Where is the error in this naive derivation of the Kalman filter?
$begingroup$
The Kalman filter is a method of predicting the future state of a linear state space system based on the previous ones.
A linear, discrete-time, stationary, state-space model is a pair of real valued stochastic processes ${X_t }_{t in mathbb{N}},{Y_t}_{t in mathbb{N}}$ that obey the recursive equations
$$
begin{cases}
X_{t+1} = F X_t + v_t &t=0,1,2dots \
Y_t = H X_t + w_t &t=0,1,2dots
end{cases}
$$
where:
$F in mathbb{R}, H in mathbb{R}$
$v_t, w_t$ are random variables (additive noise) which admit a PDF (Probability Density Function);
The noise terms are zero-mean and:
$$
forall t_1 neq t_2 : v_{t_1} perp v_{t_2}, w_{t_1} perp w_{t_2} \
forall t_1, t_2 : v_{t_1} perp w_{t_2} \
forall t : E[v_t^2] = Q, E[w_t^2] = R
$$
where the simbol $X perp Y$ means that $X$ and $Y$ are independent and $Q,R$ are assumed to be positive real numbers. The initial condition of the recursion $x_0$ is a fixed real number.
Setting $Y^t = { Y_0, dots , Y_t }$The Kalman filter tells us that
$$E[X_t | Y^{t-1}] = alpha E[X_{t-1} | Y^{t-2}] + beta Y_{t-1} $$
where $alpha$ and $beta$ are carefully chosen and depend on the parameters of the linear model under consideration.
My question:
If I where to compute $E[X_t | Y^{t-1}]$ I would notice that $X_t = F X_{t-1} + v_{t-1} $ and $X_{t-1} = Y_{t-1}/ H - w_{t-1}/H$ so
$$X_t = F X_{t-1} + v_{t-1} = F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} $$
and because $w_{t-1}, v_{t-1}$ are zero mean indipendent random variables
$$E[X_t | Y^{t-1}] = E[ F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} | Y^{t-1}] = E[F Y_{t-1}/ H | Y^{t-1}] = F Y_{t-1}/ H $$
(I have also assumed $w_t$ is symmetric) then obviously by doing the same calculations $E[X_{t+1} | Y^{t}] = F Y_{t}/ H$.
There is no way to combine these two conditional expectations to obtain the recursive formula proven by Kalman.
Where is my mistake?
probability probability-theory statistics stochastic-processes time-series
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
$endgroup$
add a comment |
$begingroup$
The Kalman filter is a method of predicting the future state of a linear state space system based on the previous ones.
A linear, discrete-time, stationary, state-space model is a pair of real valued stochastic processes ${X_t }_{t in mathbb{N}},{Y_t}_{t in mathbb{N}}$ that obey the recursive equations
$$
begin{cases}
X_{t+1} = F X_t + v_t &t=0,1,2dots \
Y_t = H X_t + w_t &t=0,1,2dots
end{cases}
$$
where:
$F in mathbb{R}, H in mathbb{R}$
$v_t, w_t$ are random variables (additive noise) which admit a PDF (Probability Density Function);
The noise terms are zero-mean and:
$$
forall t_1 neq t_2 : v_{t_1} perp v_{t_2}, w_{t_1} perp w_{t_2} \
forall t_1, t_2 : v_{t_1} perp w_{t_2} \
forall t : E[v_t^2] = Q, E[w_t^2] = R
$$
where the simbol $X perp Y$ means that $X$ and $Y$ are independent and $Q,R$ are assumed to be positive real numbers. The initial condition of the recursion $x_0$ is a fixed real number.
Setting $Y^t = { Y_0, dots , Y_t }$The Kalman filter tells us that
$$E[X_t | Y^{t-1}] = alpha E[X_{t-1} | Y^{t-2}] + beta Y_{t-1} $$
where $alpha$ and $beta$ are carefully chosen and depend on the parameters of the linear model under consideration.
My question:
If I where to compute $E[X_t | Y^{t-1}]$ I would notice that $X_t = F X_{t-1} + v_{t-1} $ and $X_{t-1} = Y_{t-1}/ H - w_{t-1}/H$ so
$$X_t = F X_{t-1} + v_{t-1} = F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} $$
and because $w_{t-1}, v_{t-1}$ are zero mean indipendent random variables
$$E[X_t | Y^{t-1}] = E[ F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} | Y^{t-1}] = E[F Y_{t-1}/ H | Y^{t-1}] = F Y_{t-1}/ H $$
(I have also assumed $w_t$ is symmetric) then obviously by doing the same calculations $E[X_{t+1} | Y^{t}] = F Y_{t}/ H$.
There is no way to combine these two conditional expectations to obtain the recursive formula proven by Kalman.
Where is my mistake?
probability probability-theory statistics stochastic-processes time-series
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
$endgroup$
1
$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11
add a comment |
$begingroup$
The Kalman filter is a method of predicting the future state of a linear state space system based on the previous ones.
A linear, discrete-time, stationary, state-space model is a pair of real valued stochastic processes ${X_t }_{t in mathbb{N}},{Y_t}_{t in mathbb{N}}$ that obey the recursive equations
$$
begin{cases}
X_{t+1} = F X_t + v_t &t=0,1,2dots \
Y_t = H X_t + w_t &t=0,1,2dots
end{cases}
$$
where:
$F in mathbb{R}, H in mathbb{R}$
$v_t, w_t$ are random variables (additive noise) which admit a PDF (Probability Density Function);
The noise terms are zero-mean and:
$$
forall t_1 neq t_2 : v_{t_1} perp v_{t_2}, w_{t_1} perp w_{t_2} \
forall t_1, t_2 : v_{t_1} perp w_{t_2} \
forall t : E[v_t^2] = Q, E[w_t^2] = R
$$
where the simbol $X perp Y$ means that $X$ and $Y$ are independent and $Q,R$ are assumed to be positive real numbers. The initial condition of the recursion $x_0$ is a fixed real number.
Setting $Y^t = { Y_0, dots , Y_t }$The Kalman filter tells us that
$$E[X_t | Y^{t-1}] = alpha E[X_{t-1} | Y^{t-2}] + beta Y_{t-1} $$
where $alpha$ and $beta$ are carefully chosen and depend on the parameters of the linear model under consideration.
My question:
If I where to compute $E[X_t | Y^{t-1}]$ I would notice that $X_t = F X_{t-1} + v_{t-1} $ and $X_{t-1} = Y_{t-1}/ H - w_{t-1}/H$ so
$$X_t = F X_{t-1} + v_{t-1} = F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} $$
and because $w_{t-1}, v_{t-1}$ are zero mean indipendent random variables
$$E[X_t | Y^{t-1}] = E[ F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} | Y^{t-1}] = E[F Y_{t-1}/ H | Y^{t-1}] = F Y_{t-1}/ H $$
(I have also assumed $w_t$ is symmetric) then obviously by doing the same calculations $E[X_{t+1} | Y^{t}] = F Y_{t}/ H$.
There is no way to combine these two conditional expectations to obtain the recursive formula proven by Kalman.
Where is my mistake?
probability probability-theory statistics stochastic-processes time-series
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
$endgroup$
The Kalman filter is a method of predicting the future state of a linear state space system based on the previous ones.
A linear, discrete-time, stationary, state-space model is a pair of real valued stochastic processes ${X_t }_{t in mathbb{N}},{Y_t}_{t in mathbb{N}}$ that obey the recursive equations
$$
begin{cases}
X_{t+1} = F X_t + v_t &t=0,1,2dots \
Y_t = H X_t + w_t &t=0,1,2dots
end{cases}
$$
where:
$F in mathbb{R}, H in mathbb{R}$
$v_t, w_t$ are random variables (additive noise) which admit a PDF (Probability Density Function);
The noise terms are zero-mean and:
$$
forall t_1 neq t_2 : v_{t_1} perp v_{t_2}, w_{t_1} perp w_{t_2} \
forall t_1, t_2 : v_{t_1} perp w_{t_2} \
forall t : E[v_t^2] = Q, E[w_t^2] = R
$$
where the simbol $X perp Y$ means that $X$ and $Y$ are independent and $Q,R$ are assumed to be positive real numbers. The initial condition of the recursion $x_0$ is a fixed real number.
Setting $Y^t = { Y_0, dots , Y_t }$The Kalman filter tells us that
$$E[X_t | Y^{t-1}] = alpha E[X_{t-1} | Y^{t-2}] + beta Y_{t-1} $$
where $alpha$ and $beta$ are carefully chosen and depend on the parameters of the linear model under consideration.
My question:
If I where to compute $E[X_t | Y^{t-1}]$ I would notice that $X_t = F X_{t-1} + v_{t-1} $ and $X_{t-1} = Y_{t-1}/ H - w_{t-1}/H$ so
$$X_t = F X_{t-1} + v_{t-1} = F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} $$
and because $w_{t-1}, v_{t-1}$ are zero mean indipendent random variables
$$E[X_t | Y^{t-1}] = E[ F (Y_{t-1}/ H - w_{t-1}/H) + v_{t-1} | Y^{t-1}] = E[F Y_{t-1}/ H | Y^{t-1}] = F Y_{t-1}/ H $$
(I have also assumed $w_t$ is symmetric) then obviously by doing the same calculations $E[X_{t+1} | Y^{t}] = F Y_{t}/ H$.
There is no way to combine these two conditional expectations to obtain the recursive formula proven by Kalman.
Where is my mistake?
probability probability-theory statistics stochastic-processes time-series
probability probability-theory statistics stochastic-processes time-series
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
asked Dec 22 '18 at 0:34
MonoliteMonolite
1,5922926
1,5922926
1
$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11
add a comment |
1
$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11
1
1
$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11
add a comment |
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$begingroup$
$mathsf{E}[w_{t-1}mid Y^{t-1}]=0$?
$endgroup$
– d.k.o.
Dec 22 '18 at 6:59
$begingroup$
@d.k.o. thanks, of course!
$endgroup$
– Monolite
Dec 22 '18 at 11:11