Mixing
Conditional expectation and sampling without replacement
Clash Royale CLAN TAG#URR8PPP
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Suppose I have $N$ numbers: $x_1,cdots,x_N$ such that $N$ is even and $sum_i=1^N x_i = 0$. Let $X_j$ be the $j$-th number I get when I sample from $leftx_1,cdots,x_Nright$ without replacement. How do I find
$$mathbbEleft(X_k+1 + cdots + X_M mid X_k, X_k-1,cdots,X_1right),?$$
I can find $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$, but I don't know how to find the rest conditional expectations. For example, given $X_k,cdots,X_1$, how do I know the distribution $X_k+2$? I feel like it would have something to do with $X_k+1$...
probability combinatorics conditional-expectation
asked Aug 1 at 3:29
Yuki Kawabata
30419
add a comment |Â
up vote
0
down vote
favorite
Suppose I have $N$ numbers: $x_1,cdots,x_N$ such that $N$ is even and $sum_i=1^N x_i = 0$. Let $X_j$ be the $j$-th number I get when I sample from $leftx_1,cdots,x_Nright$ without replacement. How do I find
$$mathbbEleft(X_k+1 + cdots + X_M mid X_k, X_k-1,cdots,X_1right),?$$
I can find $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$, but I don't know how to find the rest conditional expectations. For example, given $X_k,cdots,X_1$, how do I know the distribution $X_k+2$? I feel like it would have something to do with $X_k+1$...
probability combinatorics conditional-expectation
asked Aug 1 at 3:29
Yuki Kawabata
30419
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose I have $N$ numbers: $x_1,cdots,x_N$ such that $N$ is even and $sum_i=1^N x_i = 0$. Let $X_j$ be the $j$-th number I get when I sample from $leftx_1,cdots,x_Nright$ without replacement. How do I find
$$mathbbEleft(X_k+1 + cdots + X_M mid X_k, X_k-1,cdots,X_1right),?$$
I can find $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$, but I don't know how to find the rest conditional expectations. For example, given $X_k,cdots,X_1$, how do I know the distribution $X_k+2$? I feel like it would have something to do with $X_k+1$...
probability combinatorics conditional-expectation
asked Aug 1 at 3:29
Yuki Kawabata
30419
Suppose I have $N$ numbers: $x_1,cdots,x_N$ such that $N$ is even and $sum_i=1^N x_i = 0$. Let $X_j$ be the $j$-th number I get when I sample from $leftx_1,cdots,x_Nright$ without replacement. How do I find
$$mathbbEleft(X_k+1 + cdots + X_M mid X_k, X_k-1,cdots,X_1right),?$$
I can find $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$, but I don't know how to find the rest conditional expectations. For example, given $X_k,cdots,X_1$, how do I know the distribution $X_k+2$? I feel like it would have something to do with $X_k+1$...
probability combinatorics conditional-expectation
asked Aug 1 at 3:29
Yuki Kawabata
30419
asked Aug 1 at 3:29
Yuki Kawabata
30419
asked Aug 1 at 3:29
Yuki Kawabata
30419
asked Aug 1 at 3:29
Yuki Kawabata
30419
30419
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The marginal distributions, and thus the expectations, for $X_k+1$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$ that you already know how to find.
answered Aug 1 at 3:35
joriki
164k10179328
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The marginal distributions, and thus the expectations, for $X_k+1$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$ that you already know how to find.
answered Aug 1 at 3:35
joriki
164k10179328
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
add a comment |Â
up vote
1
down vote
accepted
The marginal distributions, and thus the expectations, for $X_k+1$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$ that you already know how to find.
answered Aug 1 at 3:35
joriki
164k10179328
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The marginal distributions, and thus the expectations, for $X_k+1$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$ that you already know how to find.
answered Aug 1 at 3:35
joriki
164k10179328
The marginal distributions, and thus the expectations, for $X_k+1$ through $X_M$ conditional on $X_1$ through $X_k$ are all the same. Thus, the expectation you want is just $M-k$ times the expectation $mathbbEleft(X_k+1 mid X_k,cdots,X_1right)$ that you already know how to find.
answered Aug 1 at 3:35
joriki
164k10179328
answered Aug 1 at 3:35
joriki
164k10179328
answered Aug 1 at 3:35
joriki
164k10179328
answered Aug 1 at 3:35
joriki
164k10179328
164k10179328
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
add a comment |Â
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
1
1
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
PS: ... Because you do know exactly what $sumlimits_j=k+1^N X_j$ will be when you know the first $k$ samples , you may use this method to find to find $mathsf Eleft(X_k+1~middle|~ (X_i)_i=1^kright)$
– Graham Kemp
Aug 1 at 4:51
add a comment |Â
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