高斯分布与多元高斯分布

2022-04-11  本文已影响0人  臻甄

均值 \mu, 标准差 \sigma
一元高斯分布和标准高斯分布绘图:https://www.cnblogs.com/bingjianing/p/9117330.html

一元高斯分布

多元高斯分布

可以简化为:
\mathcal{N}( \mathbf{x | \mu,\Sigma } ) = {1 \over (2\pi)^{D/2}}{1 \over |\mathbf{\Sigma}|^{1/2} } exp \left\{ - \frac 12 (\mathbf{x}- \mathbf{\mu}) \mathbf{\Sigma}^{-1} (\mathbf{x}- \mathbf{\mu}) \right\}

\mathbf{\Sigma}是一个对称矩阵:
\begin{pmatrix} \sigma_1^2 & 0 & \cdots & 0 \\ 0 & \sigma_2^2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \sigma_d^2 \\ \end{pmatrix}

最大似然估计

两个多元高斯分布之间的KL散度

\begin{align} D_{KL}(q(z|x)||p(z)) & = \int q(z|x) \log \frac {q(z|x)}{p(z)}dz \\ &= \int q(z|x) \{ \log q(z|x) - \log p(z) \} dz \\ &= \int q(z|x) \log q(z|x)dz - \int q(z|x) \log p(z)dz \\ &= \left( - \frac J2 \log(2\pi) - \frac 12 \sum_{j=1}^{J}(\log \sigma_{1,j}^2 +1) \right) - \left( - \frac J2 \log (2\pi) - \frac 12 \sum_{j=1}^{J} \log \sigma_{2,j}^2 - \frac 12 \sum_{j=1}^{J} \left[ {\sigma_{1,j}^2 \over \sigma_{2,j}^2} + {(\mu_{1,j} - \mu_{2,j})^2 \over \sigma_{2,j}^2} \right] \right) \\ &= - \frac 12 \sum_{j=1}^{J} \left( \log {\sigma_{1,j}^2 \over \sigma_{2,j}^2} - {\sigma_{1,j}^2 \over \sigma_{2,j}^2} - {(\mu_{1,j} - \mu_{2,j})^2 \over \sigma_{2,j}^2} + 1 \right) \\ \end{align}

在变分自编码中,q(z) \sim N (\mu, \sigma), p(z) \sim N(0,1),则有

\begin{align} D_{KL}(q(z|x)||p(z)) & = \int q(z|x) \log \frac {q(z|x)}{p(z)}dz \\ &= \int q(z|x) \{ \log q(z|x) - \log p(z) \} dz \\ &= \int q(z|x) \log q(z|x)dz - \int q(z|x) \log p(z)dz \\ &= \left( - \frac J2 \log(2\pi) - \frac 12 \sum_{j=1}^{J}(\log \sigma_j^2 +1) \right) - \left( - \frac J2 \log (2\pi) - \frac 12 \sum_{j=1}^{J}(\sigma_j^2 + \mu_j^2) \right) \\ &= - \frac 12 \sum_{j=1}^{J} \left( 1 + \log \sigma^2 - \sigma_j^2 - \mu_j^2 \right) \\ \end{align}

两个多元高斯分布之间的 对数概率 logπ

log \pi_{\theta}(x) = - \frac 12 \left( \sum_{i=1}^{i=d} \left( {(x_i - \mu_i)^2 \over \sigma_i^2} \right) + k log(2\pi) \right)

多元高斯分布的

h(x_1, x_2, ...,x_n) = h(\mathcal{N}_n(\mu, K)) = \frac 12 \log (2\pi e)^{n}|K| 其中 K 是协方差

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