Consider a simple state space model $$x_{t+1}=Ax_t+C\omega_t\\ y_t=Gx_t+v_t $$ Besides the orthogonal assumption for $w_{t+1}$ and $v_t$, we assume that $w_{t+1}\sim N(0,I)$ and $v_t\sim N(0,R)$. We observe $y_t$ while we want to estimate the sequence of $x_t$ using Kalman filter. During the process we will derive the following two conditional expectations: \begin{align*} \mathbb{E}[x_t|y_{t-1}]\\ \mathbb{E}[x_t|y_{t}] \end{align*} When applying Kalman filter, we use the first expression as our estimation of $x_t$. I want to know why we do not use the second conditional expectation instead. My intuition is that we start the algorithm with a prior $\hat{x}_0=\mathbb{E}[x_0]$, but $\mathbb{E}[x_0|y_0]$ clearly contains more information than the trivial prior. That is, at the beginning, we assume that \begin{align*} x_0\sim N(\mu_0,\Sigma_0) \end{align*} while it's not difficult to obtain the following result: \begin{align*} \mathbb{E}[x_0|y_0]=\mu_0+L_0(y_0-G\mu_0) \end{align*} where \begin{align*} L_0=\Sigma_0G'(G\Sigma_0G'+R)^{-1} \end{align*} Then, which one should we take as the estimation $\hat{x}_{t-1}$? $\mu_0$ (which could be regarded as $\mathbb{E}[x_0|y_{-1}]$) or $\mathbb{E}[x_0|y_0]$?

  • $\begingroup$ Can you give a reference on which particular filter you are using (the form of the equations). In optimal filtering, two filters can be derived (both by Kalman) depending on if $y_t$ is avaiable at the moment you want to estimate $x_t$. I mean, do you wan't to just estimate, or also predict (using $\mathbb{E}[x_t|y_t]$ vs $\mathbb{E}[x_t|y_{t-1}]$)?. (See "Introduction to Stochastic Control" by Astrom pg 228) $\endgroup$ Aug 23, 2021 at 11:42
  • $\begingroup$ @FeedbackLooper : this setup is taken from Recursive Macroeconomic Theory (4th Edition) by ‎Ljungqvist and Sargent Section 2.7, where the sequence of $y_t$ is observed and we want to estimate $x_t$. I have complemented some details in the question. $\endgroup$
    – QLin
    Aug 24, 2021 at 2:18

1 Answer 1


I would like to see the referece for

When applying Kalman filter, we use the first expression as our estimation of 𝑥,

because this claim sounds incomplete. It is either wrong or mis-interpreted.

In Kalman filters, we are really aiming at $\mathbb{E}[x_t | y_{1:t}]$. This expectation is mostly termed as our estimation of $x_t$, as it encodes the best estimate of the random variable $x_t$ given all the information from data up until time $t$ (recall the definition of conditonal expectation). Here I denote $y_{1:t} = \{ y_1, y_2, \ldots, y_t\}$.

The core of Kalman filters is to recursivly compute $\mathbb{E}[x_t | y_{1:t}]$ by using its previous estimate $\mathbb{E}[x_{t-1} | y_{1:t-1}]$ for $t=1,2,\ldots$. Note that we define $\mathbb{E}[x_{0} | y_{1:0}]= \mathbb{E}[x_{0}]$.

The quantity $\mathbb{E}[x_t | y_{1:t-1}]$ is called the predictive mean which is an essential intermediate quantity to obtain the filtering distributions.

Your intuition is correct, but $\mathbb{E}[x_0 | y_0]$ is not useful, as we, in practice, don't have $y_0$. You start the Kalman filtering equations from $x_0$ (i.e., the initial condition).


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