{X}(y))^{n}$ .

Thus you can now calculate the expectation of $Y$ in terms of the cdf of $X_{1}$ alone if you substitute the value of $F_{Y}(y)$ in the formula for expectation I gave above.

In particular for $X_{i}\sim \text{unif}(0,a)$ the cdf $$F_{X}(y)=\begin{cases}0\,,y<0\\\frac{y}{a}\,,y\in (0,a)\\ 1\,,y\geq a\end{cases}$$ .

Thus after substituting you get

$$E(Y)=\int_{0}^{a}\bigg(1-\bigg(\frac{y}{a}\bigg)^{n}\bigg)\,dy$$

Mr.Gandalf Sauron
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