Open WGook opened 1 year ago
Hi, I have a question about the equation in the paper. I'm wondering how the equality in Eq.(6) could be derived. With my knowledge, $p(x)$ becomes:
$$ \begin{align} p(x) &= \int p(x|z_0)p(z_0)dz_0, \ &\mbox{ and } \ p(z0) &= \int p(z{0:1})dz{t(1):1},\ p(z{0:1}) &= p(z1) \prod{i = 1}^Tp(z{s(i)}|z{t(i)}), \end{align} $$
if $p(z_{0:1})$ is a Markov chain.
Then, how is it same to $\int_z p(z_1)p(x|z0) \prod{i = 1}^T p(z{s(i)}|z{t(i)})$? Also, is $dz_1$ not needed to calculate the marginal distribution with integral in Eq.(6)?
Hi, I have a question about the equation in the paper. I'm wondering how the equality in Eq.(6) could be derived. With my knowledge, $p(x)$ becomes:
$$ \begin{align} p(x) &= \int p(x|z_0)p(z_0)dz_0, \ &\mbox{ and } \ p(z0) &= \int p(z{0:1})dz{t(1):1},\ p(z{0:1}) &= p(z1) \prod{i = 1}^Tp(z{s(i)}|z{t(i)}), \end{align} $$
if $p(z_{0:1})$ is a Markov chain.
Then, how is it same to $\int_z p(z_1)p(x|z0) \prod{i = 1}^T p(z{s(i)}|z{t(i)})$? Also, is $dz_1$ not needed to calculate the marginal distribution with integral in Eq.(6)?