What probability distribution represents the maximum of multiple of the same normal distribution?
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Given a normal distribution $N(mu,sigma^2)$, if we define a new distribution $M_n$ based on the maximum of $n$ random samples from the normal distribution $N$, what formula represents $M_n$. I'm certain the increasing the value of $n$ will continue to increase both $M$'s mean as well as decrease its variance, with diminishing returns for both, I'm also pretty certain that all $M_n$ will be normal distributions, but I remain unsure about how to model the distribution of the $M$'s.
statistics normal-distribution
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
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show 5 more comments
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Given a normal distribution $N(mu,sigma^2)$, if we define a new distribution $M_n$ based on the maximum of $n$ random samples from the normal distribution $N$, what formula represents $M_n$. I'm certain the increasing the value of $n$ will continue to increase both $M$'s mean as well as decrease its variance, with diminishing returns for both, I'm also pretty certain that all $M_n$ will be normal distributions, but I remain unsure about how to model the distribution of the $M$'s.
statistics normal-distribution
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
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1
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No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
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– Did
Dec 30 '18 at 2:14
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Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
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– Aaron Quitta
Dec 30 '18 at 2:17
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en.wikipedia.org/wiki/…
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– Did
Dec 30 '18 at 2:19
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Got it, thanks.
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– Aaron Quitta
Dec 30 '18 at 2:20
1
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My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51
|
show 5 more comments
$begingroup$
Given a normal distribution $N(mu,sigma^2)$, if we define a new distribution $M_n$ based on the maximum of $n$ random samples from the normal distribution $N$, what formula represents $M_n$. I'm certain the increasing the value of $n$ will continue to increase both $M$'s mean as well as decrease its variance, with diminishing returns for both, I'm also pretty certain that all $M_n$ will be normal distributions, but I remain unsure about how to model the distribution of the $M$'s.
statistics normal-distribution
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
$endgroup$
Given a normal distribution $N(mu,sigma^2)$, if we define a new distribution $M_n$ based on the maximum of $n$ random samples from the normal distribution $N$, what formula represents $M_n$. I'm certain the increasing the value of $n$ will continue to increase both $M$'s mean as well as decrease its variance, with diminishing returns for both, I'm also pretty certain that all $M_n$ will be normal distributions, but I remain unsure about how to model the distribution of the $M$'s.
statistics normal-distribution
statistics normal-distribution
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
asked Dec 30 '18 at 2:10
Aaron QuittaAaron Quitta
340214
340214
1
$begingroup$
No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
$endgroup$
– Did
Dec 30 '18 at 2:14
$begingroup$
Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:17
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Did
Dec 30 '18 at 2:19
$begingroup$
Got it, thanks.
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:20
1
$begingroup$
My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51
|
show 5 more comments
1
$begingroup$
No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
$endgroup$
– Did
Dec 30 '18 at 2:14
$begingroup$
Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:17
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Did
Dec 30 '18 at 2:19
$begingroup$
Got it, thanks.
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:20
1
$begingroup$
My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51
1
1
$begingroup$
No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
$endgroup$
– Did
Dec 30 '18 at 2:14
$begingroup$
No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
$endgroup$
– Did
Dec 30 '18 at 2:14
$begingroup$
Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:17
$begingroup$
Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:17
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Did
Dec 30 '18 at 2:19
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Did
Dec 30 '18 at 2:19
$begingroup$
Got it, thanks.
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:20
$begingroup$
Got it, thanks.
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:20
1
1
$begingroup$
My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51
$begingroup$
My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51
|
show 5 more comments
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1
$begingroup$
No $M_n$ for $ngeqslant2$ is normal. All your other questions are solved by the remark that, for every $x$, $$P(M_nleqslant x)=P(X_1leqslant x)^n=Phi((x-mu)/sigma)^n$$
$endgroup$
– Did
Dec 30 '18 at 2:14
$begingroup$
Huh. I'm not totally familiar with notation what does $Phi$ represent in this context?
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:17
$begingroup$
en.wikipedia.org/wiki/…
$endgroup$
– Did
Dec 30 '18 at 2:19
$begingroup$
Got it, thanks.
$endgroup$
– Aaron Quitta
Dec 30 '18 at 2:20
1
$begingroup$
My first comment provides everything you need to compute it.
$endgroup$
– Did
Dec 30 '18 at 2:51