Mixing
A Question about Weak Convergence of a Ratio
$begingroup$
I have a question regarding convergence in distribution, which arose during one of my research works. First, let me describe the setup a bit:
$textbf{Description of the Problem}$:
Suppose that $X_n$ and $Y_n$ are two $textbf{independent}$ sequences of random variables, such that $sqrt{n} X_n xrightarrow{d} F$ and $sqrt{n}Y_n xrightarrow{d} F$, where $F$ is a continuous probability distribution. As a consequence, $frac{X_n}{Y_n} xrightarrow{d} frac{X}{Y}$, where $X$ and $Y$ are independent observations drawn from $F$.
Now, suppose that $W_n$ and $Z_n$ are two sequences of random variables such that the joint sequences $(X_n,W_n)$ and $(Y_n,Z_n)$ are independent with each other ($X_n$ and $W_n$ may be dependent, and similarly, $Y_n$ and $Z_n$ may be dependent). Also, assume that $X_n - W_n xrightarrow{d} 0$ and $Y_n - Z_n xrightarrow{d} 0$.
$textbf{My Question}$:
Does the above setting imply that:
$$frac{W_n}{Z_n}xrightarrow{d} frac{X}{Y}~?$$
In my research, $F$ is the standard normal distribution $N(0,1)$, so I would also be happy if you can prove the claim with this additional assumption. And in case the claim is false, are there other technical assumptions which make it true?
Many thanks.
probability-theory weak-convergence
asked Jan 2 at 9:16
UsermathUsermath
1306
$endgroup$
add a comment |
$begingroup$
I have a question regarding convergence in distribution, which arose during one of my research works. First, let me describe the setup a bit:
$textbf{Description of the Problem}$:
Suppose that $X_n$ and $Y_n$ are two $textbf{independent}$ sequences of random variables, such that $sqrt{n} X_n xrightarrow{d} F$ and $sqrt{n}Y_n xrightarrow{d} F$, where $F$ is a continuous probability distribution. As a consequence, $frac{X_n}{Y_n} xrightarrow{d} frac{X}{Y}$, where $X$ and $Y$ are independent observations drawn from $F$.
Now, suppose that $W_n$ and $Z_n$ are two sequences of random variables such that the joint sequences $(X_n,W_n)$ and $(Y_n,Z_n)$ are independent with each other ($X_n$ and $W_n$ may be dependent, and similarly, $Y_n$ and $Z_n$ may be dependent). Also, assume that $X_n - W_n xrightarrow{d} 0$ and $Y_n - Z_n xrightarrow{d} 0$.
$textbf{My Question}$:
Does the above setting imply that:
$$frac{W_n}{Z_n}xrightarrow{d} frac{X}{Y}~?$$
In my research, $F$ is the standard normal distribution $N(0,1)$, so I would also be happy if you can prove the claim with this additional assumption. And in case the claim is false, are there other technical assumptions which make it true?
Many thanks.
probability-theory weak-convergence
asked Jan 2 at 9:16
UsermathUsermath
1306
$endgroup$
add a comment |
$begingroup$
I have a question regarding convergence in distribution, which arose during one of my research works. First, let me describe the setup a bit:
$textbf{Description of the Problem}$:
Suppose that $X_n$ and $Y_n$ are two $textbf{independent}$ sequences of random variables, such that $sqrt{n} X_n xrightarrow{d} F$ and $sqrt{n}Y_n xrightarrow{d} F$, where $F$ is a continuous probability distribution. As a consequence, $frac{X_n}{Y_n} xrightarrow{d} frac{X}{Y}$, where $X$ and $Y$ are independent observations drawn from $F$.
Now, suppose that $W_n$ and $Z_n$ are two sequences of random variables such that the joint sequences $(X_n,W_n)$ and $(Y_n,Z_n)$ are independent with each other ($X_n$ and $W_n$ may be dependent, and similarly, $Y_n$ and $Z_n$ may be dependent). Also, assume that $X_n - W_n xrightarrow{d} 0$ and $Y_n - Z_n xrightarrow{d} 0$.
$textbf{My Question}$:
Does the above setting imply that:
$$frac{W_n}{Z_n}xrightarrow{d} frac{X}{Y}~?$$
In my research, $F$ is the standard normal distribution $N(0,1)$, so I would also be happy if you can prove the claim with this additional assumption. And in case the claim is false, are there other technical assumptions which make it true?
Many thanks.
probability-theory weak-convergence
asked Jan 2 at 9:16
UsermathUsermath
1306
$endgroup$
I have a question regarding convergence in distribution, which arose during one of my research works. First, let me describe the setup a bit:
$textbf{Description of the Problem}$:
Suppose that $X_n$ and $Y_n$ are two $textbf{independent}$ sequences of random variables, such that $sqrt{n} X_n xrightarrow{d} F$ and $sqrt{n}Y_n xrightarrow{d} F$, where $F$ is a continuous probability distribution. As a consequence, $frac{X_n}{Y_n} xrightarrow{d} frac{X}{Y}$, where $X$ and $Y$ are independent observations drawn from $F$.
Now, suppose that $W_n$ and $Z_n$ are two sequences of random variables such that the joint sequences $(X_n,W_n)$ and $(Y_n,Z_n)$ are independent with each other ($X_n$ and $W_n$ may be dependent, and similarly, $Y_n$ and $Z_n$ may be dependent). Also, assume that $X_n - W_n xrightarrow{d} 0$ and $Y_n - Z_n xrightarrow{d} 0$.
$textbf{My Question}$:
Does the above setting imply that:
$$frac{W_n}{Z_n}xrightarrow{d} frac{X}{Y}~?$$
In my research, $F$ is the standard normal distribution $N(0,1)$, so I would also be happy if you can prove the claim with this additional assumption. And in case the claim is false, are there other technical assumptions which make it true?
Many thanks.
probability-theory weak-convergence
probability-theory weak-convergence
asked Jan 2 at 9:16
UsermathUsermath
1306
asked Jan 2 at 9:16
UsermathUsermath
1306
asked Jan 2 at 9:16
UsermathUsermath
1306
asked Jan 2 at 9:16
UsermathUsermath
1306
asked Jan 2 at 9:16
UsermathUsermath
1306
1306
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Then one can show that (by e.g. Portmanteau lemma)
$$
sqrt{n}(X_n - W_n) to_d 0 Leftrightarrow sqrt{n}(X_n - W_n) to_p 0.
$$ Thus, one can conclude that
$$
sqrt{n}W_n = sqrt{n}X_n +sqrt{n} (W_n - X_n) to_d F.
$$ This is because $A_n to_d A$ and $B_n to_p c$ implies $(A_n,B_n) to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n to_d A+c$. In the same spirit, we have $Z_n to_d F$, and hence $(W_n, Z_n) to_d (X,Y)$ where $X,Ysim F$ and $Xperp !!!perp Y.$ Continuous mapping theorem gives
$$
frac{W_n}{Z_n}to_d frac{X}{Y}
$$as wanted.
answered Jan 2 at 9:26
SongSong
7,578423
$endgroup$
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
add a comment |
$begingroup$
Sorry for the confusion. I think one simple counterexample is $W_n = frac{1}{n}$ and $Z_n = frac{2}{n}$. Take $X_n$ and $Y_n$ to be the means of iid random variables with mean $0$ and finite variance.
answered Jan 2 at 9:38
UsermathUsermath
1306
$endgroup$
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
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oldest
votes
$begingroup$
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Then one can show that (by e.g. Portmanteau lemma)
$$
sqrt{n}(X_n - W_n) to_d 0 Leftrightarrow sqrt{n}(X_n - W_n) to_p 0.
$$ Thus, one can conclude that
$$
sqrt{n}W_n = sqrt{n}X_n +sqrt{n} (W_n - X_n) to_d F.
$$ This is because $A_n to_d A$ and $B_n to_p c$ implies $(A_n,B_n) to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n to_d A+c$. In the same spirit, we have $Z_n to_d F$, and hence $(W_n, Z_n) to_d (X,Y)$ where $X,Ysim F$ and $Xperp !!!perp Y.$ Continuous mapping theorem gives
$$
frac{W_n}{Z_n}to_d frac{X}{Y}
$$as wanted.
answered Jan 2 at 9:26
SongSong
7,578423
$endgroup$
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
add a comment |
$begingroup$
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Then one can show that (by e.g. Portmanteau lemma)
$$
sqrt{n}(X_n - W_n) to_d 0 Leftrightarrow sqrt{n}(X_n - W_n) to_p 0.
$$ Thus, one can conclude that
$$
sqrt{n}W_n = sqrt{n}X_n +sqrt{n} (W_n - X_n) to_d F.
$$ This is because $A_n to_d A$ and $B_n to_p c$ implies $(A_n,B_n) to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n to_d A+c$. In the same spirit, we have $Z_n to_d F$, and hence $(W_n, Z_n) to_d (X,Y)$ where $X,Ysim F$ and $Xperp !!!perp Y.$ Continuous mapping theorem gives
$$
frac{W_n}{Z_n}to_d frac{X}{Y}
$$as wanted.
answered Jan 2 at 9:26
SongSong
7,578423
$endgroup$
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
add a comment |
$begingroup$
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Then one can show that (by e.g. Portmanteau lemma)
$$
sqrt{n}(X_n - W_n) to_d 0 Leftrightarrow sqrt{n}(X_n - W_n) to_p 0.
$$ Thus, one can conclude that
$$
sqrt{n}W_n = sqrt{n}X_n +sqrt{n} (W_n - X_n) to_d F.
$$ This is because $A_n to_d A$ and $B_n to_p c$ implies $(A_n,B_n) to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n to_d A+c$. In the same spirit, we have $Z_n to_d F$, and hence $(W_n, Z_n) to_d (X,Y)$ where $X,Ysim F$ and $Xperp !!!perp Y.$ Continuous mapping theorem gives
$$
frac{W_n}{Z_n}to_d frac{X}{Y}
$$as wanted.
answered Jan 2 at 9:26
SongSong
7,578423
$endgroup$
The statement is false given your assumptions. A simple counterexample is already given. To induce the desired result, one can strengthen the assumption to $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Then one can show that (by e.g. Portmanteau lemma)
$$
sqrt{n}(X_n - W_n) to_d 0 Leftrightarrow sqrt{n}(X_n - W_n) to_p 0.
$$ Thus, one can conclude that
$$
sqrt{n}W_n = sqrt{n}X_n +sqrt{n} (W_n - X_n) to_d F.
$$ This is because $A_n to_d A$ and $B_n to_p c$ implies $(A_n,B_n) to_d (A,c)$. Then, continuous mapping theorem gives $A_n + B_n to_d A+c$. In the same spirit, we have $Z_n to_d F$, and hence $(W_n, Z_n) to_d (X,Y)$ where $X,Ysim F$ and $Xperp !!!perp Y.$ Continuous mapping theorem gives
$$
frac{W_n}{Z_n}to_d frac{X}{Y}
$$as wanted.
answered Jan 2 at 9:26
SongSong
7,578423
edited Jan 2 at 9:42
answered Jan 2 at 9:26
SongSong
7,578423
answered Jan 2 at 9:26
SongSong
7,578423
answered Jan 2 at 9:26
SongSong
7,578423
7,578423
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
add a comment |
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
$begingroup$
Thanks for your help, but note that I am assuming that $sqrt{n}X_n$ and $sqrt{n}Y_n$ are converging weakly to $F$, not $X_n$ and $Y_n$.
$endgroup$
– Usermath
Jan 2 at 9:32
1
1
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
$begingroup$
Oops, I misread it. Then you need stronger assumption that $sqrt{n}(W_n-X_n) to_d 0$ and $sqrt{n}(Z_n-Y_n) to_d 0$. Otherwise, It can even fail that $sqrt{n}W_n to_d F.$
$endgroup$
– Song
Jan 2 at 9:35
1
1
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
I'm saying that you should strengthen the assumption. A simple example shows that it fails given your original assumption. Take $W_n = 5X_n$ and $Z_n = 3Y_n$ for example.
$endgroup$
– Song
Jan 2 at 9:38
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
Thanks, Song! I got it.
$endgroup$
– Usermath
Jan 2 at 9:39
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
$begingroup$
I hope it will help you :)
$endgroup$
– Song
Jan 2 at 9:46
add a comment |
$begingroup$
Sorry for the confusion. I think one simple counterexample is $W_n = frac{1}{n}$ and $Z_n = frac{2}{n}$. Take $X_n$ and $Y_n$ to be the means of iid random variables with mean $0$ and finite variance.
answered Jan 2 at 9:38
UsermathUsermath
1306
$endgroup$
add a comment |
$begingroup$
Sorry for the confusion. I think one simple counterexample is $W_n = frac{1}{n}$ and $Z_n = frac{2}{n}$. Take $X_n$ and $Y_n$ to be the means of iid random variables with mean $0$ and finite variance.
answered Jan 2 at 9:38
UsermathUsermath
1306
$endgroup$
add a comment |
$begingroup$
Sorry for the confusion. I think one simple counterexample is $W_n = frac{1}{n}$ and $Z_n = frac{2}{n}$. Take $X_n$ and $Y_n$ to be the means of iid random variables with mean $0$ and finite variance.
answered Jan 2 at 9:38
UsermathUsermath
1306
$endgroup$
Sorry for the confusion. I think one simple counterexample is $W_n = frac{1}{n}$ and $Z_n = frac{2}{n}$. Take $X_n$ and $Y_n$ to be the means of iid random variables with mean $0$ and finite variance.
answered Jan 2 at 9:38
UsermathUsermath
1306
answered Jan 2 at 9:38
UsermathUsermath
1306
answered Jan 2 at 9:38
UsermathUsermath
1306
answered Jan 2 at 9:38
UsermathUsermath
1306
1306
add a comment |
add a comment |
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