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DefineLet $\mathbf z = [z_1, z_2]^T$ be the observations and $\mathbf x = [x_1, x_2]^T$ be the hidden states, and the observation model is $z_i = x_i + n_i$ for $i=1,2$. Here $n_i$'s are independent (but not identical, since the noise has different distributions).

So I would like to know under which restrictions the first step is allowed (I for example know that it its correct for a multivariate normal distribution)

By definition, the density function of a random vector is given by the joint density of its components. So $f(\mathbf z | \mathbf x) = f(z_1, z_2 | \mathbf x)$ is always true.

and also how I could enhance the second approximation

The second approximation is actually an equality under the observation model with independent components. Since $z_2 = x_2 + n_2$, conditioned on $x_2$, we have $z_2 \perp (z_1, x_1)$. This implies that $f(z_2 | \mathbf x, z_1) = f(z_2 | x_1, x_2, z_1) = f(z_2|x_2)$. And similarly, $f(z_1|\mathbf x) = f(z_1 | x_1)$.

Define $\mathbf z = [z_1, z_2]^T$ and $\mathbf x = [x_1, x_2]^T$ and the observation model $z_i = x_i + n_i$ for $i=1,2$. Here $n_i$'s are independent (but not identical, since the noise has different distributions).

So I would like to know under which restrictions the first step is allowed (I for example know that it its correct for a multivariate normal distribution)

By definition, the density function of a random vector is given by the joint density of its components. So $f(\mathbf z | \mathbf x) = f(z_1, z_2 | \mathbf x)$ is always true.

and also how I could enhance the second approximation

The second approximation is actually an equality under the observation model with independent components. Since $z_2 = x_2 + n_2$, conditioned on $x_2$, we have $z_2 \perp (z_1, x_1)$. This implies that $f(z_2 | \mathbf x, z_1) = f(z_2 | x_1, x_2, z_1) = f(z_2|x_2)$. And similarly, $f(z_1|\mathbf x) = f(z_1 | x_1)$.

Let $\mathbf z = [z_1, z_2]^T$ be the observations and $\mathbf x = [x_1, x_2]^T$ be the hidden states, and the observation model is $z_i = x_i + n_i$ for $i=1,2$. Here $n_i$'s are independent (but not identical, since the noise has different distributions).

So I would like to know under which restrictions the first step is allowed (I for example know that it its correct for a multivariate normal distribution)

By definition, the density function of a random vector is given by the joint density of its components. So $f(\mathbf z | \mathbf x) = f(z_1, z_2 | \mathbf x)$ is always true.

and also how I could enhance the second approximation

The second approximation is actually an equality under the observation model with independent components. Since $z_2 = x_2 + n_2$, conditioned on $x_2$, we have $z_2 \perp (z_1, x_1)$. This implies that $f(z_2 | \mathbf x, z_1) = f(z_2 | x_1, x_2, z_1) = f(z_2|x_2)$. And similarly, $f(z_1|\mathbf x) = f(z_1 | x_1)$.

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Define $\mathbf z = [z_1, z_2]^T$ and $\mathbf x = [x_1, x_2]^T$ and the observation model $z_i = x_i + n_i$ for $i=1,2$. Here $n_i$'s are independent (but not identical, since the noise has different distributions).

So I would like to know under which restrictions the first step is allowed (I for example know that it its correct for a multivariate normal distribution)

By definition, the density function of a random vector is given by the joint density of its components. So $f(\mathbf z | \mathbf x) = f(z_1, z_2 | \mathbf x)$ is always true.

and also how I could enhance the second approximation

The second approximation is actually an equality under the observation model with independent components. Since $z_2 = x_2 + n_2$, conditioned on $x_2$, we have $z_2 \perp (z_1, x_1)$. This implies that $f(z_2 | \mathbf x, z_1) = f(z_2 | x_1, x_2, z_1) = f(z_2|x_2)$. And similarly, $f(z_1|\mathbf x) = f(z_1 | x_1)$.