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Law of large numbers for sequence of running minima of i.i.d. Uniform (0,1) random variables
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Let $(X_i)$ be i.i.d. Unif$(0,1)$. Define $M_n=minX_1,ldots,X_n$ and $T_n=sumlimits_k=1^n M_k$ for every $nge1$. Show that $$dfracT_nE(T_n)stackrelPto 1$$
The problem I am encountering is that the random variables $M_n$ are not independent or uncorrelated so $operatornameVar(T_n)$ does not seem to have a nice expression.
probability-theory uniform-distribution probability-limit-theorems law-of-large-numbers
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
asked Jul 24 at 7:37
Andrew Richards
1139
add a comment |Â
up vote
3
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Let $(X_i)$ be i.i.d. Unif$(0,1)$. Define $M_n=minX_1,ldots,X_n$ and $T_n=sumlimits_k=1^n M_k$ for every $nge1$. Show that $$dfracT_nE(T_n)stackrelPto 1$$
The problem I am encountering is that the random variables $M_n$ are not independent or uncorrelated so $operatornameVar(T_n)$ does not seem to have a nice expression.
probability-theory uniform-distribution probability-limit-theorems law-of-large-numbers
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
asked Jul 24 at 7:37
Andrew Richards
1139
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $(X_i)$ be i.i.d. Unif$(0,1)$. Define $M_n=minX_1,ldots,X_n$ and $T_n=sumlimits_k=1^n M_k$ for every $nge1$. Show that $$dfracT_nE(T_n)stackrelPto 1$$
The problem I am encountering is that the random variables $M_n$ are not independent or uncorrelated so $operatornameVar(T_n)$ does not seem to have a nice expression.
probability-theory uniform-distribution probability-limit-theorems law-of-large-numbers
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
asked Jul 24 at 7:37
Andrew Richards
1139
Let $(X_i)$ be i.i.d. Unif$(0,1)$. Define $M_n=minX_1,ldots,X_n$ and $T_n=sumlimits_k=1^n M_k$ for every $nge1$. Show that $$dfracT_nE(T_n)stackrelPto 1$$
The problem I am encountering is that the random variables $M_n$ are not independent or uncorrelated so $operatornameVar(T_n)$ does not seem to have a nice expression.
probability-theory uniform-distribution probability-limit-theorems law-of-large-numbers
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
asked Jul 24 at 7:37
Andrew Richards
1139
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
edited Jul 31 at 12:48
Davide Giraudo
121k15146249
121k15146249
asked Jul 24 at 7:37
Andrew Richards
1139
asked Jul 24 at 7:37
Andrew Richards
1139
asked Jul 24 at 7:37
Andrew Richards
1139
1139
add a comment |Â
add a comment |Â
1 Answer
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This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $mathbb L^2$ hence uniformly integrable, hence it should also converge in $mathbb L^1$ to $0$.
First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $nleft(1-uright)^n-1mathbf 1_(0,1)(u)$. As a consequence, $mathbb Eleft[T_nright]=sum_i=1^n1/(i+1)$ and it would suffices to prove that
$$lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[left(sum_i=1^nM_i-mathbb Eleft[M_iright]right)^2right]=0.
$$
To this aim, we would like to show that
$$tag1
lim_nto +infty
frac 1left(ln nright)^2sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=0;
$$
$$tag2lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[ sum_1leqslant ilt jleqslant nM_iM_j-mathbb Eleft[M_iM_jright] right]=0.
$$
Let us show (1). Using the expression for the density of $M_i$, we get
$$
mathbb Eleft[M_i^2right]=int_0^1u^2ileft(1-uright)^i-1mathrm du
=int_0^1left(1-vright)^2iv^i-1mathrm dv=1-2fracii+1+fracii+2
$$
which can be simplified as
$$
mathbb Eleft[M_i^2right]=frac 2i+1-frac 2i+2,
$$
hence $$sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=1-frac2n+2-sum_i=1^nfrac 1(i+1)^2leqslant 1,
$$
from which (1) follows.Let us look at (2). Observe that (2) can be rewritten as
$$
lim_nto +infty
frac 1left(ln nright)^2 sum_i=1^n sum_j=1^n-ic_i,j =0,
$$
where $c_i,j:=mathbb Eleft[M_iM_i+jright]-mathbb Eleft[M_iright]mathbb Eleft[M_i+jright] $.
Observing that $M_i+j$ has the same law as $minleftM_i,M'_jright$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get
$$
c_i,j=int_0^1int_0^1uminleftu,vrightileft(1-uright)^i-1jleft(1-vright)^j-1mathrm dumathrm dv-frac 1left(i+1right)left(j+1right).
$$
A computation shows that
$$
c_i,j=frac 1j+1left(fracii+j+2-fracii+j+1right),
$$
which can be transformed as
$$
c_i,j=frac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1+frac 1(j+1)(i+j+2).
$$
For the first two terms, we switch the sum and use telescoping in $i$ to find
$$sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1=sum_j=1^n-1frac 1j+1fracn-j+1n+2-frac 1j+1frac2j+3
$$
hence
$$
leftlvert sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1rightrvertleqslant 3sum_j=1^n-1frac 1j+1
$$
which is negligible with respect to $(ln n)^2$.
For the last term, it behaves like
$$sum_j=1^n-1sum_i=j+1^nfrac 1ij=sum_i=2^nfrac 1i sum_j=1^i-1frac 1j,$$
which is exactly the failure of (2), since the latter quantity is of order $(ln n)^2$.
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $mathbb L^2$ hence uniformly integrable, hence it should also converge in $mathbb L^1$ to $0$.
First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $nleft(1-uright)^n-1mathbf 1_(0,1)(u)$. As a consequence, $mathbb Eleft[T_nright]=sum_i=1^n1/(i+1)$ and it would suffices to prove that
$$lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[left(sum_i=1^nM_i-mathbb Eleft[M_iright]right)^2right]=0.
$$
To this aim, we would like to show that
$$tag1
lim_nto +infty
frac 1left(ln nright)^2sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=0;
$$
$$tag2lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[ sum_1leqslant ilt jleqslant nM_iM_j-mathbb Eleft[M_iM_jright] right]=0.
$$
Let us show (1). Using the expression for the density of $M_i$, we get
$$
mathbb Eleft[M_i^2right]=int_0^1u^2ileft(1-uright)^i-1mathrm du
=int_0^1left(1-vright)^2iv^i-1mathrm dv=1-2fracii+1+fracii+2
$$
which can be simplified as
$$
mathbb Eleft[M_i^2right]=frac 2i+1-frac 2i+2,
$$
hence $$sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=1-frac2n+2-sum_i=1^nfrac 1(i+1)^2leqslant 1,
$$
from which (1) follows.Let us look at (2). Observe that (2) can be rewritten as
$$
lim_nto +infty
frac 1left(ln nright)^2 sum_i=1^n sum_j=1^n-ic_i,j =0,
$$
where $c_i,j:=mathbb Eleft[M_iM_i+jright]-mathbb Eleft[M_iright]mathbb Eleft[M_i+jright] $.
Observing that $M_i+j$ has the same law as $minleftM_i,M'_jright$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get
$$
c_i,j=int_0^1int_0^1uminleftu,vrightileft(1-uright)^i-1jleft(1-vright)^j-1mathrm dumathrm dv-frac 1left(i+1right)left(j+1right).
$$
A computation shows that
$$
c_i,j=frac 1j+1left(fracii+j+2-fracii+j+1right),
$$
which can be transformed as
$$
c_i,j=frac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1+frac 1(j+1)(i+j+2).
$$
For the first two terms, we switch the sum and use telescoping in $i$ to find
$$sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1=sum_j=1^n-1frac 1j+1fracn-j+1n+2-frac 1j+1frac2j+3
$$
hence
$$
leftlvert sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1rightrvertleqslant 3sum_j=1^n-1frac 1j+1
$$
which is negligible with respect to $(ln n)^2$.
For the last term, it behaves like
$$sum_j=1^n-1sum_i=j+1^nfrac 1ij=sum_i=2^nfrac 1i sum_j=1^i-1frac 1j,$$
which is exactly the failure of (2), since the latter quantity is of order $(ln n)^2$.
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
add a comment |Â
up vote
1
down vote
This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $mathbb L^2$ hence uniformly integrable, hence it should also converge in $mathbb L^1$ to $0$.
First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $nleft(1-uright)^n-1mathbf 1_(0,1)(u)$. As a consequence, $mathbb Eleft[T_nright]=sum_i=1^n1/(i+1)$ and it would suffices to prove that
$$lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[left(sum_i=1^nM_i-mathbb Eleft[M_iright]right)^2right]=0.
$$
To this aim, we would like to show that
$$tag1
lim_nto +infty
frac 1left(ln nright)^2sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=0;
$$
$$tag2lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[ sum_1leqslant ilt jleqslant nM_iM_j-mathbb Eleft[M_iM_jright] right]=0.
$$
Let us show (1). Using the expression for the density of $M_i$, we get
$$
mathbb Eleft[M_i^2right]=int_0^1u^2ileft(1-uright)^i-1mathrm du
=int_0^1left(1-vright)^2iv^i-1mathrm dv=1-2fracii+1+fracii+2
$$
which can be simplified as
$$
mathbb Eleft[M_i^2right]=frac 2i+1-frac 2i+2,
$$
hence $$sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=1-frac2n+2-sum_i=1^nfrac 1(i+1)^2leqslant 1,
$$
from which (1) follows.Let us look at (2). Observe that (2) can be rewritten as
$$
lim_nto +infty
frac 1left(ln nright)^2 sum_i=1^n sum_j=1^n-ic_i,j =0,
$$
where $c_i,j:=mathbb Eleft[M_iM_i+jright]-mathbb Eleft[M_iright]mathbb Eleft[M_i+jright] $.
Observing that $M_i+j$ has the same law as $minleftM_i,M'_jright$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get
$$
c_i,j=int_0^1int_0^1uminleftu,vrightileft(1-uright)^i-1jleft(1-vright)^j-1mathrm dumathrm dv-frac 1left(i+1right)left(j+1right).
$$
A computation shows that
$$
c_i,j=frac 1j+1left(fracii+j+2-fracii+j+1right),
$$
which can be transformed as
$$
c_i,j=frac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1+frac 1(j+1)(i+j+2).
$$
For the first two terms, we switch the sum and use telescoping in $i$ to find
$$sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1=sum_j=1^n-1frac 1j+1fracn-j+1n+2-frac 1j+1frac2j+3
$$
hence
$$
leftlvert sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1rightrvertleqslant 3sum_j=1^n-1frac 1j+1
$$
which is negligible with respect to $(ln n)^2$.
For the last term, it behaves like
$$sum_j=1^n-1sum_i=j+1^nfrac 1ij=sum_i=2^nfrac 1i sum_j=1^i-1frac 1j,$$
which is exactly the failure of (2), since the latter quantity is of order $(ln n)^2$.
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $mathbb L^2$ hence uniformly integrable, hence it should also converge in $mathbb L^1$ to $0$.
First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $nleft(1-uright)^n-1mathbf 1_(0,1)(u)$. As a consequence, $mathbb Eleft[T_nright]=sum_i=1^n1/(i+1)$ and it would suffices to prove that
$$lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[left(sum_i=1^nM_i-mathbb Eleft[M_iright]right)^2right]=0.
$$
To this aim, we would like to show that
$$tag1
lim_nto +infty
frac 1left(ln nright)^2sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=0;
$$
$$tag2lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[ sum_1leqslant ilt jleqslant nM_iM_j-mathbb Eleft[M_iM_jright] right]=0.
$$
Let us show (1). Using the expression for the density of $M_i$, we get
$$
mathbb Eleft[M_i^2right]=int_0^1u^2ileft(1-uright)^i-1mathrm du
=int_0^1left(1-vright)^2iv^i-1mathrm dv=1-2fracii+1+fracii+2
$$
which can be simplified as
$$
mathbb Eleft[M_i^2right]=frac 2i+1-frac 2i+2,
$$
hence $$sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=1-frac2n+2-sum_i=1^nfrac 1(i+1)^2leqslant 1,
$$
from which (1) follows.Let us look at (2). Observe that (2) can be rewritten as
$$
lim_nto +infty
frac 1left(ln nright)^2 sum_i=1^n sum_j=1^n-ic_i,j =0,
$$
where $c_i,j:=mathbb Eleft[M_iM_i+jright]-mathbb Eleft[M_iright]mathbb Eleft[M_i+jright] $.
Observing that $M_i+j$ has the same law as $minleftM_i,M'_jright$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get
$$
c_i,j=int_0^1int_0^1uminleftu,vrightileft(1-uright)^i-1jleft(1-vright)^j-1mathrm dumathrm dv-frac 1left(i+1right)left(j+1right).
$$
A computation shows that
$$
c_i,j=frac 1j+1left(fracii+j+2-fracii+j+1right),
$$
which can be transformed as
$$
c_i,j=frac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1+frac 1(j+1)(i+j+2).
$$
For the first two terms, we switch the sum and use telescoping in $i$ to find
$$sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1=sum_j=1^n-1frac 1j+1fracn-j+1n+2-frac 1j+1frac2j+3
$$
hence
$$
leftlvert sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1rightrvertleqslant 3sum_j=1^n-1frac 1j+1
$$
which is negligible with respect to $(ln n)^2$.
For the last term, it behaves like
$$sum_j=1^n-1sum_i=j+1^nfrac 1ij=sum_i=2^nfrac 1i sum_j=1^i-1frac 1j,$$
which is exactly the failure of (2), since the latter quantity is of order $(ln n)^2$.
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
This is not the answer; it is just to save the time of a person who would like to try the approach with correlation. It turns out that the sequence we want to prove the convergence in probability is bounded in $mathbb L^2$ hence uniformly integrable, hence it should also converge in $mathbb L^1$ to $0$.
First let us extract some useful information from this thread: the expectation of $M_n$ is $1/(n+1)$ and its density is $nleft(1-uright)^n-1mathbf 1_(0,1)(u)$. As a consequence, $mathbb Eleft[T_nright]=sum_i=1^n1/(i+1)$ and it would suffices to prove that
$$lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[left(sum_i=1^nM_i-mathbb Eleft[M_iright]right)^2right]=0.
$$
To this aim, we would like to show that
$$tag1
lim_nto +infty
frac 1left(ln nright)^2sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=0;
$$
$$tag2lim_nto +infty
frac 1left(ln nright)^2mathbb Eleft[ sum_1leqslant ilt jleqslant nM_iM_j-mathbb Eleft[M_iM_jright] right]=0.
$$
Let us show (1). Using the expression for the density of $M_i$, we get
$$
mathbb Eleft[M_i^2right]=int_0^1u^2ileft(1-uright)^i-1mathrm du
=int_0^1left(1-vright)^2iv^i-1mathrm dv=1-2fracii+1+fracii+2
$$
which can be simplified as
$$
mathbb Eleft[M_i^2right]=frac 2i+1-frac 2i+2,
$$
hence $$sum_i=1^nmathbb Eleft[left(M_i-mathbb Eleft[M_iright]right)^2right]=1-frac2n+2-sum_i=1^nfrac 1(i+1)^2leqslant 1,
$$
from which (1) follows.Let us look at (2). Observe that (2) can be rewritten as
$$
lim_nto +infty
frac 1left(ln nright)^2 sum_i=1^n sum_j=1^n-ic_i,j =0,
$$
where $c_i,j:=mathbb Eleft[M_iM_i+jright]-mathbb Eleft[M_iright]mathbb Eleft[M_i+jright] $.
Observing that $M_i+j$ has the same law as $minleftM_i,M'_jright$, where $M'_j$ has the same law as $M_j$ and is independent of $M_i$, we get
$$
c_i,j=int_0^1int_0^1uminleftu,vrightileft(1-uright)^i-1jleft(1-vright)^j-1mathrm dumathrm dv-frac 1left(i+1right)left(j+1right).
$$
A computation shows that
$$
c_i,j=frac 1j+1left(fracii+j+2-fracii+j+1right),
$$
which can be transformed as
$$
c_i,j=frac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1+frac 1(j+1)(i+j+2).
$$
For the first two terms, we switch the sum and use telescoping in $i$ to find
$$sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1=sum_j=1^n-1frac 1j+1fracn-j+1n+2-frac 1j+1frac2j+3
$$
hence
$$
leftlvert sum_j=1^n-1sum_i=1^n-jfrac 1j+1fraci+1i+j+2-frac 1j+1fracii+j+1rightrvertleqslant 3sum_j=1^n-1frac 1j+1
$$
which is negligible with respect to $(ln n)^2$.
For the last term, it behaves like
$$sum_j=1^n-1sum_i=j+1^nfrac 1ij=sum_i=2^nfrac 1i sum_j=1^i-1frac 1j,$$
which is exactly the failure of (2), since the latter quantity is of order $(ln n)^2$.
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
edited Jul 31 at 12:52
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
answered Jul 31 at 10:52
Davide Giraudo
121k15146249
121k15146249
add a comment |Â
add a comment |Â
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