2 added 42 characters in body edited Sep 14 '17 at 15:50 Atul Ingle 3,72811 gold badge1010 silver badges2222 bronze badges 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, $$f(z_1|\mathbf x) = f(z_1 | x_1)$$. 1 answered Sep 13 '17 at 15:58 Atul Ingle 3,72811 gold badge1010 silver badges2222 bronze badges 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, $$f(z_1|\mathbf x) = f(z_1 | x_1)$$.