Kalman smoother equations
$begingroup$
I'm trying to find the Kalman filter equations, i.e $mathbb{E}[X_k mid y_0,...,y_n]$ with $k lt n$ by figuring out the law of the the density $p(x_k mid y_0,...,y_n)$ under the Hidden Markov Chain Model ($X_k$ is a markov chain and conditionally on $(X_{0:n}=(X_0,...,X_n))$, the observations $(Y_{0:n})$ are independent and the observations $i$ only depends on the hidden state $X_i$ :
begin{equation}
p(x_{0:n},y_{0:n})=p(x_0)prod_{i=1}^n p(x_i mid x_{i-1}) prod_{i=0}^n p(x_i mid y_i)
end{equation}
The thing is, i'm struggling with a preliminary question: that is to prove :
begin{equation}
p(x_{0:k} mid x_{k+1}, y_{0:n})=p(x_{0:k} mid x_{k+1}, y_{0:k})
end{equation}
Anyone know how to prove it?
statistics kalman-filter
asked Dec 17 '18 at 10:07
yjntyjnt
113
$endgroup$
add a comment |
$begingroup$
I'm trying to find the Kalman filter equations, i.e $mathbb{E}[X_k mid y_0,...,y_n]$ with $k lt n$ by figuring out the law of the the density $p(x_k mid y_0,...,y_n)$ under the Hidden Markov Chain Model ($X_k$ is a markov chain and conditionally on $(X_{0:n}=(X_0,...,X_n))$, the observations $(Y_{0:n})$ are independent and the observations $i$ only depends on the hidden state $X_i$ :
begin{equation}
p(x_{0:n},y_{0:n})=p(x_0)prod_{i=1}^n p(x_i mid x_{i-1}) prod_{i=0}^n p(x_i mid y_i)
end{equation}
The thing is, i'm struggling with a preliminary question: that is to prove :
begin{equation}
p(x_{0:k} mid x_{k+1}, y_{0:n})=p(x_{0:k} mid x_{k+1}, y_{0:k})
end{equation}
Anyone know how to prove it?
statistics kalman-filter
asked Dec 17 '18 at 10:07
yjntyjnt
113
$endgroup$
add a comment |
$begingroup$
I'm trying to find the Kalman filter equations, i.e $mathbb{E}[X_k mid y_0,...,y_n]$ with $k lt n$ by figuring out the law of the the density $p(x_k mid y_0,...,y_n)$ under the Hidden Markov Chain Model ($X_k$ is a markov chain and conditionally on $(X_{0:n}=(X_0,...,X_n))$, the observations $(Y_{0:n})$ are independent and the observations $i$ only depends on the hidden state $X_i$ :
begin{equation}
p(x_{0:n},y_{0:n})=p(x_0)prod_{i=1}^n p(x_i mid x_{i-1}) prod_{i=0}^n p(x_i mid y_i)
end{equation}
The thing is, i'm struggling with a preliminary question: that is to prove :
begin{equation}
p(x_{0:k} mid x_{k+1}, y_{0:n})=p(x_{0:k} mid x_{k+1}, y_{0:k})
end{equation}
Anyone know how to prove it?
statistics kalman-filter
asked Dec 17 '18 at 10:07
yjntyjnt
113
$endgroup$
I'm trying to find the Kalman filter equations, i.e $mathbb{E}[X_k mid y_0,...,y_n]$ with $k lt n$ by figuring out the law of the the density $p(x_k mid y_0,...,y_n)$ under the Hidden Markov Chain Model ($X_k$ is a markov chain and conditionally on $(X_{0:n}=(X_0,...,X_n))$, the observations $(Y_{0:n})$ are independent and the observations $i$ only depends on the hidden state $X_i$ :
begin{equation}
p(x_{0:n},y_{0:n})=p(x_0)prod_{i=1}^n p(x_i mid x_{i-1}) prod_{i=0}^n p(x_i mid y_i)
end{equation}
The thing is, i'm struggling with a preliminary question: that is to prove :
begin{equation}
p(x_{0:k} mid x_{k+1}, y_{0:n})=p(x_{0:k} mid x_{k+1}, y_{0:k})
end{equation}
Anyone know how to prove it?
statistics kalman-filter
statistics kalman-filter
asked Dec 17 '18 at 10:07
yjntyjnt
113
asked Dec 17 '18 at 10:07
yjntyjnt
113
asked Dec 17 '18 at 10:07
yjntyjnt
113
asked Dec 17 '18 at 10:07
yjntyjnt
113
asked Dec 17 '18 at 10:07
yjntyjnt
113
113
add a comment |
add a comment |
1 Answer
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I don't have the reputation to put this as a comment, so I post as an answer.
It follows from the Markov assumptions. The past, $x_{0:k}$, is independent of the future given the present, $x_{k+1}$.
If you draw the graphical model it may be easier to see. Given $x_{k+1}$, $x_{0:k}$ is "blocked" from $y_{k+1}$ and every observation afterwards.
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
$endgroup$
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I don't have the reputation to put this as a comment, so I post as an answer.
It follows from the Markov assumptions. The past, $x_{0:k}$, is independent of the future given the present, $x_{k+1}$.
If you draw the graphical model it may be easier to see. Given $x_{k+1}$, $x_{0:k}$ is "blocked" from $y_{k+1}$ and every observation afterwards.
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
$endgroup$
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
add a comment |
$begingroup$
I don't have the reputation to put this as a comment, so I post as an answer.
It follows from the Markov assumptions. The past, $x_{0:k}$, is independent of the future given the present, $x_{k+1}$.
If you draw the graphical model it may be easier to see. Given $x_{k+1}$, $x_{0:k}$ is "blocked" from $y_{k+1}$ and every observation afterwards.
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
$endgroup$
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
add a comment |
$begingroup$
I don't have the reputation to put this as a comment, so I post as an answer.
It follows from the Markov assumptions. The past, $x_{0:k}$, is independent of the future given the present, $x_{k+1}$.
If you draw the graphical model it may be easier to see. Given $x_{k+1}$, $x_{0:k}$ is "blocked" from $y_{k+1}$ and every observation afterwards.
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
$endgroup$
I don't have the reputation to put this as a comment, so I post as an answer.
It follows from the Markov assumptions. The past, $x_{0:k}$, is independent of the future given the present, $x_{k+1}$.
If you draw the graphical model it may be easier to see. Given $x_{k+1}$, $x_{0:k}$ is "blocked" from $y_{k+1}$ and every observation afterwards.
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
answered Feb 8 at 19:26
grxxvytonygrxxvytony
442
442
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
add a comment |
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
$begingroup$
@Leucippus: I believe it does answer the OP's question on how to prove the result. If the OP requires clarification, then the OP should tell me -- not you.
$endgroup$
– grxxvytony
Feb 8 at 23:58
add a comment |
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