Hints find $Z$ so that $n min{X_{1},…,X_{n}}xrightarrow{d} Z$
$begingroup$
Let $n in mathbb N$ while $(X_{i})_{i=1}^{n}$ are independent and uniformly distributed random variables on $[0,1]$ and define $M_{n}:=min{X_{1},...,X_{n}}$.
Find a $Z$ so that $n M_{n}xrightarrow{d} Z$
Solution:
If we can show $lim_{n to infty}P(nM_{n}>z)=P(Z>z)$ then it immediately follows by the definition of the cdf that $lim_{n to infty}F_{X_{n}}=F_{Z}$
$P(nM_{n}> z)=P(nX_{1} > z,...nX_{n}>z)=prod_{i=1}^{n}P(nX_{1}>z)=(P(nX_{i}>z))^{n}=(int_{frac{z}{n}}^{1}1dx)^n=(1-frac{z}{n})^{n}1_{z in [0,n]}$
For $lim_{n to infty}(1-frac{z}{n})^{n}=e^{-z}$
So we can choose $Z$~$exp({-z})$ where $z in [0, n]$ such that
$nM_{n} xrightarrow{d} Z$
Is this correct?
probability probability-theory random-variables independence
asked Jan 14 at 20:50
SABOYSABOY
683311
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb N$ while $(X_{i})_{i=1}^{n}$ are independent and uniformly distributed random variables on $[0,1]$ and define $M_{n}:=min{X_{1},...,X_{n}}$.
Find a $Z$ so that $n M_{n}xrightarrow{d} Z$
Solution:
If we can show $lim_{n to infty}P(nM_{n}>z)=P(Z>z)$ then it immediately follows by the definition of the cdf that $lim_{n to infty}F_{X_{n}}=F_{Z}$
$P(nM_{n}> z)=P(nX_{1} > z,...nX_{n}>z)=prod_{i=1}^{n}P(nX_{1}>z)=(P(nX_{i}>z))^{n}=(int_{frac{z}{n}}^{1}1dx)^n=(1-frac{z}{n})^{n}1_{z in [0,n]}$
For $lim_{n to infty}(1-frac{z}{n})^{n}=e^{-z}$
So we can choose $Z$~$exp({-z})$ where $z in [0, n]$ such that
$nM_{n} xrightarrow{d} Z$
Is this correct?
probability probability-theory random-variables independence
asked Jan 14 at 20:50
SABOYSABOY
683311
$endgroup$
$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58
add a comment |
$begingroup$
Let $n in mathbb N$ while $(X_{i})_{i=1}^{n}$ are independent and uniformly distributed random variables on $[0,1]$ and define $M_{n}:=min{X_{1},...,X_{n}}$.
Find a $Z$ so that $n M_{n}xrightarrow{d} Z$
Solution:
If we can show $lim_{n to infty}P(nM_{n}>z)=P(Z>z)$ then it immediately follows by the definition of the cdf that $lim_{n to infty}F_{X_{n}}=F_{Z}$
$P(nM_{n}> z)=P(nX_{1} > z,...nX_{n}>z)=prod_{i=1}^{n}P(nX_{1}>z)=(P(nX_{i}>z))^{n}=(int_{frac{z}{n}}^{1}1dx)^n=(1-frac{z}{n})^{n}1_{z in [0,n]}$
For $lim_{n to infty}(1-frac{z}{n})^{n}=e^{-z}$
So we can choose $Z$~$exp({-z})$ where $z in [0, n]$ such that
$nM_{n} xrightarrow{d} Z$
Is this correct?
probability probability-theory random-variables independence
asked Jan 14 at 20:50
SABOYSABOY
683311
$endgroup$
Let $n in mathbb N$ while $(X_{i})_{i=1}^{n}$ are independent and uniformly distributed random variables on $[0,1]$ and define $M_{n}:=min{X_{1},...,X_{n}}$.
Find a $Z$ so that $n M_{n}xrightarrow{d} Z$
Solution:
If we can show $lim_{n to infty}P(nM_{n}>z)=P(Z>z)$ then it immediately follows by the definition of the cdf that $lim_{n to infty}F_{X_{n}}=F_{Z}$
$P(nM_{n}> z)=P(nX_{1} > z,...nX_{n}>z)=prod_{i=1}^{n}P(nX_{1}>z)=(P(nX_{i}>z))^{n}=(int_{frac{z}{n}}^{1}1dx)^n=(1-frac{z}{n})^{n}1_{z in [0,n]}$
For $lim_{n to infty}(1-frac{z}{n})^{n}=e^{-z}$
So we can choose $Z$~$exp({-z})$ where $z in [0, n]$ such that
$nM_{n} xrightarrow{d} Z$
Is this correct?
probability probability-theory random-variables independence
probability probability-theory random-variables independence
asked Jan 14 at 20:50
SABOYSABOY
683311
asked Jan 14 at 20:50
SABOYSABOY
683311
edited Jan 14 at 21:59
SABOY
asked Jan 14 at 20:50
SABOYSABOY
683311
asked Jan 14 at 20:50
SABOYSABOY
683311
asked Jan 14 at 20:50
SABOYSABOY
683311
683311
$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58
add a comment |
$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58
$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each $n$, compute $P(nM_n > z)$.
$$P(nM_n > z) = (1 - frac{z}{n})^n cdot 1_{z in [0, n]}.$$
Then
$$P(Z > z) = lim_{z to infty} P(n M_n > z) = cdots$$
and hopefully you recognize the resulting distribution.
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
$endgroup$
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
|
show 5 more comments
Your Answer
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1 Answer
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1 Answer
1
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oldest
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votes
$begingroup$
For each $n$, compute $P(nM_n > z)$.
$$P(nM_n > z) = (1 - frac{z}{n})^n cdot 1_{z in [0, n]}.$$
Then
$$P(Z > z) = lim_{z to infty} P(n M_n > z) = cdots$$
and hopefully you recognize the resulting distribution.
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
$endgroup$
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
|
show 5 more comments
$begingroup$
For each $n$, compute $P(nM_n > z)$.
$$P(nM_n > z) = (1 - frac{z}{n})^n cdot 1_{z in [0, n]}.$$
Then
$$P(Z > z) = lim_{z to infty} P(n M_n > z) = cdots$$
and hopefully you recognize the resulting distribution.
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
$endgroup$
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
|
show 5 more comments
$begingroup$
For each $n$, compute $P(nM_n > z)$.
$$P(nM_n > z) = (1 - frac{z}{n})^n cdot 1_{z in [0, n]}.$$
Then
$$P(Z > z) = lim_{z to infty} P(n M_n > z) = cdots$$
and hopefully you recognize the resulting distribution.
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
$endgroup$
For each $n$, compute $P(nM_n > z)$.
$$P(nM_n > z) = (1 - frac{z}{n})^n cdot 1_{z in [0, n]}.$$
Then
$$P(Z > z) = lim_{z to infty} P(n M_n > z) = cdots$$
and hopefully you recognize the resulting distribution.
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
answered Jan 14 at 21:01
angryavianangryavian
42.4k23481
42.4k23481
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
|
show 5 more comments
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
How did you get $P(nM_{n} > z)=(1-frac{z}{n})^{n}1_{z in [0,n]}$? The cdf of $P(M_{n} > frac{z}{n})=int_{ frac{z}{n}}^{infty}1dx$
$endgroup$
– SABOY
Jan 14 at 21:34
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
@SABOY $M_n$ is the minimum, it is not uniform on $[0,1]$.
$endgroup$
– angryavian
Jan 14 at 21:37
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
I corrected my question/answer. Is it correct?
$endgroup$
– SABOY
Jan 14 at 22:00
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
@SABOY "$Z sim exp(-z)$ where $z in [0, n]$" does not make sense for a number of reasons: is $exp(-z)$ a CDF? PDF? we usually don't use the $sim$ notation with a CDF or PDF anyway. Also, your answer should not depend on $n$ since you are taking the limit as $n to infty$. But you are on the right track.
$endgroup$
– angryavian
Jan 14 at 22:05
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
$begingroup$
No it's supposed to show $Z$ is exponentially distributed
$endgroup$
– SABOY
Jan 14 at 22:26
|
show 5 more comments
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$begingroup$
Check: stats.stackexchange.com/questions/18433/…
$endgroup$
– lightxbulb
Jan 14 at 20:52
$begingroup$
Unfortunately, since you do not have a sum/average, you cannot use central limit theorem here. Try computing the distribution function directly and take $n to infty$.
$endgroup$
– angryavian
Jan 14 at 20:58