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Suppose

$$X_1, X_2, \dots, X_n\sim Unif(0, \theta), iid$$

and suppose

$$\hat\theta = \max\{X_1, X_2, \dots, X_n\}$$

How would I find the probability density of $\hat\theta$?

Thank you!

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    $\begingroup$ Here is a question for you: did you ask all your questions on MSE with no indication whatsoever on what you understood of the problem or what you tried before asking? $\endgroup$
    – Did
    Feb 24, 2013 at 22:19
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    $\begingroup$ @Did, I am not sure what you are referring to. I am sorry if I left out what I have tried in my question. I knew that I was supposed to find the CDF of $\hat\theta$ and then differentiate it, but I got confused at the step of calculating $P\{max\{X_1, X_2, \dots, X_n\} < x\}$. $\endgroup$
    – Enzo
    Feb 24, 2013 at 23:08
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    $\begingroup$ @Did, no seriously I don't know what you are talking about. What do you mean by all the MSE questions? This is my first question that's related to MSE and I just told in in the comment what I did, so I am not sure what you are accusing me of. $\endgroup$
    – Enzo
    Feb 25, 2013 at 14:07
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    $\begingroup$ I see. I thought you were talking about mean squared error. Well, I apologize if you didn't see what I have tried before asking, but I did indicate them somewhere in the questions, either in comments, or as a feedback of my own work. For example, on one of the proofs involving differential forms, I provided a copy of my own solution in the end. So I don't think you are in a right position to accuse me, and I don't feel comfortable about that. Thank you for helping me before though. $\endgroup$
    – Enzo
    Feb 25, 2013 at 16:35
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    $\begingroup$ Well thanks for your respond. I'll keep that in mind. If all you would like to do is questioning my intellectual effort, then be it. $\endgroup$
    – Enzo
    Feb 27, 2013 at 7:10

3 Answers 3

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\begin{align} P(Y\leq x) &= P(\max(X_1,X_2 ,\cdots,X_n)\leq x)\\ &= P(X_1\leq x,X_2\leq x,\cdots,X_n\leq x)\\ &\stackrel{ind}{=} \prod_{i=1}^nP(X_i\leq x )\\ &= \prod_{i=1}^n\dfrac{x}{\theta}\\&=\left(\dfrac{x}{\theta}\right)^n \end{align}

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    $\begingroup$ In line 2, are the ns supposed to be xs or is this correct as is? $\endgroup$
    – kram1032
    Jul 26, 2014 at 20:50
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Let random variable $W$ denote the maximum of the $X_i$. We will assume that the $X_i$ are independent, else we can say very little about the distribution of $W$.

Note that the maximum of the $X_i$ is $\le w$ if and only if all the $X_i$ are $\le w$. For $w$ in the interval $[0,\theta]$, the probability that $X_i\le w$ is $\frac{w}{\theta}$. It follows by independence that the probability that $W\le w$ is $\left(\frac{w}{\theta}\right)^n$.

Thus, in our interval, the cumulative distribution function $F_W(w)$ of $W$ is given by $$F_W(w)= \left(\frac{w}{\theta}\right)^n.$$ Differentiate to get the density function of $W$.

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  • $\begingroup$ There appears to be a TeX error. I can't edit it because it's just a single backslash that needs removal but math.SE asks me to change the post by at least 6 symbols. $\endgroup$
    – kram1032
    Jul 26, 2014 at 20:51
  • $\begingroup$ Thank you for telling me, yes I had written \w for w. $\endgroup$ Jul 26, 2014 at 21:02
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The general formula for the probability density of the maximum of any $iid$ sample set of the random variable $x$, $M = max\{x_1,x_2,…,x_n\}$ is: $$f_M(M = x) = n * F_x(x)^{n-1} * f_x(x)$$ where $f_x(x)$ is the probability density of $x$, and $F_x(x)$ is the cumulative distribution function of same.

In this case we have: $f_x(x) = \frac{1}{\theta}$ , $F_x(x) = \frac{x}{\theta}$, so we get: $$f_M(M = x) = n * (\frac{x}{\theta})^{n-1} * \frac{1}{\theta}$$ $$= \frac{n * x^{n-1}}{\theta^n}$$

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