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Sum of identically distributed but not independent Bernoulli's is non-uniform
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Let $X_1,X_2,dots$ denote a sequence of identically distributed, exchangeable, but not independent, Bernoulli$(p)$ random variables. If there exists $n>0$ such that
$$
sum_i=1^nX_i
$$
is not uniformly distributed on $0,1,dots,n$, how can we show that this implies
$$
sum_i=1^nX_i + X_n+1
$$
is not uniformly distributed on $0,1,dots,n+1$? If $p ne 1/2$ then the claim follows by expectations, so we can consider $p=1/2$.
probability sequences-and-series probability-theory random-variables
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jesterII
1,10411123
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up vote
6
down vote
favorite
Let $X_1,X_2,dots$ denote a sequence of identically distributed, exchangeable, but not independent, Bernoulli$(p)$ random variables. If there exists $n>0$ such that
$$
sum_i=1^nX_i
$$
is not uniformly distributed on $0,1,dots,n$, how can we show that this implies
$$
sum_i=1^nX_i + X_n+1
$$
is not uniformly distributed on $0,1,dots,n+1$? If $p ne 1/2$ then the claim follows by expectations, so we can consider $p=1/2$.
probability sequences-and-series probability-theory random-variables
asked 3 hours ago
jesterII
1,10411123
What is the source for this exercise?
– Did
1 hour ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $X_1,X_2,dots$ denote a sequence of identically distributed, exchangeable, but not independent, Bernoulli$(p)$ random variables. If there exists $n>0$ such that
$$
sum_i=1^nX_i
$$
is not uniformly distributed on $0,1,dots,n$, how can we show that this implies
$$
sum_i=1^nX_i + X_n+1
$$
is not uniformly distributed on $0,1,dots,n+1$? If $p ne 1/2$ then the claim follows by expectations, so we can consider $p=1/2$.
probability sequences-and-series probability-theory random-variables
asked 3 hours ago
jesterII
1,10411123
Let $X_1,X_2,dots$ denote a sequence of identically distributed, exchangeable, but not independent, Bernoulli$(p)$ random variables. If there exists $n>0$ such that
$$
sum_i=1^nX_i
$$
is not uniformly distributed on $0,1,dots,n$, how can we show that this implies
$$
sum_i=1^nX_i + X_n+1
$$
is not uniformly distributed on $0,1,dots,n+1$? If $p ne 1/2$ then the claim follows by expectations, so we can consider $p=1/2$.
probability sequences-and-series probability-theory random-variables
probability sequences-and-series probability-theory random-variables
asked 3 hours ago
jesterII
1,10411123
asked 3 hours ago
jesterII
1,10411123
edited 3 hours ago
asked 3 hours ago
jesterII
1,10411123
asked 3 hours ago
jesterII
1,10411123
asked 3 hours ago
jesterII
1,10411123
1,10411123
What is the source for this exercise?
– Did
1 hour ago
add a comment |Â
What is the source for this exercise?
– Did
1 hour ago
What is the source for this exercise?
– Did
1 hour ago
What is the source for this exercise?
– Did
1 hour ago
add a comment |Â
2 Answers
2
active
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up vote
3
down vote
First, some notations: for every $n$, let $X_1:n=(X_1,X_2,ldots,X_n)$, and, for every word $w=(w_1,w_2,ldots,w_n)$ in $0,1^n$, let $|w|=w_1+w_2+cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.
Assume that, for some $ngeqslant1$, $|X_1:n+1|$ is uniformly distributed on $0,1,ldots,n+1$.
Then, by exchangeability, for every word $w$ in $0,1^n+1$, $P(X_1:n+1=w)$ depends only on $|w|$. For each $k$ in $0,1,ldots,n+1$, there are $n+1choose k$ words $w$ in $0,1^n+1$ such that $|w|=k$, hence, for every word $w$ in $0,1^n+1$ such that $|w|=k$, $$P(X_1:n+1=w)=n+1choose k^-1P(|X_1:n+1|=k)=(n+2)^-1n+1choose k^-1$$
For every word $v$ in $0,1^n$, the event $[X_1:n=v]$ is the disjoint union of the events $[X_1:n=v,X_n+1=0]$ and $[X_1:n=v,X_n+1=1]$.
If $|v|=k$ for some $k$ in $0,1,ldots,n$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_1:n=v)=(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1$$
Now, it happens that $$(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1=(n+1)^-1nchoose k^-1tag$ast$$$ hence $$P(X_1:n=v)=(n+1)^-1nchoose k^-1$$ Summing these over the $nchoose k$ words $v$ in $0,1^n$ such that $|v|=k$, one gets, for every $k$ in $0,1,ldots,n$, $$P(|X_1:n|=k)=(n+1)^-1$$
Thus, if $|X_1:n+1|=X_1+X_2+cdots+X_n+1$ is uniformly distributed on $0,1,ldots,n+1$, then $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$.
By contraposition, this proves the desired statement -- and also that, as soon as $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$ for some $ngeqslant1$, then $|X_1:1|=X_1$ is uniformly distributed on $0,1$, and, again by exchangeability, every $X_n$ is uniformly distributed on $0,1$, that is, necessarily, $p=frac12$.
Exercise: Prove $(ast)$.
answered 2 hours ago
Did
243k23209445
add a comment |Â
up vote
1
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Here is a large hint/outline.
A good method is to prove the contrapositive. Letting $S_n=sum_i=1^n X_i$, assume that $S_n+1$ is uniformly distributed over $0,1,dots,n+1$, and prove that $S_n$ is uniform as well.
To prove this, use the decomposition
$$
P(S_n=k) = P(S_n=kcap X_n+1=0)+P(S_n=kcap X_n+1=1)tag*
$$
We need to somehow relate the events on the RHS to events of the form $S_n+1=h$, whose probabilities are known to be $frac1n+2$.
If we consider our sample space to be the set of sequences of $n+1$ zeroes and ones, then $S_n=kcap X_n+1=0$ consists of all $binomnk$ sequences ending in a $0$ and consisting of $k$ ones. By exchangeability, these events are all equally likely. In particular, letting $E$ be the event that $X_i=1$ for $i=1,2,dots,k$ and $X_i=0$ for $i=k+1,k+2,dots,n+1$ (the string $11dots100dots0$), then
$$
P(S_n=kcap X_n+1=0) = binomnkcdot P(E)
$$
On the other hand, $E$ is a subset of the event $S_n+1=k$, which consists of $binomn+1k$ equally likely sequences, so we also have
$$
frac1n+2=P(S_n+1=k)=binomn+1kcdot P(E)
$$
The last two equations allow you to eliminate $P(S_n=kcap X_n+1=0)$ in $(*)$. You can do something similar to eliminate $P(S_n=kcap X_n+1=1)$.
answered 2 hours ago
Mike Earnest
18.7k11950
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
First, some notations: for every $n$, let $X_1:n=(X_1,X_2,ldots,X_n)$, and, for every word $w=(w_1,w_2,ldots,w_n)$ in $0,1^n$, let $|w|=w_1+w_2+cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.
Assume that, for some $ngeqslant1$, $|X_1:n+1|$ is uniformly distributed on $0,1,ldots,n+1$.
Then, by exchangeability, for every word $w$ in $0,1^n+1$, $P(X_1:n+1=w)$ depends only on $|w|$. For each $k$ in $0,1,ldots,n+1$, there are $n+1choose k$ words $w$ in $0,1^n+1$ such that $|w|=k$, hence, for every word $w$ in $0,1^n+1$ such that $|w|=k$, $$P(X_1:n+1=w)=n+1choose k^-1P(|X_1:n+1|=k)=(n+2)^-1n+1choose k^-1$$
For every word $v$ in $0,1^n$, the event $[X_1:n=v]$ is the disjoint union of the events $[X_1:n=v,X_n+1=0]$ and $[X_1:n=v,X_n+1=1]$.
If $|v|=k$ for some $k$ in $0,1,ldots,n$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_1:n=v)=(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1$$
Now, it happens that $$(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1=(n+1)^-1nchoose k^-1tag$ast$$$ hence $$P(X_1:n=v)=(n+1)^-1nchoose k^-1$$ Summing these over the $nchoose k$ words $v$ in $0,1^n$ such that $|v|=k$, one gets, for every $k$ in $0,1,ldots,n$, $$P(|X_1:n|=k)=(n+1)^-1$$
Thus, if $|X_1:n+1|=X_1+X_2+cdots+X_n+1$ is uniformly distributed on $0,1,ldots,n+1$, then $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$.
By contraposition, this proves the desired statement -- and also that, as soon as $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$ for some $ngeqslant1$, then $|X_1:1|=X_1$ is uniformly distributed on $0,1$, and, again by exchangeability, every $X_n$ is uniformly distributed on $0,1$, that is, necessarily, $p=frac12$.
Exercise: Prove $(ast)$.
answered 2 hours ago
Did
243k23209445
add a comment |Â
up vote
3
down vote
First, some notations: for every $n$, let $X_1:n=(X_1,X_2,ldots,X_n)$, and, for every word $w=(w_1,w_2,ldots,w_n)$ in $0,1^n$, let $|w|=w_1+w_2+cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.
Assume that, for some $ngeqslant1$, $|X_1:n+1|$ is uniformly distributed on $0,1,ldots,n+1$.
Then, by exchangeability, for every word $w$ in $0,1^n+1$, $P(X_1:n+1=w)$ depends only on $|w|$. For each $k$ in $0,1,ldots,n+1$, there are $n+1choose k$ words $w$ in $0,1^n+1$ such that $|w|=k$, hence, for every word $w$ in $0,1^n+1$ such that $|w|=k$, $$P(X_1:n+1=w)=n+1choose k^-1P(|X_1:n+1|=k)=(n+2)^-1n+1choose k^-1$$
For every word $v$ in $0,1^n$, the event $[X_1:n=v]$ is the disjoint union of the events $[X_1:n=v,X_n+1=0]$ and $[X_1:n=v,X_n+1=1]$.
If $|v|=k$ for some $k$ in $0,1,ldots,n$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_1:n=v)=(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1$$
Now, it happens that $$(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1=(n+1)^-1nchoose k^-1tag$ast$$$ hence $$P(X_1:n=v)=(n+1)^-1nchoose k^-1$$ Summing these over the $nchoose k$ words $v$ in $0,1^n$ such that $|v|=k$, one gets, for every $k$ in $0,1,ldots,n$, $$P(|X_1:n|=k)=(n+1)^-1$$
Thus, if $|X_1:n+1|=X_1+X_2+cdots+X_n+1$ is uniformly distributed on $0,1,ldots,n+1$, then $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$.
By contraposition, this proves the desired statement -- and also that, as soon as $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$ for some $ngeqslant1$, then $|X_1:1|=X_1$ is uniformly distributed on $0,1$, and, again by exchangeability, every $X_n$ is uniformly distributed on $0,1$, that is, necessarily, $p=frac12$.
Exercise: Prove $(ast)$.
answered 2 hours ago
Did
243k23209445
add a comment |Â
up vote
3
down vote
up vote
3
down vote
First, some notations: for every $n$, let $X_1:n=(X_1,X_2,ldots,X_n)$, and, for every word $w=(w_1,w_2,ldots,w_n)$ in $0,1^n$, let $|w|=w_1+w_2+cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.
Assume that, for some $ngeqslant1$, $|X_1:n+1|$ is uniformly distributed on $0,1,ldots,n+1$.
Then, by exchangeability, for every word $w$ in $0,1^n+1$, $P(X_1:n+1=w)$ depends only on $|w|$. For each $k$ in $0,1,ldots,n+1$, there are $n+1choose k$ words $w$ in $0,1^n+1$ such that $|w|=k$, hence, for every word $w$ in $0,1^n+1$ such that $|w|=k$, $$P(X_1:n+1=w)=n+1choose k^-1P(|X_1:n+1|=k)=(n+2)^-1n+1choose k^-1$$
For every word $v$ in $0,1^n$, the event $[X_1:n=v]$ is the disjoint union of the events $[X_1:n=v,X_n+1=0]$ and $[X_1:n=v,X_n+1=1]$.
If $|v|=k$ for some $k$ in $0,1,ldots,n$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_1:n=v)=(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1$$
Now, it happens that $$(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1=(n+1)^-1nchoose k^-1tag$ast$$$ hence $$P(X_1:n=v)=(n+1)^-1nchoose k^-1$$ Summing these over the $nchoose k$ words $v$ in $0,1^n$ such that $|v|=k$, one gets, for every $k$ in $0,1,ldots,n$, $$P(|X_1:n|=k)=(n+1)^-1$$
Thus, if $|X_1:n+1|=X_1+X_2+cdots+X_n+1$ is uniformly distributed on $0,1,ldots,n+1$, then $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$.
By contraposition, this proves the desired statement -- and also that, as soon as $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$ for some $ngeqslant1$, then $|X_1:1|=X_1$ is uniformly distributed on $0,1$, and, again by exchangeability, every $X_n$ is uniformly distributed on $0,1$, that is, necessarily, $p=frac12$.
Exercise: Prove $(ast)$.
answered 2 hours ago
Did
243k23209445
First, some notations: for every $n$, let $X_1:n=(X_1,X_2,ldots,X_n)$, and, for every word $w=(w_1,w_2,ldots,w_n)$ in $0,1^n$, let $|w|=w_1+w_2+cdots+w_n$. Now, to the proof, which uses crucially the exchangeability hypothesis.
Assume that, for some $ngeqslant1$, $|X_1:n+1|$ is uniformly distributed on $0,1,ldots,n+1$.
Then, by exchangeability, for every word $w$ in $0,1^n+1$, $P(X_1:n+1=w)$ depends only on $|w|$. For each $k$ in $0,1,ldots,n+1$, there are $n+1choose k$ words $w$ in $0,1^n+1$ such that $|w|=k$, hence, for every word $w$ in $0,1^n+1$ such that $|w|=k$, $$P(X_1:n+1=w)=n+1choose k^-1P(|X_1:n+1|=k)=(n+2)^-1n+1choose k^-1$$
For every word $v$ in $0,1^n$, the event $[X_1:n=v]$ is the disjoint union of the events $[X_1:n=v,X_n+1=0]$ and $[X_1:n=v,X_n+1=1]$.
If $|v|=k$ for some $k$ in $0,1,ldots,n$, then $|v0|=k$ and $|v1|=k+1$, hence $$P(X_1:n=v)=(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1$$
Now, it happens that $$(n+2)^-1n+1choose k^-1+(n+2)^-1n+1choose k+1^-1=(n+1)^-1nchoose k^-1tag$ast$$$ hence $$P(X_1:n=v)=(n+1)^-1nchoose k^-1$$ Summing these over the $nchoose k$ words $v$ in $0,1^n$ such that $|v|=k$, one gets, for every $k$ in $0,1,ldots,n$, $$P(|X_1:n|=k)=(n+1)^-1$$
Thus, if $|X_1:n+1|=X_1+X_2+cdots+X_n+1$ is uniformly distributed on $0,1,ldots,n+1$, then $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$.
By contraposition, this proves the desired statement -- and also that, as soon as $|X_1:n|=X_1+X_2+cdots+X_n$ is uniformly distributed on $0,1,ldots,n$ for some $ngeqslant1$, then $|X_1:1|=X_1$ is uniformly distributed on $0,1$, and, again by exchangeability, every $X_n$ is uniformly distributed on $0,1$, that is, necessarily, $p=frac12$.
Exercise: Prove $(ast)$.
answered 2 hours ago
Did
243k23209445
edited 1 hour ago
answered 2 hours ago
Did
243k23209445
answered 2 hours ago
Did
243k23209445
answered 2 hours ago
Did
243k23209445
243k23209445
add a comment |Â
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up vote
1
down vote
Here is a large hint/outline.
A good method is to prove the contrapositive. Letting $S_n=sum_i=1^n X_i$, assume that $S_n+1$ is uniformly distributed over $0,1,dots,n+1$, and prove that $S_n$ is uniform as well.
To prove this, use the decomposition
$$
P(S_n=k) = P(S_n=kcap X_n+1=0)+P(S_n=kcap X_n+1=1)tag*
$$
We need to somehow relate the events on the RHS to events of the form $S_n+1=h$, whose probabilities are known to be $frac1n+2$.
If we consider our sample space to be the set of sequences of $n+1$ zeroes and ones, then $S_n=kcap X_n+1=0$ consists of all $binomnk$ sequences ending in a $0$ and consisting of $k$ ones. By exchangeability, these events are all equally likely. In particular, letting $E$ be the event that $X_i=1$ for $i=1,2,dots,k$ and $X_i=0$ for $i=k+1,k+2,dots,n+1$ (the string $11dots100dots0$), then
$$
P(S_n=kcap X_n+1=0) = binomnkcdot P(E)
$$
On the other hand, $E$ is a subset of the event $S_n+1=k$, which consists of $binomn+1k$ equally likely sequences, so we also have
$$
frac1n+2=P(S_n+1=k)=binomn+1kcdot P(E)
$$
The last two equations allow you to eliminate $P(S_n=kcap X_n+1=0)$ in $(*)$. You can do something similar to eliminate $P(S_n=kcap X_n+1=1)$.
answered 2 hours ago
Mike Earnest
18.7k11950
add a comment |Â
up vote
1
down vote
Here is a large hint/outline.
A good method is to prove the contrapositive. Letting $S_n=sum_i=1^n X_i$, assume that $S_n+1$ is uniformly distributed over $0,1,dots,n+1$, and prove that $S_n$ is uniform as well.
To prove this, use the decomposition
$$
P(S_n=k) = P(S_n=kcap X_n+1=0)+P(S_n=kcap X_n+1=1)tag*
$$
We need to somehow relate the events on the RHS to events of the form $S_n+1=h$, whose probabilities are known to be $frac1n+2$.
If we consider our sample space to be the set of sequences of $n+1$ zeroes and ones, then $S_n=kcap X_n+1=0$ consists of all $binomnk$ sequences ending in a $0$ and consisting of $k$ ones. By exchangeability, these events are all equally likely. In particular, letting $E$ be the event that $X_i=1$ for $i=1,2,dots,k$ and $X_i=0$ for $i=k+1,k+2,dots,n+1$ (the string $11dots100dots0$), then
$$
P(S_n=kcap X_n+1=0) = binomnkcdot P(E)
$$
On the other hand, $E$ is a subset of the event $S_n+1=k$, which consists of $binomn+1k$ equally likely sequences, so we also have
$$
frac1n+2=P(S_n+1=k)=binomn+1kcdot P(E)
$$
The last two equations allow you to eliminate $P(S_n=kcap X_n+1=0)$ in $(*)$. You can do something similar to eliminate $P(S_n=kcap X_n+1=1)$.
answered 2 hours ago
Mike Earnest
18.7k11950
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here is a large hint/outline.
A good method is to prove the contrapositive. Letting $S_n=sum_i=1^n X_i$, assume that $S_n+1$ is uniformly distributed over $0,1,dots,n+1$, and prove that $S_n$ is uniform as well.
To prove this, use the decomposition
$$
P(S_n=k) = P(S_n=kcap X_n+1=0)+P(S_n=kcap X_n+1=1)tag*
$$
We need to somehow relate the events on the RHS to events of the form $S_n+1=h$, whose probabilities are known to be $frac1n+2$.
If we consider our sample space to be the set of sequences of $n+1$ zeroes and ones, then $S_n=kcap X_n+1=0$ consists of all $binomnk$ sequences ending in a $0$ and consisting of $k$ ones. By exchangeability, these events are all equally likely. In particular, letting $E$ be the event that $X_i=1$ for $i=1,2,dots,k$ and $X_i=0$ for $i=k+1,k+2,dots,n+1$ (the string $11dots100dots0$), then
$$
P(S_n=kcap X_n+1=0) = binomnkcdot P(E)
$$
On the other hand, $E$ is a subset of the event $S_n+1=k$, which consists of $binomn+1k$ equally likely sequences, so we also have
$$
frac1n+2=P(S_n+1=k)=binomn+1kcdot P(E)
$$
The last two equations allow you to eliminate $P(S_n=kcap X_n+1=0)$ in $(*)$. You can do something similar to eliminate $P(S_n=kcap X_n+1=1)$.
answered 2 hours ago
Mike Earnest
18.7k11950
Here is a large hint/outline.
A good method is to prove the contrapositive. Letting $S_n=sum_i=1^n X_i$, assume that $S_n+1$ is uniformly distributed over $0,1,dots,n+1$, and prove that $S_n$ is uniform as well.
To prove this, use the decomposition
$$
P(S_n=k) = P(S_n=kcap X_n+1=0)+P(S_n=kcap X_n+1=1)tag*
$$
We need to somehow relate the events on the RHS to events of the form $S_n+1=h$, whose probabilities are known to be $frac1n+2$.
If we consider our sample space to be the set of sequences of $n+1$ zeroes and ones, then $S_n=kcap X_n+1=0$ consists of all $binomnk$ sequences ending in a $0$ and consisting of $k$ ones. By exchangeability, these events are all equally likely. In particular, letting $E$ be the event that $X_i=1$ for $i=1,2,dots,k$ and $X_i=0$ for $i=k+1,k+2,dots,n+1$ (the string $11dots100dots0$), then
$$
P(S_n=kcap X_n+1=0) = binomnkcdot P(E)
$$
On the other hand, $E$ is a subset of the event $S_n+1=k$, which consists of $binomn+1k$ equally likely sequences, so we also have
$$
frac1n+2=P(S_n+1=k)=binomn+1kcdot P(E)
$$
The last two equations allow you to eliminate $P(S_n=kcap X_n+1=0)$ in $(*)$. You can do something similar to eliminate $P(S_n=kcap X_n+1=1)$.
answered 2 hours ago
Mike Earnest
18.7k11950
answered 2 hours ago
Mike Earnest
18.7k11950
answered 2 hours ago
Mike Earnest
18.7k11950
answered 2 hours ago
Mike Earnest
18.7k11950
18.7k11950
add a comment |Â
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1 hour ago