Mixing
Show that $n-nmaxX_i$ equals exponential distribution for $X_i$ i.i.d. and uniformly distributed on $[0,1]$
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Say, we have a sequence of random variables $(X_i)_1leq i leq n$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.
Why is $n-ncdotmaxX_1,cdots,X_n$ distributed like an exponential distribution for real and positive $n$?
I know that $lim_nrightarrowinftye(x)=(1+fracxn)^n$...
I also know that $maxX_1,cdots,X_n=P(X_1)cdotcdotscdot P(X_n)$ for independent variables...
I have:
beginalign
F(x) & =P(n-ncdotmaxX_1,cdots,X_n<x)=P(maxldots \[10pt]
& geq1-x/n)=1-P(maxldots<1-x/n).
endalign
Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.
But how do I proceed now? Or are there any mistakes so far?
probability
edited Jul 25 at 12:43
Did
242k23208442
asked Jul 25 at 11:13
IceFire
199212
add a comment |Â
up vote
1
down vote
favorite
Say, we have a sequence of random variables $(X_i)_1leq i leq n$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.
Why is $n-ncdotmaxX_1,cdots,X_n$ distributed like an exponential distribution for real and positive $n$?
I know that $lim_nrightarrowinftye(x)=(1+fracxn)^n$...
I also know that $maxX_1,cdots,X_n=P(X_1)cdotcdotscdot P(X_n)$ for independent variables...
I have:
beginalign
F(x) & =P(n-ncdotmaxX_1,cdots,X_n<x)=P(maxldots \[10pt]
& geq1-x/n)=1-P(maxldots<1-x/n).
endalign
Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.
But how do I proceed now? Or are there any mistakes so far?
probability
edited Jul 25 at 12:43
Did
242k23208442
asked Jul 25 at 11:13
IceFire
199212
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Say, we have a sequence of random variables $(X_i)_1leq i leq n$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.
Why is $n-ncdotmaxX_1,cdots,X_n$ distributed like an exponential distribution for real and positive $n$?
I know that $lim_nrightarrowinftye(x)=(1+fracxn)^n$...
I also know that $maxX_1,cdots,X_n=P(X_1)cdotcdotscdot P(X_n)$ for independent variables...
I have:
beginalign
F(x) & =P(n-ncdotmaxX_1,cdots,X_n<x)=P(maxldots \[10pt]
& geq1-x/n)=1-P(maxldots<1-x/n).
endalign
Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.
But how do I proceed now? Or are there any mistakes so far?
probability
edited Jul 25 at 12:43
Did
242k23208442
asked Jul 25 at 11:13
IceFire
199212
Say, we have a sequence of random variables $(X_i)_1leq i leq n$ that are i.i.d. and distributed according to a uniform distribution defined on interval $[0,1]$.
Why is $n-ncdotmaxX_1,cdots,X_n$ distributed like an exponential distribution for real and positive $n$?
I know that $lim_nrightarrowinftye(x)=(1+fracxn)^n$...
I also know that $maxX_1,cdots,X_n=P(X_1)cdotcdotscdot P(X_n)$ for independent variables...
I have:
beginalign
F(x) & =P(n-ncdotmaxX_1,cdots,X_n<x)=P(maxldots \[10pt]
& geq1-x/n)=1-P(maxldots<1-x/n).
endalign
Now, due to the cdf of $U(0,1)$ which is just $x$, this is equivalent to $1-(1-x/n)^n$.
But how do I proceed now? Or are there any mistakes so far?
probability
edited Jul 25 at 12:43
Did
242k23208442
asked Jul 25 at 11:13
IceFire
199212
edited Jul 25 at 12:43
Did
242k23208442
edited Jul 25 at 12:43
Did
242k23208442
edited Jul 25 at 12:43
Did
242k23208442
242k23208442
asked Jul 25 at 11:13
IceFire
199212
asked Jul 25 at 11:13
IceFire
199212
asked Jul 25 at 11:13
IceFire
199212
199212
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40
add a comment |Â
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Let $M_n = max_1leqslant ileqslant nX_i$ for each positive integer $n$. For $0<t<1$ we have $$M_n leqslant t = bigcap_i=1^n X_ileqslant t,$$
so by independence it follows that
$$
mathbb P(M_nleqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $ngeqslant N$ we have
beginalign
mathbb P(Y_nleqslant t) &= mathbb P(n(1-M_n)leqslant t)\
&=mathbb P(1-M_nleqslant t/n)\
&=mathbb P(M_ngeqslant 1-t/n)\
&=1-mathbb P(M_nleqslant 1-t/n)\
&=1 - (1-t/n)^n.
endalign
Since $$lim_ntoinfty1 - (1-t/n)^n = 1 - e^-t,$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
answered Jul 25 at 11:36
Math1000
18.4k31444
add a comment |Â
up vote
-1
down vote
You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 leq 1-x/n leq 1$
$2.$ You have to consider $lim nrightarrowinfty$. Then you get $exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
answered Jul 25 at 11:47
me me and me
63
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $M_n = max_1leqslant ileqslant nX_i$ for each positive integer $n$. For $0<t<1$ we have $$M_n leqslant t = bigcap_i=1^n X_ileqslant t,$$
so by independence it follows that
$$
mathbb P(M_nleqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $ngeqslant N$ we have
beginalign
mathbb P(Y_nleqslant t) &= mathbb P(n(1-M_n)leqslant t)\
&=mathbb P(1-M_nleqslant t/n)\
&=mathbb P(M_ngeqslant 1-t/n)\
&=1-mathbb P(M_nleqslant 1-t/n)\
&=1 - (1-t/n)^n.
endalign
Since $$lim_ntoinfty1 - (1-t/n)^n = 1 - e^-t,$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
answered Jul 25 at 11:36
Math1000
18.4k31444
add a comment |Â
up vote
2
down vote
accepted
Let $M_n = max_1leqslant ileqslant nX_i$ for each positive integer $n$. For $0<t<1$ we have $$M_n leqslant t = bigcap_i=1^n X_ileqslant t,$$
so by independence it follows that
$$
mathbb P(M_nleqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $ngeqslant N$ we have
beginalign
mathbb P(Y_nleqslant t) &= mathbb P(n(1-M_n)leqslant t)\
&=mathbb P(1-M_nleqslant t/n)\
&=mathbb P(M_ngeqslant 1-t/n)\
&=1-mathbb P(M_nleqslant 1-t/n)\
&=1 - (1-t/n)^n.
endalign
Since $$lim_ntoinfty1 - (1-t/n)^n = 1 - e^-t,$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
answered Jul 25 at 11:36
Math1000
18.4k31444
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $M_n = max_1leqslant ileqslant nX_i$ for each positive integer $n$. For $0<t<1$ we have $$M_n leqslant t = bigcap_i=1^n X_ileqslant t,$$
so by independence it follows that
$$
mathbb P(M_nleqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $ngeqslant N$ we have
beginalign
mathbb P(Y_nleqslant t) &= mathbb P(n(1-M_n)leqslant t)\
&=mathbb P(1-M_nleqslant t/n)\
&=mathbb P(M_ngeqslant 1-t/n)\
&=1-mathbb P(M_nleqslant 1-t/n)\
&=1 - (1-t/n)^n.
endalign
Since $$lim_ntoinfty1 - (1-t/n)^n = 1 - e^-t,$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
answered Jul 25 at 11:36
Math1000
18.4k31444
Let $M_n = max_1leqslant ileqslant nX_i$ for each positive integer $n$. For $0<t<1$ we have $$M_n leqslant t = bigcap_i=1^n X_ileqslant t,$$
so by independence it follows that
$$
mathbb P(M_nleqslant t) = t^n.
$$
Let $Y_n = n(1-M_n)$. Let $t>0$ and choose $N>1/t$. Then for $ngeqslant N$ we have
beginalign
mathbb P(Y_nleqslant t) &= mathbb P(n(1-M_n)leqslant t)\
&=mathbb P(1-M_nleqslant t/n)\
&=mathbb P(M_ngeqslant 1-t/n)\
&=1-mathbb P(M_nleqslant 1-t/n)\
&=1 - (1-t/n)^n.
endalign
Since $$lim_ntoinfty1 - (1-t/n)^n = 1 - e^-t,$$
it follows that $Y_n$ converges in distribution to the exponential distribution with mean $1$.
answered Jul 25 at 11:36
Math1000
18.4k31444
answered Jul 25 at 11:36
Math1000
18.4k31444
answered Jul 25 at 11:36
Math1000
18.4k31444
answered Jul 25 at 11:36
Math1000
18.4k31444
18.4k31444
add a comment |Â
add a comment |Â
up vote
-1
down vote
You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 leq 1-x/n leq 1$
$2.$ You have to consider $lim nrightarrowinfty$. Then you get $exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
answered Jul 25 at 11:47
me me and me
63
add a comment |Â
up vote
-1
down vote
You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 leq 1-x/n leq 1$
$2.$ You have to consider $lim nrightarrowinfty$. Then you get $exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
answered Jul 25 at 11:47
me me and me
63
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 leq 1-x/n leq 1$
$2.$ You have to consider $lim nrightarrowinfty$. Then you get $exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
answered Jul 25 at 11:47
me me and me
63
You are right in your calculations. There are two points and you are done:
$1.$ From uniform to exponential you have the condition $0 leq 1-x/n leq 1$
$2.$ You have to consider $lim nrightarrowinfty$. Then you get $exp(-x)$. Your formula about this is also correct.
So in conclusion you get an exponential distribution with parameter $+1$ and it is $exp(-1)$ safely enough it doesnt blow up.
answered Jul 25 at 11:47
me me and me
63
edited Jul 25 at 11:58
answered Jul 25 at 11:47
me me and me
63
answered Jul 25 at 11:47
me me and me
63
answered Jul 25 at 11:47
me me and me
63
63
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862318%2fshow-that-n-n-max-x-i-equals-exponential-distribution-for-x-i-i-i-d-and%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Of course, $n-ncdotmaxX_1,cdots,X_n$ is not distributed like an exponential distribution, for any positive integer $n$.
– Did
Jul 25 at 11:40