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Let X,Y,Z be independent, identically distributed random variables,each with density $f(x)=6x^5$ for $0\leq x\leq1$ and 0 elsewhere.I want to find the distribution and density functions of the maximum of X,Y and Z.

Answer:-I know by integrating density function, distribution function can be obtained.I computed CDF of maximum of $X,Y,Z=X^6$ and its PDF=$18*X^{17}$.

CDF of minimum of $X,Y,Z=1-(1-X^6)^3$ and its PDF is $6*X^5*3*(1-X^6)^2$


marked as duplicate by whuber Feb 26 '16 at 15:58

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    $\begingroup$ Answers to this appear in many threads. They can be hard to find because "maximum" occurs thousands of times as part of "maximum likelihood." Searches on "minimum" and "order" tend to be more fruitful. Some of the hits I found are at stats.stackexchange.com/questions/92845 and stats.stackexchange.com/questions/18433 . Here's one search: stats.stackexchange.com/… $\endgroup$ – whuber Feb 26 '16 at 15:58
  • $\begingroup$ I have calculated distribution function of the max(X,Y,Z) which is $3x^{12}$. Whereas density function of the maximum of iid random variables X,Y,Z is $18x^{17}$.I want to give detailed answer, but I didn't find 'ANSWER' tab below this question. $\endgroup$ – Dhamnekar Winod May 28 '16 at 14:14
  • $\begingroup$ Your calculation cannot be correct, because the integral of $3x^{12}$ over the interval $[0,1]$ is only $3/13$ rather than $1$. $\endgroup$ – whuber May 28 '16 at 18:17
  • $\begingroup$ @whuber You are correct.You are rightly pointed out my mistake. Howsoever, after rectifying my mistake, the distribution function of max iid random variables is 3 and its density function is $18x^5$. I hope these would be correct answers. $\endgroup$ – Dhamnekar Winod May 29 '16 at 14:43
  • $\begingroup$ Those values obviously are incorrect, for the same reason given before (and because no distribution function can equal $3$, since probabilities lie between $0$ and $1$). You should study the related threads more closely. $\endgroup$ – whuber May 29 '16 at 14:58