Mixing
Convergence of maximum of random variables converging to zero
$begingroup$
Suppose that for a sample of identically distributed but non-independent uniformly bounded non-negative random variables we can show that $X_i = O_P(b_n)$ with $b_n$ a deterministic sequence converging to zero. Would it be possible to say something about the maximum of these random variables, namely $Y_n = max_{i leq n} X_i$? Intuitively one would expect that $Y_n = O_P(b_n)$ but with such things the devil is always in the details. Thank you.
probability convergence random-variables
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
$endgroup$
add a comment |
$begingroup$
Suppose that for a sample of identically distributed but non-independent uniformly bounded non-negative random variables we can show that $X_i = O_P(b_n)$ with $b_n$ a deterministic sequence converging to zero. Would it be possible to say something about the maximum of these random variables, namely $Y_n = max_{i leq n} X_i$? Intuitively one would expect that $Y_n = O_P(b_n)$ but with such things the devil is always in the details. Thank you.
probability convergence random-variables
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
$endgroup$
$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30
add a comment |
$begingroup$
Suppose that for a sample of identically distributed but non-independent uniformly bounded non-negative random variables we can show that $X_i = O_P(b_n)$ with $b_n$ a deterministic sequence converging to zero. Would it be possible to say something about the maximum of these random variables, namely $Y_n = max_{i leq n} X_i$? Intuitively one would expect that $Y_n = O_P(b_n)$ but with such things the devil is always in the details. Thank you.
probability convergence random-variables
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
$endgroup$
Suppose that for a sample of identically distributed but non-independent uniformly bounded non-negative random variables we can show that $X_i = O_P(b_n)$ with $b_n$ a deterministic sequence converging to zero. Would it be possible to say something about the maximum of these random variables, namely $Y_n = max_{i leq n} X_i$? Intuitively one would expect that $Y_n = O_P(b_n)$ but with such things the devil is always in the details. Thank you.
probability convergence random-variables
probability convergence random-variables
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
edited Feb 4 at 14:32
JohnK
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
asked Jan 31 at 12:50
JohnKJohnK
2,88211739
2,88211739
$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30
add a comment |
$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30
$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30
add a comment |
1 Answer
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$begingroup$
Still not completely sure about your question setting but here's a shot:
For any $n$, the following are true:
There are random variables $X_{i,n}$ for $i in {1,2,...,n}$.
Every $X_{i,n}$ is uniformly bounded and non-negative.
$forall i neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.
$X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} over b_n}$ is probabilistically bounded for large enough $n$.
Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.
For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $forall epsilon>0, delta>0: exists N: forall n > N: Prob(X_{i,n}/b_n > delta) < epsilon$.
If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $forall n: Y_n equiv 1$ since exactly one of the $X_{i,n} = 1$.
== ADDENDUM ==
In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 iff forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1over n})^n approx {1 over e}$ and $Prob(Y_n=1) approx 1 - {1 over e}$ and does not tend to $0$.
answered Feb 8 at 17:25
antkamantkam
2,737312
$endgroup$
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Still not completely sure about your question setting but here's a shot:
For any $n$, the following are true:
There are random variables $X_{i,n}$ for $i in {1,2,...,n}$.
Every $X_{i,n}$ is uniformly bounded and non-negative.
$forall i neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.
$X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} over b_n}$ is probabilistically bounded for large enough $n$.
Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.
For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $forall epsilon>0, delta>0: exists N: forall n > N: Prob(X_{i,n}/b_n > delta) < epsilon$.
If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $forall n: Y_n equiv 1$ since exactly one of the $X_{i,n} = 1$.
== ADDENDUM ==
In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 iff forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1over n})^n approx {1 over e}$ and $Prob(Y_n=1) approx 1 - {1 over e}$ and does not tend to $0$.
answered Feb 8 at 17:25
antkamantkam
2,737312
$endgroup$
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
add a comment |
$begingroup$
Still not completely sure about your question setting but here's a shot:
For any $n$, the following are true:
There are random variables $X_{i,n}$ for $i in {1,2,...,n}$.
Every $X_{i,n}$ is uniformly bounded and non-negative.
$forall i neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.
$X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} over b_n}$ is probabilistically bounded for large enough $n$.
Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.
For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $forall epsilon>0, delta>0: exists N: forall n > N: Prob(X_{i,n}/b_n > delta) < epsilon$.
If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $forall n: Y_n equiv 1$ since exactly one of the $X_{i,n} = 1$.
== ADDENDUM ==
In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 iff forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1over n})^n approx {1 over e}$ and $Prob(Y_n=1) approx 1 - {1 over e}$ and does not tend to $0$.
answered Feb 8 at 17:25
antkamantkam
2,737312
$endgroup$
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
add a comment |
$begingroup$
Still not completely sure about your question setting but here's a shot:
For any $n$, the following are true:
There are random variables $X_{i,n}$ for $i in {1,2,...,n}$.
Every $X_{i,n}$ is uniformly bounded and non-negative.
$forall i neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.
$X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} over b_n}$ is probabilistically bounded for large enough $n$.
Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.
For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $forall epsilon>0, delta>0: exists N: forall n > N: Prob(X_{i,n}/b_n > delta) < epsilon$.
If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $forall n: Y_n equiv 1$ since exactly one of the $X_{i,n} = 1$.
== ADDENDUM ==
In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 iff forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1over n})^n approx {1 over e}$ and $Prob(Y_n=1) approx 1 - {1 over e}$ and does not tend to $0$.
answered Feb 8 at 17:25
antkamantkam
2,737312
$endgroup$
Still not completely sure about your question setting but here's a shot:
For any $n$, the following are true:
There are random variables $X_{i,n}$ for $i in {1,2,...,n}$.
Every $X_{i,n}$ is uniformly bounded and non-negative.
$forall i neq j: X_{i,n}$ and $X_{j,n}$ are identically distributed, but not independent.
$X_{i,n} = O_P(b_n)$ or in other words, ${X_{i,n} over b_n}$ is probabilistically bounded for large enough $n$.
Am I correct above? If so, your conjecture is false, with this counterexample: For any $n$, imagine there are $n$ balls and you pick one of them uniformly at random, and $X_{i,n}$ is the indicator for picking the $i$th ball. I.e., $X_{i,n} = 1$ with probabililty $1/n$ and $= 0$ otherwise. These satisfy the first 3 bullets above.
For the 4th bullet, pick $b_n = 1/n$. Then $X_{i,n}/b_n$ is either $n$ (with prob $1/n$) or $0$ (otherwise). Thus $forall epsilon>0, delta>0: exists N: forall n > N: Prob(X_{i,n}/b_n > delta) < epsilon$.
If these $X_{i,n}$ satisfy your prerequisites, then it is a counterexample because $forall n: Y_n equiv 1$ since exactly one of the $X_{i,n} = 1$.
== ADDENDUM ==
In fact, a slightly modified example shows that independence wouldn't help. Consider $X_{i,n}$ binary with $Prob(X_{i,n} = 1) = 1/n$, and all the $X_{i,n}$ are independent. You still have $Y_n = 0 iff forall i: X_{i,n} = 0$, and so $Prob(Y_n=0) = (1 - {1over n})^n approx {1 over e}$ and $Prob(Y_n=1) approx 1 - {1 over e}$ and does not tend to $0$.
answered Feb 8 at 17:25
antkamantkam
2,737312
edited Feb 8 at 20:31
answered Feb 8 at 17:25
antkamantkam
2,737312
answered Feb 8 at 17:25
antkamantkam
2,737312
answered Feb 8 at 17:25
antkamantkam
2,737312
2,737312
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
add a comment |
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
That's a nice counterexample, thank you.
$endgroup$
– JohnK
Feb 11 at 8:53
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
$begingroup$
you're welcome, and indeed "the devil is always in the details" :)
$endgroup$
– antkam
Feb 11 at 12:31
add a comment |
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$begingroup$
Not sure I understand your question... (1) you mean $X_n = O_P(b_n)$, not $X_i = O_P(b_n)$, right? (2) My understanding is that $O_P(.)$ means $X_n/b_n$ is bounded (probabilistically, for large $n$). However you also said $X_n$ are identically distributed. So how can they be bounded to a sequence $b_n rightarrow 0$? Maybe I'm missing something?
$endgroup$
– antkam
Feb 6 at 20:24
$begingroup$
It is implied that the $X_i$s tend to zero at a rate faster than equal to $b_n$, this is pretty standard notation.
$endgroup$
– JohnK
Feb 8 at 15:15
$begingroup$
ok thanks for clarifying, but still, if $X_i$ are identically distributed, how can they also tend to zero (unless they are all zero to begin with)? also, if $X_i rightarrow 0$, then $Y_n$ would be dominated by the initial $X_i$ values, e.g. if the sequence achieves max at $i=5$ then $Y_n = X_5 forall n ge 5$, and so $Y_n$ will certainly not tend to 0.
$endgroup$
– antkam
Feb 8 at 15:20
$begingroup$
There is no contradiction, their distribution depends on $n$. So there is reason to believe that the maximum will also tend to zero.
$endgroup$
– JohnK
Feb 8 at 15:30