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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 $\hat\theta$?

kjetil b halvorsen
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Enzo
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  • are the Xs Independent? – Inquest Feb 24 '13 at 22:12
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    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? – Did Feb 24 '13 at 22:19
  • @Inquest, yes they are! Sorry that I forgot to point that out. – Enzo Feb 24 '13 at 23:06
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    @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 $P{max{X_1, X_2, \dots, X_n} < x}$. – Enzo Feb 24 '13 at 23:08
  • The question is crystal clear. That you evade it is telling. I suggest to stop being sorry and to start putting some personal input in your questions. – Did Feb 25 '13 at 07:23
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    @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. – Enzo Feb 25 '13 at 14:07
  • MSE = Math Stack Exchange. You asked 6 questions so far. Is this so difficult to provide some indications about what you tried before asking, what you understand of the problem, and what is blocking you? – Did Feb 25 '13 at 16:16
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    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. – Enzo Feb 25 '13 at 16:35
  • You may cut on the drama, nobody is accusing you, I am simply making the obvious remark that this question (as well as some previous ones of yours, despite what you say) is delivered on the site without any personal input. If you feel comfortable about THAT, so be it. – Did Feb 26 '13 at 16:47
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    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. – Enzo Feb 27 '13 at 07:10

3 Answers3

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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}

sligocki
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Inquest
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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 $W$.

André Nicolas
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4

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: $$f_M(M = x) = n * (\frac{x}{\theta})^{n-1} * \frac{1}{\theta}$$ $$= \frac{n * x^{n-1}}{\theta^n}$$

אלימלך שרייבר
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