Mixing
Rate of convergence of the product of two random variable sequences
Clash Royale CLAN TAG#URR8PPP
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Given that
$$forall epsilon_1 exists delta_epsilon_1>0, N_epsilon_1>0,text s.t. Pr n^alpha <epsilon_1 forall n>N_epsilon_1$$
$$forall epsilon_2 exists delta_epsilon_2>0, N_epsilon_2>0,text s.t. Pr n^beta <epsilon_2 forall n>N_epsilon_2 $$
How to prove that
$$forall epsilon exists delta_epsilon>0, N_epsilon>0,text s.t. Pr n^alpha+beta <epsilon forall n>N_epsilon $$
In other word, can we get $X_nY_n=O_p(n^-(alpha+beta))$ based on $X_n=O_p(n^-alpha)$ and $Y_n=O_p(n^-beta)$? what conditions should be added? Thanks!
probability statistics asymptotics concentration-of-measure
asked Jul 23 at 23:54
Carlton Chen
62
add a comment |Â
up vote
1
down vote
favorite
Given that
$$forall epsilon_1 exists delta_epsilon_1>0, N_epsilon_1>0,text s.t. Pr n^alpha <epsilon_1 forall n>N_epsilon_1$$
$$forall epsilon_2 exists delta_epsilon_2>0, N_epsilon_2>0,text s.t. Pr n^beta <epsilon_2 forall n>N_epsilon_2 $$
How to prove that
$$forall epsilon exists delta_epsilon>0, N_epsilon>0,text s.t. Pr n^alpha+beta <epsilon forall n>N_epsilon $$
In other word, can we get $X_nY_n=O_p(n^-(alpha+beta))$ based on $X_n=O_p(n^-alpha)$ and $Y_n=O_p(n^-beta)$? what conditions should be added? Thanks!
probability statistics asymptotics concentration-of-measure
asked Jul 23 at 23:54
Carlton Chen
62
1
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
1
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given that
$$forall epsilon_1 exists delta_epsilon_1>0, N_epsilon_1>0,text s.t. Pr n^alpha <epsilon_1 forall n>N_epsilon_1$$
$$forall epsilon_2 exists delta_epsilon_2>0, N_epsilon_2>0,text s.t. Pr n^beta <epsilon_2 forall n>N_epsilon_2 $$
How to prove that
$$forall epsilon exists delta_epsilon>0, N_epsilon>0,text s.t. Pr n^alpha+beta <epsilon forall n>N_epsilon $$
In other word, can we get $X_nY_n=O_p(n^-(alpha+beta))$ based on $X_n=O_p(n^-alpha)$ and $Y_n=O_p(n^-beta)$? what conditions should be added? Thanks!
probability statistics asymptotics concentration-of-measure
asked Jul 23 at 23:54
Carlton Chen
62
Given that
$$forall epsilon_1 exists delta_epsilon_1>0, N_epsilon_1>0,text s.t. Pr n^alpha <epsilon_1 forall n>N_epsilon_1$$
$$forall epsilon_2 exists delta_epsilon_2>0, N_epsilon_2>0,text s.t. Pr n^beta <epsilon_2 forall n>N_epsilon_2 $$
How to prove that
$$forall epsilon exists delta_epsilon>0, N_epsilon>0,text s.t. Pr n^alpha+beta <epsilon forall n>N_epsilon $$
In other word, can we get $X_nY_n=O_p(n^-(alpha+beta))$ based on $X_n=O_p(n^-alpha)$ and $Y_n=O_p(n^-beta)$? what conditions should be added? Thanks!
probability statistics asymptotics concentration-of-measure
asked Jul 23 at 23:54
Carlton Chen
62
edited Jul 24 at 9:55
asked Jul 23 at 23:54
Carlton Chen
62
asked Jul 23 at 23:54
Carlton Chen
62
asked Jul 23 at 23:54
Carlton Chen
62
62
1
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
1
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52
add a comment |Â
1
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
1
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52
1
1
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
1
1
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Thank you @Michael. I guess the proof shuold be: $forall epsilon_1<fracepsilon2, epsilon_2<fracepsilon2, exists delta_1gedelta_epsilon_1gesqrtdelta, delta_2gedelta_epsilon_2gesqrtdelta,
N_epsilon_1,delta_1ge 0, N_epsilon_2,delta_2ge 0$ s.t.
beginalign*
P n^alpha+beta &le P ge delta_1 + P n^beta \
&le fracepsilon2 + fracepsilon2 \
&= epsilon textfor forall n>maxN_epsilon_1,delta_1, N_epsilon_2,delta_2
endalign*
Am I right?
answered Jul 24 at 13:54
Carlton Chen
62
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Thank you @Michael. I guess the proof shuold be: $forall epsilon_1<fracepsilon2, epsilon_2<fracepsilon2, exists delta_1gedelta_epsilon_1gesqrtdelta, delta_2gedelta_epsilon_2gesqrtdelta,
N_epsilon_1,delta_1ge 0, N_epsilon_2,delta_2ge 0$ s.t.
beginalign*
P n^alpha+beta &le P ge delta_1 + P n^beta \
&le fracepsilon2 + fracepsilon2 \
&= epsilon textfor forall n>maxN_epsilon_1,delta_1, N_epsilon_2,delta_2
endalign*
Am I right?
answered Jul 24 at 13:54
Carlton Chen
62
add a comment |Â
up vote
0
down vote
Thank you @Michael. I guess the proof shuold be: $forall epsilon_1<fracepsilon2, epsilon_2<fracepsilon2, exists delta_1gedelta_epsilon_1gesqrtdelta, delta_2gedelta_epsilon_2gesqrtdelta,
N_epsilon_1,delta_1ge 0, N_epsilon_2,delta_2ge 0$ s.t.
beginalign*
P n^alpha+beta &le P ge delta_1 + P n^beta \
&le fracepsilon2 + fracepsilon2 \
&= epsilon textfor forall n>maxN_epsilon_1,delta_1, N_epsilon_2,delta_2
endalign*
Am I right?
answered Jul 24 at 13:54
Carlton Chen
62
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Thank you @Michael. I guess the proof shuold be: $forall epsilon_1<fracepsilon2, epsilon_2<fracepsilon2, exists delta_1gedelta_epsilon_1gesqrtdelta, delta_2gedelta_epsilon_2gesqrtdelta,
N_epsilon_1,delta_1ge 0, N_epsilon_2,delta_2ge 0$ s.t.
beginalign*
P n^alpha+beta &le P ge delta_1 + P n^beta \
&le fracepsilon2 + fracepsilon2 \
&= epsilon textfor forall n>maxN_epsilon_1,delta_1, N_epsilon_2,delta_2
endalign*
Am I right?
answered Jul 24 at 13:54
Carlton Chen
62
Thank you @Michael. I guess the proof shuold be: $forall epsilon_1<fracepsilon2, epsilon_2<fracepsilon2, exists delta_1gedelta_epsilon_1gesqrtdelta, delta_2gedelta_epsilon_2gesqrtdelta,
N_epsilon_1,delta_1ge 0, N_epsilon_2,delta_2ge 0$ s.t.
beginalign*
P n^alpha+beta &le P ge delta_1 + P n^beta \
&le fracepsilon2 + fracepsilon2 \
&= epsilon textfor forall n>maxN_epsilon_1,delta_1, N_epsilon_2,delta_2
endalign*
Am I right?
answered Jul 24 at 13:54
Carlton Chen
62
answered Jul 24 at 13:54
Carlton Chen
62
answered Jul 24 at 13:54
Carlton Chen
62
answered Jul 24 at 13:54
Carlton Chen
62
62
add a comment |Â
add a comment |Â
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1
Likely the second inequality should have $epsilon_2$ inside the $Pr$. Also, should that have $Y_n$ instead of $X_n$? Else, $Y_n$ is not introduced anywhere.
– Michael
Jul 23 at 23:56
Yes, I will correct it, thanks!
– Carlton Chen
Jul 23 at 23:59
1
What if you use the following (assuming $delta>0$): $$>deltasubseteq >sqrtdelta cup $$
– Michael
Jul 24 at 0:00
You didn't actually correct it; it still says $X_n$ in the second inequality.
– joriki
Jul 24 at 3:30
Thank you @Michael, I guess the complete proof shoyld be:
– Carlton Chen
Jul 24 at 13:52