Does the Bayes-Filter perform a convolution in the prediction step? I am watching the (fantastic) SLAM lectures of Claus Brenner, where he introduces the Bayes-Filter (Kalman-Filter, Particle-Filter, Histogram-Filter).
He says, that the prediction step involves the convolution of distributions and the correction step a multiplication of distributions (link).
$$\text{prediction: }p(x)=\sum_y p(x\mid y)p(y)$$
$$\text{correction: }p(x\mid z)=\alpha p(z\mid x)p(x)$$
My problem is with the convolution. It makes sense the way he derives it, but I cannot make the connection to the standard definition of convolution as give, e.g. at Wikipedia:
$$p(Z=z) = \sum_{k=-\infty}^\infty p(X=k)p(Y=z-k)$$
Is this a mistake in the video, or am I missing something? It just looks like the law of total probability.
 A: As you said, given certain assumptions, the prediction step could be thought of as convolution in the classic sense. For example if $p(x|y)=\mathcal N(x;y,\Sigma)$ i.e. the conditional probability of $x$ given $y$ is a Gaussian distribution centered at $y$, with fixed covariance. This is not a crazy assumption if your $y$ is real position and $x$ is the sensor reading of the position, then the noise (emission distribution) may be spatially invariant.
In that case, the transformation from $p(y)$ to $p(x)$ is exactly convolution with Gaussian kernel
$$
p(x)=\int_{-\infty}^{\infty} dk p(x|y=k)p(y=k)\\
=\int_{-\infty}^{\infty} dk \mathcal N(x;k,\Sigma)p(y=k)\\
=\int_{-\infty}^{\infty} dk \mathcal N(x-k;0,\Sigma)p(y=k)\\
$$
In this form you can see the similarity to your final convolution formula: random variable $p(y)$ is your $p(X)$, the centered normal distribution $\mathcal N(.;0,\Sigma)$ is the $p(Y)$
More generally if the probability density $p(x|y)$ is not in the form of $g(x-y)$ (translation invariant) you may conceptualize this process as convolution with a changing kernel.

BTW, on implementation level, you may not be able to integrate from $(-\infty,\infty)$, also $x,y$ may be discretized which could make the correspondance not exact.
