Mixing
Find $(a_n)$ such that $frac1{a_n}sumlimits_{k=1}^nX_k^2to1$ in probability, for $(X_n)$ independent and...
$begingroup$
Let $X_1, X_2, ..., X_n, ...$ be independent random variables. Assume that for each $n$, the random variable $X_n$ is distributed uniformly on $[0,n]$. Find a sequence $a_n$ such that $(X_1^2 + ... + X_n^2) / a_n$ converges to $1$ in probability.
I honestly have no idea how to start this problem. The only things that come to mind is to make a Law of Large Numbers type argument or to try to brute force it by using the definition of convergence in probability.
probability-theory convergence law-of-large-numbers
edited Jan 28 at 12:02
Did
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, ..., X_n, ...$ be independent random variables. Assume that for each $n$, the random variable $X_n$ is distributed uniformly on $[0,n]$. Find a sequence $a_n$ such that $(X_1^2 + ... + X_n^2) / a_n$ converges to $1$ in probability.
I honestly have no idea how to start this problem. The only things that come to mind is to make a Law of Large Numbers type argument or to try to brute force it by using the definition of convergence in probability.
probability-theory convergence law-of-large-numbers
edited Jan 28 at 12:02
Did
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
$endgroup$
add a comment |
$begingroup$
Let $X_1, X_2, ..., X_n, ...$ be independent random variables. Assume that for each $n$, the random variable $X_n$ is distributed uniformly on $[0,n]$. Find a sequence $a_n$ such that $(X_1^2 + ... + X_n^2) / a_n$ converges to $1$ in probability.
I honestly have no idea how to start this problem. The only things that come to mind is to make a Law of Large Numbers type argument or to try to brute force it by using the definition of convergence in probability.
probability-theory convergence law-of-large-numbers
edited Jan 28 at 12:02
Did
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
$endgroup$
Let $X_1, X_2, ..., X_n, ...$ be independent random variables. Assume that for each $n$, the random variable $X_n$ is distributed uniformly on $[0,n]$. Find a sequence $a_n$ such that $(X_1^2 + ... + X_n^2) / a_n$ converges to $1$ in probability.
I honestly have no idea how to start this problem. The only things that come to mind is to make a Law of Large Numbers type argument or to try to brute force it by using the definition of convergence in probability.
probability-theory convergence law-of-large-numbers
probability-theory convergence law-of-large-numbers
edited Jan 28 at 12:02
Did
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
edited Jan 28 at 12:02
Did
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
edited Jan 28 at 12:02
Did
249k23226466
edited Jan 28 at 12:02
Did
249k23226466
edited Jan 28 at 12:02
Did
249k23226466
249k23226466
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
asked Aug 1 '13 at 17:30
ZelarethZelareth
973
973
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If ${Y_n}$ are non-negative random variables converging in distribution to $Y$, then $$mathbb E(Y)leqslant liminf_{nto infty}mathbb E(Y_n),$$
hence we should have
$$1leqslant liminf_{nto infty}frac 1{a_n}sum_{j=1}^nfrac{j^2}3.$$
If we take $a_n:=sum_{j=1}^nfrac{j^2}3$, we can compute
$$frac 1{a_n^2}mathbb Eleft(sum_{j=1}^n{X_j^2-mathbb E(X_j^2)}right)^2.$$
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
add a comment |
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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votes
$begingroup$
If ${Y_n}$ are non-negative random variables converging in distribution to $Y$, then $$mathbb E(Y)leqslant liminf_{nto infty}mathbb E(Y_n),$$
hence we should have
$$1leqslant liminf_{nto infty}frac 1{a_n}sum_{j=1}^nfrac{j^2}3.$$
If we take $a_n:=sum_{j=1}^nfrac{j^2}3$, we can compute
$$frac 1{a_n^2}mathbb Eleft(sum_{j=1}^n{X_j^2-mathbb E(X_j^2)}right)^2.$$
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
add a comment |
$begingroup$
If ${Y_n}$ are non-negative random variables converging in distribution to $Y$, then $$mathbb E(Y)leqslant liminf_{nto infty}mathbb E(Y_n),$$
hence we should have
$$1leqslant liminf_{nto infty}frac 1{a_n}sum_{j=1}^nfrac{j^2}3.$$
If we take $a_n:=sum_{j=1}^nfrac{j^2}3$, we can compute
$$frac 1{a_n^2}mathbb Eleft(sum_{j=1}^n{X_j^2-mathbb E(X_j^2)}right)^2.$$
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
add a comment |
$begingroup$
If ${Y_n}$ are non-negative random variables converging in distribution to $Y$, then $$mathbb E(Y)leqslant liminf_{nto infty}mathbb E(Y_n),$$
hence we should have
$$1leqslant liminf_{nto infty}frac 1{a_n}sum_{j=1}^nfrac{j^2}3.$$
If we take $a_n:=sum_{j=1}^nfrac{j^2}3$, we can compute
$$frac 1{a_n^2}mathbb Eleft(sum_{j=1}^n{X_j^2-mathbb E(X_j^2)}right)^2.$$
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
$endgroup$
If ${Y_n}$ are non-negative random variables converging in distribution to $Y$, then $$mathbb E(Y)leqslant liminf_{nto infty}mathbb E(Y_n),$$
hence we should have
$$1leqslant liminf_{nto infty}frac 1{a_n}sum_{j=1}^nfrac{j^2}3.$$
If we take $a_n:=sum_{j=1}^nfrac{j^2}3$, we can compute
$$frac 1{a_n^2}mathbb Eleft(sum_{j=1}^n{X_j^2-mathbb E(X_j^2)}right)^2.$$
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
answered Aug 1 '13 at 18:09
Davide GiraudoDavide Giraudo
128k17154268
128k17154268
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
add a comment |
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
1
1
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
Not sure the OP will be able to transform this (which might have been written in a hurry) into a solution. (Indeed, $a_n=n^3/9$ works.)
$endgroup$
– Did
Aug 1 '13 at 19:28
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
I understand the first two steps are valid by Fatou's lemma, but I'm still kind of confused by the third step or what that buys me.
$endgroup$
– Zelareth
Aug 1 '13 at 23:59
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
$begingroup$
Have you computed the term in the last line?
$endgroup$
– Davide Giraudo
Aug 2 '13 at 11:07
add a comment |
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