我本该高兴的，却差点儿红了眼眶。

我记得，圣女吹笛方能召唤出漫天蝴蝶，在一年前的庆典之上，我还曾在圣女用笛音的伴奏下跳过一曲霓裳舞，那日的万毒窟祭祀台，就似一方绝美的蝴蝶谷。

圣女同我的生辰相隔不久，尤川同她如今的关系不似从前，大抵有两相对峙之意，我想，这簪子本该是要送给圣女的吧？

不过，只要是他送的，那就是好的，我接过簪子，恍若无事般扬头乐呵呵一笑：“在中原买的？”

“对。”他一如既往地言简意赅。

“豁！那你这是打算拿替我买礼物这事儿，给自个儿做掩咯！”我故意皱了皱鼻子，假意生气，“尤川，你好不厚道呢！”

“我……”

果不其然，他终究还是没“我”出个所以然来。

于是我率先面露惬意地朝他莞尔：“好了，逗你玩的。”

“多谢，我很喜欢！”

第 3 章

Now, we can ignore the constant terms since they won't affect the maximization result. Therefore, we only need to focus on the terms that are related to $\boldsymbol{\Sigma}_k$, which is

$$

-\frac{1}{2} \sum_{n=1}^N \gamma\left(z_{n k}\right) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k)+ \frac{1}{2} \sum_{n=1}^N \gamma\left(z_{n k}\right) \ln \boldsymbol{\Sigma}_k^{-1}.

$$

To maximize this expression, we can take the derivative with respect to $\boldsymbol{\Sigma}_k$ and set it equal to zero. This will give us the solution for maximization. Specifically, we have

$$

\frac{\partial}{\partial \boldsymbol{\Sigma}_k} \left(-\frac{1}{2} \sum_{n=1}^N \gamma\left(z_{n k}\right) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k) + \frac{1}{2} \sum_{n=1}^N \gamma\left(z_{n k}\right) \ln \boldsymbol{\Sigma}_k^{-1}\right) = 0.

$$

First, let's consider the first term:

\[

-\frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k).

\]

Using matrix calculus, the derivative of \(\mathbf{a}^T \mathbf{A}^{-1} \mathbf{a}\) with respect to \(\mathbf{A}\) is:

\[

\frac{\partial}{\partial \mathbf{A}} (\mathbf{a}^T \mathbf{A}^{-1} \mathbf{a}) = -\mathbf{A}^{-1} \mathbf{a} \mathbf{a}^T \mathbf{A}^{-1}.

\]

Therefore, for the first term:

\[

\frac{\partial}{\partial \boldsymbol{\Sigma}_k} \left( -\frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k) \right) = \frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1}.

\]

Next, consider the second term:

\[

\frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \ln \boldsymbol{\Sigma}_k^{-1}.

\]

We know that \(\ln \mathbf{A}^{-1} = -\ln \mathbf{A}\), and the derivative of \(\ln \mathbf{A}\) with respect to \(\mathbf{A}\) is \(\mathbf{A}^{-1}\). Thus,

\[

\frac{\partial}{\partial \boldsymbol{\Sigma}_k} \left( \frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \ln \boldsymbol{\Sigma}_k^{-1} \right) = -\frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \boldsymbol{\Sigma}_k^{-1}.

\]

bining the derivatives of both terms, we get:

\[

\frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \boldsymbol{\Sigma}_k^{-1} (\mathbf{x}_n - \boldsymbol{\mu}_k) (\mathbf{x}_n - \boldsymbol{\mu}_k)^T \boldsymbol{\Sigma}_k^{-1} - \frac{1}{2} \sum_{n=1}^N \gamma(z_{nk}) \boldsymbol{\Sigma}_k^{-1} = 0.

\]

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To find the optimal value of \(\pi_k\), we can use the Lagrange multiplier method.

Consider the maximization of the following expression with respect to \(\pi_k\) while keeping the responsibilities \(\gamma(z_{nk})\) fixed:

\[

\mathbb{E}_{\mathbf{Z}}[\ln p(\mathbf{X}, \mathbf{Z} \mid \boldsymbol{\mu}, \boldsymbol{\Sigma}, \boldsymbol{\pi})] = \sum_{n=1}^N \sum_{k=1}^K \gamma(z_{nk}) \left\{\ln \pi_k + \ln \mathcal{N}(\mathbf{x}_n \mid \boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)\right\}.

\]

Let's define the Lagrangian function as:

\[

\mathcal{L}(\boldsymbol{\pi}, \lambda) = \sum_{n=1}^N \sum_{k=1}^K \gamma(z_{nk}) \left\{\ln \pi_k + \ln \mathcal{N}(\mathbf{x}_n \mid \boldsymbol{\mu}_k, \boldsymbol{\Sigma}_k)\right\} + \lambda \left(\sum_{k=1}^K \pi_k - 1\right).