Mixing
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Find the limiting probability
Clash Royale CLAN TAG#URR8PPP
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Let $U_n$ be a sequence of i.i.d Uniform$(0,1)$ random variables and $a_k$ be a sequence of i.i.d random variables such that, $P(a_k = pm1) = frac12$ and are independent of the $U_n$. Define the quantities,
$$X_n = sumlimits_k=1^nU_ka_k, ;;;; Y_n = sumlimits_k=1^n(U_k^2 - 1/2)a_k$$
Find the limit, $limlimits_nrightarrowinftyPleft(|X_n| > Y_nsqrtfrac207right)$.
We can easily calculate that $$E[U_ka_k] = 0, ; E[U_k^2] = frac13$$ and similarly, $$E[(U_k^2-1/2)a_k] = 0, ; E[(U_k^2-1/2)^2] = frac760$$ So by the Central Limit Theorem, we get,
$$ frac1sqrtnX_n xrightarrowD N(0,1/3), ;;;; frac1sqrtnY_n xrightarrowD N(0,7/60)$$
Ideally, I can use something like the Continuous Mapping Theorem to deduce the limiting distribution of $dfracY_n$, but the question makes no additional assumptions on the joint distribution of $(X_n, Y_n)$. Is there another approach I can take from here?
EDIT: I went ahead and calculated the covariance of $U_ka_k$ and $(U_k^2-1/2)a_k$,
$$ E[U_ka_k(U_k^2-1/2)a_k] = E[U_k(U_k^2-1/2)] =int_0^1(u^3-1/2u)du = 0$$
since $a_k^2 = 1$ with probability $1$. Now we can apply the multivariate CLT like @LandonCarter says.
probability probability-distributions central-limit-theorem
asked yesterday
Flowsnake
1,4581217
add a comment |Â
up vote
1
down vote
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Let $U_n$ be a sequence of i.i.d Uniform$(0,1)$ random variables and $a_k$ be a sequence of i.i.d random variables such that, $P(a_k = pm1) = frac12$ and are independent of the $U_n$. Define the quantities,
$$X_n = sumlimits_k=1^nU_ka_k, ;;;; Y_n = sumlimits_k=1^n(U_k^2 - 1/2)a_k$$
Find the limit, $limlimits_nrightarrowinftyPleft(|X_n| > Y_nsqrtfrac207right)$.
We can easily calculate that $$E[U_ka_k] = 0, ; E[U_k^2] = frac13$$ and similarly, $$E[(U_k^2-1/2)a_k] = 0, ; E[(U_k^2-1/2)^2] = frac760$$ So by the Central Limit Theorem, we get,
$$ frac1sqrtnX_n xrightarrowD N(0,1/3), ;;;; frac1sqrtnY_n xrightarrowD N(0,7/60)$$
Ideally, I can use something like the Continuous Mapping Theorem to deduce the limiting distribution of $dfracY_n$, but the question makes no additional assumptions on the joint distribution of $(X_n, Y_n)$. Is there another approach I can take from here?
EDIT: I went ahead and calculated the covariance of $U_ka_k$ and $(U_k^2-1/2)a_k$,
$$ E[U_ka_k(U_k^2-1/2)a_k] = E[U_k(U_k^2-1/2)] =int_0^1(u^3-1/2u)du = 0$$
since $a_k^2 = 1$ with probability $1$. Now we can apply the multivariate CLT like @LandonCarter says.
probability probability-distributions central-limit-theorem
asked yesterday
Flowsnake
1,4581217
1
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $U_n$ be a sequence of i.i.d Uniform$(0,1)$ random variables and $a_k$ be a sequence of i.i.d random variables such that, $P(a_k = pm1) = frac12$ and are independent of the $U_n$. Define the quantities,
$$X_n = sumlimits_k=1^nU_ka_k, ;;;; Y_n = sumlimits_k=1^n(U_k^2 - 1/2)a_k$$
Find the limit, $limlimits_nrightarrowinftyPleft(|X_n| > Y_nsqrtfrac207right)$.
We can easily calculate that $$E[U_ka_k] = 0, ; E[U_k^2] = frac13$$ and similarly, $$E[(U_k^2-1/2)a_k] = 0, ; E[(U_k^2-1/2)^2] = frac760$$ So by the Central Limit Theorem, we get,
$$ frac1sqrtnX_n xrightarrowD N(0,1/3), ;;;; frac1sqrtnY_n xrightarrowD N(0,7/60)$$
Ideally, I can use something like the Continuous Mapping Theorem to deduce the limiting distribution of $dfracY_n$, but the question makes no additional assumptions on the joint distribution of $(X_n, Y_n)$. Is there another approach I can take from here?
EDIT: I went ahead and calculated the covariance of $U_ka_k$ and $(U_k^2-1/2)a_k$,
$$ E[U_ka_k(U_k^2-1/2)a_k] = E[U_k(U_k^2-1/2)] =int_0^1(u^3-1/2u)du = 0$$
since $a_k^2 = 1$ with probability $1$. Now we can apply the multivariate CLT like @LandonCarter says.
probability probability-distributions central-limit-theorem
asked yesterday
Flowsnake
1,4581217
Let $U_n$ be a sequence of i.i.d Uniform$(0,1)$ random variables and $a_k$ be a sequence of i.i.d random variables such that, $P(a_k = pm1) = frac12$ and are independent of the $U_n$. Define the quantities,
$$X_n = sumlimits_k=1^nU_ka_k, ;;;; Y_n = sumlimits_k=1^n(U_k^2 - 1/2)a_k$$
Find the limit, $limlimits_nrightarrowinftyPleft(|X_n| > Y_nsqrtfrac207right)$.
We can easily calculate that $$E[U_ka_k] = 0, ; E[U_k^2] = frac13$$ and similarly, $$E[(U_k^2-1/2)a_k] = 0, ; E[(U_k^2-1/2)^2] = frac760$$ So by the Central Limit Theorem, we get,
$$ frac1sqrtnX_n xrightarrowD N(0,1/3), ;;;; frac1sqrtnY_n xrightarrowD N(0,7/60)$$
Ideally, I can use something like the Continuous Mapping Theorem to deduce the limiting distribution of $dfracY_n$, but the question makes no additional assumptions on the joint distribution of $(X_n, Y_n)$. Is there another approach I can take from here?
EDIT: I went ahead and calculated the covariance of $U_ka_k$ and $(U_k^2-1/2)a_k$,
$$ E[U_ka_k(U_k^2-1/2)a_k] = E[U_k(U_k^2-1/2)] =int_0^1(u^3-1/2u)du = 0$$
since $a_k^2 = 1$ with probability $1$. Now we can apply the multivariate CLT like @LandonCarter says.
probability probability-distributions central-limit-theorem
asked yesterday
Flowsnake
1,4581217
edited 18 hours ago
asked yesterday
Flowsnake
1,4581217
asked yesterday
Flowsnake
1,4581217
asked yesterday
Flowsnake
1,4581217
1,4581217
1
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago
add a comment |Â
1
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago
1
1
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
By multivariate CLT, $$(X_n/sqrtn, Y_n/sqrtn)stackreldtoN((0,0),diag(1/3,7/60))$$so in the limit $X_n/sqrtn$ and $Y_n/sqrtn$ become jointly independent Normals.
So $P[|X_n|>Y_nsqrt20/7]=P[dfracY_n/sqrtn<sqrt7/20]to P[N_1/|N_2|<sqrt7/20]$ as $ntoinfty$ where $N_1sim N(0,7/60)$ and $N_2sim N(0,1/3)$ independent. Note $N_1=sqrt7/60Z_1$ and $N_2=sqrt1/3Z_2$ where $Z_1,Z_2sim N(0,1)$ iid.
So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.
answered yesterday
Landon Carter
7,13611540
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
By multivariate CLT, $$(X_n/sqrtn, Y_n/sqrtn)stackreldtoN((0,0),diag(1/3,7/60))$$so in the limit $X_n/sqrtn$ and $Y_n/sqrtn$ become jointly independent Normals.
So $P[|X_n|>Y_nsqrt20/7]=P[dfracY_n/sqrtn<sqrt7/20]to P[N_1/|N_2|<sqrt7/20]$ as $ntoinfty$ where $N_1sim N(0,7/60)$ and $N_2sim N(0,1/3)$ independent. Note $N_1=sqrt7/60Z_1$ and $N_2=sqrt1/3Z_2$ where $Z_1,Z_2sim N(0,1)$ iid.
So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.
answered yesterday
Landon Carter
7,13611540
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
add a comment |Â
up vote
2
down vote
accepted
By multivariate CLT, $$(X_n/sqrtn, Y_n/sqrtn)stackreldtoN((0,0),diag(1/3,7/60))$$so in the limit $X_n/sqrtn$ and $Y_n/sqrtn$ become jointly independent Normals.
So $P[|X_n|>Y_nsqrt20/7]=P[dfracY_n/sqrtn<sqrt7/20]to P[N_1/|N_2|<sqrt7/20]$ as $ntoinfty$ where $N_1sim N(0,7/60)$ and $N_2sim N(0,1/3)$ independent. Note $N_1=sqrt7/60Z_1$ and $N_2=sqrt1/3Z_2$ where $Z_1,Z_2sim N(0,1)$ iid.
So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.
answered yesterday
Landon Carter
7,13611540
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
By multivariate CLT, $$(X_n/sqrtn, Y_n/sqrtn)stackreldtoN((0,0),diag(1/3,7/60))$$so in the limit $X_n/sqrtn$ and $Y_n/sqrtn$ become jointly independent Normals.
So $P[|X_n|>Y_nsqrt20/7]=P[dfracY_n/sqrtn<sqrt7/20]to P[N_1/|N_2|<sqrt7/20]$ as $ntoinfty$ where $N_1sim N(0,7/60)$ and $N_2sim N(0,1/3)$ independent. Note $N_1=sqrt7/60Z_1$ and $N_2=sqrt1/3Z_2$ where $Z_1,Z_2sim N(0,1)$ iid.
So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.
answered yesterday
Landon Carter
7,13611540
By multivariate CLT, $$(X_n/sqrtn, Y_n/sqrtn)stackreldtoN((0,0),diag(1/3,7/60))$$so in the limit $X_n/sqrtn$ and $Y_n/sqrtn$ become jointly independent Normals.
So $P[|X_n|>Y_nsqrt20/7]=P[dfracY_n/sqrtn<sqrt7/20]to P[N_1/|N_2|<sqrt7/20]$ as $ntoinfty$ where $N_1sim N(0,7/60)$ and $N_2sim N(0,1/3)$ independent. Note $N_1=sqrt7/60Z_1$ and $N_2=sqrt1/3Z_2$ where $Z_1,Z_2sim N(0,1)$ iid.
So the limiting probability (after some simplification) is $P[Z_1/|Z_2|<1]$. Maybe you can simplify this using the fact that $Z_1/Z_2$ is $Cauchy (0,1)$.
answered yesterday
Landon Carter
7,13611540
edited yesterday
answered yesterday
Landon Carter
7,13611540
answered yesterday
Landon Carter
7,13611540
answered yesterday
Landon Carter
7,13611540
7,13611540
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
add a comment |Â
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
Ah, I was a bit confused as to why the limiting joint distribution was independent. Then I calculated that the covariance between $U_ka_k$ and $(U_k^2-1/2)a_k$ is $0$. Thank you.
– Flowsnake
18 hours ago
1
1
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
By the way note that you can indeed simplify this as follows. Observe $T=Z_1/|Z_2|$ is symmetrical about 0 so the final limiting probability is $P[T<1]=1/2+P[0<T<1]=0.5(1+P(|T|<1))$ and now use that T is standard Cauchy.
– Landon Carter
17 hours ago
add a comment |Â
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1
How could there be any assumptions on the joint distribution of $(X_n,Y_n)$? Can't you calculate it? Every piece of information is given
– mathworker21
yesterday
Are the $U_n$ independent of the $a_k$?
– angryavian
yesterday
Yes sorry I forgot to add that assumption. The question has been edited.
– Flowsnake
yesterday
@mathworker21 Sorry I just realized that now. One can easily calculate the covariance between the $U_ka_k$ and $(U_k^2-1/2)a_k$ to be $0$. Thank you.
– Flowsnake
18 hours ago