সাধারণ-উইশার্টের উত্তরোত্তর ব্যয়

আমি একটি নরমাল-উইশার্টের উত্তরোত্তর আবিষ্কারের জন্য কাজ করছি তবে আমি কোনও একটি পরামিতি (স্কেল ম্যাট্রিক্সের উত্তরবর্তী, নীচে দেখুন) এ আটকে আছি।

কেবল প্রসঙ্গ এবং সম্পূর্ণতার জন্য, এখানে মডেল এবং বাকী সমস্ত ডেরাইভেশন রয়েছে:

\begin{aligned} x_{i} & \sim N (μ, Λ) \\ μ & \sim N (μ_{0}, (κ_{0} Λ)^{- 1}) \\ Λ & \sim W (υ_{0}, W_{0}) \end{aligned}

$\begin{align} x_i &\sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Lambda})\\ \boldsymbol{\mu} &\sim \mathcal{N}(\boldsymbol{\mu_0}, (\kappa_0 \boldsymbol{\Lambda})^{-1})\\ \boldsymbol{\Lambda} &\sim \mathcal{W}(\upsilon_0, \mathbf{W}_0) \end{align}$

তিনটি উপাদানের প্রত্যেকটির প্রসারিত ফর্মগুলি হ'ল (একটি আনুপাতিকতা ধ্রুবক পর্যন্ত):

সম্ভাবনা: start
$\begin{aligned} N (x_{i} & | μ, Λ) \propto \\ | Λ |^{N / 2} \exp (- \frac{1}{2} \sum_{i = 1}^{N} (x_{i}^{T} Λ x_{i} - 2 μ^{T} Λ x_{i} + μ^{T} Λ μ)) \end{aligned}$ $\begin{align} \mathcal{N}(\mathbf{x}_i &| \boldsymbol{\mu}, \boldsymbol{\Lambda}) \propto\notag\\ &|\boldsymbol{\Lambda}|^{N/2} \exp{\left(-\frac{1}{2}\sum_{i=1}^N \left( \mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i - 2 \boldsymbol{\mu}^T \boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu}^T\boldsymbol{\Lambda}\boldsymbol{\mu}\right) \right)} \end{align}$
সাধারণ পূর্ব:
$\begin{aligned} N (μ & | (μ_{0}, κ_{0} Λ)^{- 1}) \propto \\ | Λ |^{1 / 2} \exp (- \frac{1}{2} (μ^{T} κ_{0} Λ μ - 2 μ^{T} κ_{0} Λ μ_{0} + μ_{0}^{T} κ_{0} Λ μ_{0})) \end{aligned}$ $\begin{align} \mathcal{N}(\boldsymbol{\mu} &| (\boldsymbol{\mu}_0, \kappa_0 \boldsymbol{\Lambda})^{-1}) \propto\notag\\ &|\boldsymbol{\Lambda}|^{1/2} \exp{\left(-\frac{1}{2}\left( \boldsymbol{\mu}^T\kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu} - 2 \boldsymbol{\mu}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0} + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\right) \right)} \end{align}$
উইশার্ট পূর্বে:
$\begin{aligned} W (Λ | υ_{0}, W_{0}) \propto | Λ |^{\frac{υ_{0} - D - 1}{2}} \exp (- \frac{1}{2} t r (W_{0}^{- 1} Λ)) \end{aligned}$ $\begin{align} \mathcal{W}(\boldsymbol{\Lambda} | \upsilon_0, \mathbf{W}_0) \propto |\boldsymbol{\Lambda}|^{\frac{\upsilon_0-D-1}{2}} \exp{\left(-\frac{1}{2} tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda})\right)} \end{align}$

আমরা উত্তরোত্তর-উইশার্ট চাই ( ) যা পচে যেতে পারে পাশাপাশি : $\mathbf{\mu}, \boldsymbol{\Lambda} | \boldsymbol{\mu}', \kappa', \upsilon', \mathbf{W}'$ $\mathcal{N}(\boldsymbol{\mu} | \boldsymbol{\mu}, \kappa' \boldsymbol{\Lambda}) \mathcal{W}(\boldsymbol{\Lambda} | \upsilon', \mathbf{W}')$

স্বাধীনতার ডিগ্রি $\upsilon'$

সম্ভাবনার প্রথম বিষয়গুলি এবং উইশার্টকে উইশার্ট ফ্যাক্টরের প্রথম ফ্যাক্টরটি পাই: therefore এবং সেইজন্য আমাদের প্রথম প্যারামিটার রয়েছে:

\begin{aligned} | Λ |^{\frac{υ_{0} + N - D - 1}{2}} \end{aligned}

$\begin{align} |\boldsymbol{\Lambda}|^{\frac{\upsilon_0+N-D-1}{2}} \end{align}$

\begin{aligned} υ^{'} = υ_{0} + N \end{aligned}

$\begin{align} \upsilon' = \upsilon_0 +N \end{align}$

স্কেল ফ্যাক্টর $\kappa'$

আমরা চিহ্নিত উপাদান দ্বারা বেষ্টিত এবং এটি যারা পূর্বে সম্ভাবনা দ্বারা আপডেট করা হয়: এবং তাই আমরা দ্বিতীয় প্যারামিটার পেয়েছি: $\boldsymbol{\mu}^T$ $\boldsymbol{\mu}$ $\kappa_0 \boldsymbol{\Lambda}$

\begin{aligned} μ^{T} ((κ_{0} + N) Λ) μ \end{aligned}

$\begin{align} \boldsymbol{\mu}^T\left((\kappa_0 + N) \boldsymbol{\Lambda}\right)\boldsymbol{\mu} \end{align}$

\begin{aligned} κ^{'} = κ_{0} + N \end{aligned}

$\begin{align} \kappa' = \kappa_0 + N \end{align}$

গড় $\boldsymbol{\mu}'$

তৃতীয় প্যারামিটারগুলি ভিতরে যা আছে তা সনাক্তকরণ থেকে আসে : এবং তাই আমরা তৃতীয় প্যারামিটার পেয়েছি: $2\boldsymbol{\mu}^T...$

\begin{aligned} 2 μ^{T} (Λ N \bar{x} + κ_{0} Λ μ_{0}) & = 2 μ^{T} κ^{'} Λ μ^{'} \\ (Λ N \bar{x} + κ_{0} Λ μ_{0}) & = κ^{'} Λ μ^{'} \\ (N \bar{x} + κ_{0} μ_{0}) & = κ^{'} μ^{'} \end{aligned}

$\begin{align} 2\boldsymbol{\mu}^T\left(\boldsymbol{\Lambda} N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu}_0}\right) &= 2\boldsymbol{\mu}^T \kappa'\boldsymbol{\Lambda} \boldsymbol{\mu}'\\ \left(\boldsymbol{\Lambda} N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu}_0}\right) &= \kappa'\boldsymbol{\Lambda} \boldsymbol{\mu}'\\ \left(N \mathbf{\overline{x} + \kappa_0 \boldsymbol{\mu}_0}\right) &= \kappa' \boldsymbol{\mu}' \end{align}$

\begin{aligned} μ^{'} = \frac{1}{k^{'}} (N \bar{x} + κ_{0} μ_{0}) \end{aligned}

$\begin{align} \boldsymbol{\mu}' = \frac{1}{k'}(N\mathbf{\overline{x}} + \kappa_0 \boldsymbol{\mu}_0) \end{align}$

স্কেল ম্যাট্রিক্স $\boldsymbol{W}'$

এবং চতুর্থ প্যারামিটারটি বাকী প্যারামিটারগুলিতে কাজ করার পরে আসে:

\begin{aligned} t r (W^{' - 1} Λ) & = t r (W_{0}^{- 1} Λ) + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + μ_{0}^{T} κ_{0} Λ μ_{0} \\ = t r (W_{0}^{- 1} Λ) + \sum_{i = 1}^{N} t r (x_{i}^{T} Λ x_{i}) + t r (μ_{0}^{T} κ_{0} Λ μ_{0}) \\ = t r (W_{0}^{- 1} Λ + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + μ_{0}^{T} κ_{0} Λ μ_{0}) \end{aligned}

$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) &= tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda}) + \sum_{i=1}^{N}\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\\ &= tr(\mathbf{W}_0^{-1} \boldsymbol{\Lambda}) + \sum_{i=1}^{N}tr(\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i) + tr(\boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0})\\ &= tr\bigg(\mathbf{W}_0^{-1} \boldsymbol{\Lambda} + \sum_{i=1}^{N}\mathbf{x}_i^T\boldsymbol{\Lambda}\mathbf{x}_i + \boldsymbol{\mu_0}^T \kappa_0 \boldsymbol{\Lambda}\boldsymbol{\mu_0}\bigg) \end{align}$

কীভাবে এখান থেকে যাব (যদি আমি এখনও পর্যন্ত কোনও ভুল না করি) এবং এর মানক সমাধান পেতে পারি? $\mathbf{W}'$

সম্পাদনা 1 :

এখন আমরা শর্তাদি পুনরায় সাজিয়েছি, স্ট্যান্ডার্ড সমাধান হিসাবে দুটি স্কোয়ার পেতে কিছু উপাদান যুক্ত এবং বিয়োগ করে:

\begin{aligned} t r (W^{' - 1} Λ) = & t r (W^{- 1} Λ \\ + \sum_{i = 1}^{N} (x_{i}^{T} Λ x_{i} + {\bar{x}}^{T} Λ \bar{x} - 2 x_{i}^{T} Λ \bar{x}) \\ + κ_{0} (μ_{0}^{T} Λ μ_{0} + {\bar{x}}^{T} Λ \bar{x} - 2 {\bar{x}}^{T} Λ μ_{0}) \\ - \sum_{i = 1}^{N} {\bar{x}}^{T} Λ \bar{x} + 2 \sum_{i = 1}^{N} x_{i}^{T} Λ \bar{x} - κ_{0} {\bar{x}}^{T} Λ \bar{x} + 2 κ_{0} {\bar{x}}^{T} Λ μ_{0}) \\ = & t r (W^{- 1} Λ + \sum_{i = 1}^{N} (x_{i} - \bar{x}) Λ (x_{i} - \bar{x})^{T} + κ_{0} (\bar{x} - μ_{0}) Λ (\bar{x} - μ_{0})^{T} \\ - N \bar{x} Λ {\bar{x}}^{T} + 2 N \bar{x} Λ {\bar{x}}^{T} - κ_{0} \bar{x} Λ {\bar{x}}^{T} + 2 κ_{0} \bar{x} Λ μ_{0}^{T}) \end{aligned}

$\begin{align} tr(\mathbf{W}'^{-1}\boldsymbol{\Lambda}) =\;& tr\bigg(\mathbf{W}^{-1} \boldsymbol{\Lambda}\\ &+ \sum_{i=1}^N (\mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{x}_i + \overline{\mathbf{x}}^T \boldsymbol{\Lambda} \overline{\mathbf{x}} -2 \mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{\overline{x}})\\ &+\kappa_0 (\boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \boldsymbol{\overline{x}}^T \boldsymbol{\Lambda}\boldsymbol{\overline{x}} - 2\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0) \notag\\% Substract added factors &-\sum_{i=1}^{N}\overline{\mathbf{x}}^T \boldsymbol{\Lambda} \overline{\mathbf{x}} +2\sum_{i=1}^{N}\mathbf{x}_i^T \boldsymbol{\Lambda} \mathbf{\overline{x}} - \kappa_0\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda} \boldsymbol{\overline{x}} + 2\kappa_0\boldsymbol{\overline{x}}^T \boldsymbol{\Lambda}\boldsymbol{\mu}_0\bigg)\\ =\;& tr\bigg( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})\boldsymbol{\Lambda}(\mathbf{x}_i - \overline{\mathbf{x}})^T + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)\boldsymbol{\Lambda}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \notag\\% Substract added factors &-N\overline{\mathbf{x}} \boldsymbol{\Lambda} \overline{\mathbf{x}}^T +2N\mathbf{\overline{x}} \boldsymbol{\Lambda}\mathbf{\overline{x}}^T - \kappa_0\boldsymbol{\overline{x}}\boldsymbol{\Lambda}\boldsymbol{\overline{x}}^T + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\Lambda} \boldsymbol{\mu}_0^T \bigg) \end{align}$

স্কোয়ারের বাইরে থাকা উপাদানগুলি আমরা সরলীকরণ করি:

\begin{aligned} t r (W^{' - 1} Λ) = & t r (W^{- 1} Λ + \sum_{i = 1}^{N} (x_{i} - \bar{x})^{T} Λ (x_{i} - \bar{x}) + κ_{0} (\bar{x} - μ_{0})^{T} Λ (\bar{x} - μ_{0}) \\ + (N - κ_{0}) {\bar{x}}^{T} Λ \bar{x} + 2 κ_{0} {\bar{x}}^{T} Λ μ_{0}) \end{aligned}

$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) =\;& tr( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})^T\boldsymbol{\Lambda}(\mathbf{x}_i - \overline{\mathbf{x}}) + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T\boldsymbol{\Lambda}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0) \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}}^T\boldsymbol{\Lambda} \mathbf{\overline{x}} + 2\kappa_0\boldsymbol{\overline{x}}^T\boldsymbol{\Lambda}\boldsymbol{\mu}_0 \bigg) \end{align}$

2 সম্পাদনা করুন ( @ বিডিয়নভিকের উত্তরের জন্য ধন্যবাদ অনুসরণ করুন )

ট্রেসটি চক্রযুক্ত, সুতরাং । তারপরে: এবং তারপর: $tr(ABC) = tr(BCA) = tr(CAB)$

\begin{aligned} t r (W^{' - 1} Λ) = & t r (W^{- 1} Λ + \sum_{i = 1}^{N} (x_{i} - \bar{x}) (x_{i} - \bar{x})^{T} Λ + κ_{0} (\bar{x} - μ_{0}) (\bar{x} - μ_{0})^{T} Λ \\ + (N - κ_{0}) \bar{x} {\bar{x}}^{T} Λ + 2 κ_{0} \bar{x} μ_{0}^{T} Λ) \end{aligned}

$\begin{align} tr(\mathbf{W}'^{-1} \boldsymbol{\Lambda}) =\;& tr\bigg( \mathbf{W}^{-1} \boldsymbol{\Lambda}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T\boldsymbol{\Lambda} + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T\boldsymbol{\Lambda} \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}} \mathbf{\overline{x}}^T\boldsymbol{\Lambda} + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \bigg) \end{align}$

\begin{aligned} t r (W^{' - 1}) = & t r (W^{- 1} + \sum_{i = 1}^{N} (x_{i} - \bar{x}) (x_{i} - \bar{x})^{T} + κ_{0} (\bar{x} - μ_{0}) (\bar{x} - μ_{0})^{T} \\ + (N - κ_{0}) \bar{x} {\bar{x}}^{T} + 2 κ_{0} \bar{x} μ_{0}^{T}) \end{aligned}

$\begin{align} tr(\mathbf{W}'^{-1}) =\;& tr\bigg( \mathbf{W}^{-1}+ \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T + \kappa_0(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \notag\\% Substract added factors & +(N - \kappa_0)\mathbf{\overline{x}} \mathbf{\overline{x}}^T + 2\kappa_0\boldsymbol{\overline{x}}\boldsymbol{\mu}_0^T \bigg) \end{align}$

প্রায়! তবে এখনও সেখানে নেই। লক্ষ্যটি হ'ল:

\begin{aligned} W^{- 1} + \sum_{i = 1}^{N} (x_{i} - \bar{x}) (x_{i} - \bar{x})^{T} + \frac{κ_{0} N}{κ_{0} + N} (\bar{x} - μ_{0}) (\bar{x} - μ_{0})^{T} \end{aligned}

$\begin{align} \mathbf{W}^{-1} + \sum_{i=1}^N (\mathbf{x}_i - \overline{\mathbf{x}})(\mathbf{x}_i - \overline{\mathbf{x}})^T + \frac{\kappa_0 N}{\kappa_0 + N}(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)(\boldsymbol{\overline{x}} - \boldsymbol{\mu}_0)^T \end{align}$

bayesian posterior

— আলবার্তো
সূত্র

উত্তর:

ট্রেসটি চক্রযুক্ত, সুতরাং । এছাড়াও ট্রেস অতিরিক্ত সংযোজন করে যাতে । এই তথ্যের সাহায্যে আপনার ট্রেস পদগুলিতে শব্দটি পিছনে পিছনে ঘুরতে সক্ষম হওয়া উচিত , ট্রেস শর্তাদি একসাথে একত্রিত করতে হবে। ফলাফলটি $tr(ABC) = tr(BCA) = tr(CAB)$ $tr(A+B) = tr(A) + tr(B)$ $\Lambda$

W^{' - 1} = W^{- 1} + \sum_{i = 1}^{N} x_{i} x_{i}^{⊺} + μ_{0} μ_{0}^{⊺}

$W'^{-1} = W^{-1} + \sum_{i=1}^N x_i x_i^\intercal + \mu_0\mu_0^\intercal$

— bdeonovic
সূত্র

ধন্যবাদ! এখনো, আমি কিভাবে সেখান থেকে মান ফলাফল (পেতে দেখি না en.wikipedia.org/wiki/Conjugate_prior ) যা ধারণ করে এবং । এমনকি আমার নেতিবাচক চিহ্নগুলিও নেই: ও

\sum (x_{i} - \bar{x})

$\sum(x_i-\overline{x})$

\bar{x} - μ_{0}

$\overline{x} - \mu_0$

— আলবার্তো

সম্ভাবনা আগে $\times$

| Λ |^{N / 2} \exp {- \frac{1}{2} (\sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} - N {\bar{x}}^{T} Λ μ - μ^{T} Λ N \bar{x} + N μ^{T} Λ μ)} \times | Λ |^{(ν_{0} - D - 1) / 2} \exp {- \frac{1}{2} t r (W_{0}^{- 1} Λ)} \times | Λ |^{1 / 2} \exp {- \frac{κ_{0}}{2} (μ^{T} Λ μ - μ^{T} Λ μ_{0} - μ_{0}^{T} Λ μ + μ_{0}^{T} Λ μ_{0})} .

$|\boldsymbol{\Lambda}|^{N/2} \exp\left\{-\frac{1}{2}\left(\sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - N\boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} N\boldsymbol{\bar{x}} + N\boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{(\nu_0 - D - 1)/2} \exp\left\{-\frac{1}{2} tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{1/2} \exp \left\{ -\frac{\kappa_0}{2} \left( \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T\boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) \right\}.$ এটি আমরা আবার লিখতে পারি

| Λ |^{1 / 2} | Λ |^{(ν_{0} + N - D - 1) / 2} \times \exp {- \frac{1}{2} ((κ_{0} + N) μ^{T} Λ μ - μ^{T} Λ (κ_{0} μ_{0} + N \bar{x}) - (κ_{0} μ_{0} + N \bar{x})^{T} Λ μ + κ_{0} μ_{0}^{T} Λ μ_{0} + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + t r (W_{0}^{- 1} Λ))}

$|\boldsymbol{\Lambda}|^{1/2} |\boldsymbol{\Lambda}|^{(\nu_0 + N - D - 1)/2} \\ \times \exp \left\{ -\frac{1}{2} \left( \left(\kappa_0 + N\right) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) \right) \right\}$

(κ_{0} + N) μ^{T} Λ μ - μ^{T} Λ (κ_{0} μ_{0} + N \bar{x}) - (κ_{0} μ_{0} + N \bar{x})^{T} Λ μ + κ_{0} μ_{0}^{T} Λ μ_{0} + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + t r (W_{0}^{- 1} Λ)

(κ_{0} + N) μ^{T} Λ μ - μ^{T} Λ (κ_{0} μ_{0} + N \bar{x}) - (κ_{0} μ_{0} + N \bar{x})^{T} Λ μ + \frac{1}{κ_{0} + N} (κ_{0} μ_{0} + N \bar{x})^{T} Λ (κ_{0} μ_{0} + N \bar{x}) - \frac{1}{κ_{0} + N} (κ_{0} μ_{0} + N \bar{x})^{T} Λ (κ_{0} μ_{0} + N \bar{x}) + κ_{0} μ_{0}^{T} Λ μ_{0} + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + t r (W_{0}^{- 1} Λ) .

$(\kappa_0 + N) \boldsymbol{\mu}^T \boldsymbol{\Lambda} \boldsymbol{\mu} - \boldsymbol{\mu}^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) - (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} \boldsymbol{\mu} \\ + \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) \\ + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$ শীর্ষ দুটি লাইন এখন হিসাবে ফ্যাক্টরিস

(κ_{0} + N) {(μ - \frac{κ_{0} μ + N \bar{x}}{κ_{0} + N})}^{T} Λ (μ - \frac{κ_{0} μ + N \bar{x}}{κ_{0} + N}) .

$(\kappa_0 + N) \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right).$

, যুক্ত এবং বিয়োগ করে নিম্নলিখিত: হিসাবে আবার লিখতে পারেন $N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}$

- \frac{1}{κ_{0} + N} (κ_{0} μ_{0} + N \bar{x})^{T} Λ (κ_{0} μ_{0} + N \bar{x}) + κ_{0} μ_{0}^{T} Λ μ_{0} + \sum_{i = 1}^{N} x_{i}^{T} Λ x_{i} + t r (W_{0}^{- 1} Λ)

$- \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + \sum_{i=1}^N \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right)$

\sum_{i = 1}^{N} (x_{i}^{T} Λ x_{i} - x_{i}^{T} Λ \bar{x} - {\bar{x}}^{T} Λ x_{i} + {\bar{x}}^{T} Λ \bar{x}) + N {\bar{x}}^{T} Λ \bar{x} + κ_{0} μ_{0}^{T} Λ μ_{0} - \frac{1}{κ_{0} + N} (κ_{0} μ_{0} + N \bar{x})^{T} Λ (κ_{0} μ_{0} + N \bar{x}) + t r (W_{0}^{- 1} Λ) .

$\sum_{i=1}^N \left( \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{x}_i - \boldsymbol{x}_i^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{x}_i + \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} \right) + N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}})^T \boldsymbol{\Lambda} (\kappa_0 \boldsymbol{\mu}_0 + N \boldsymbol{\bar{x}}) + tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right).$ যোগফল

\sum_{i = 1}^{N} (x_{i}^{T} Λ x_{i} - x_{i}^{T} Λ \bar{x} - {\bar{x}}^{T} Λ x_{i} + {\bar{x}}^{T} Λ \bar{x})

\sum_{i = 1}^{N} {(x_{i} - \bar{x})}^{T} Λ (x_{i} - \bar{x}) .

$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right).$

N {\bar{x}}^{T} Λ \bar{x} + κ_{0} μ_{0}^{T} Λ μ_{0} - \frac{1}{κ_{0} + N} (κ_{0} μ_{0} + N \bar{x})^{T} Λ (κ_{0} μ_{0} + N \bar{x})

N {\bar{x}}^{T} Λ \bar{x} + κ_{0} μ_{0}^{T} Λ μ_{0} - \frac{1}{κ_{0} + N} (κ_{0}^{2} μ_{0}^{T} Λ μ_{0} + N κ_{0} μ_{0}^{T} Λ {\bar{x}}_{0} + N κ_{0} {\bar{x}}^{T} Λ μ_{0} + N^{2} {\bar{x}}^{T} Λ \bar{x}),

$N \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \frac{1}{\kappa_0 + N}(\kappa_0^2 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N \kappa_0 \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}_0 + N \kappa_0 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 + N^2 \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}}),$ যা সমান

\frac{N κ_{0}}{κ_{0} + N} ({\bar{x}}^{T} Λ \bar{x} - {\bar{x}}^{T} Λ μ_{0} - μ_{0}^{T} Λ \bar{x} + μ_{0}^{T} Λ μ_{0}) = \frac{N κ_{0}}{κ_{0} + N} {(\bar{x} - μ_{0})}^{T} Λ (\bar{x} - μ_{0}) .

$\frac{N\kappa_0}{\kappa_0 + N} \left( \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} - \boldsymbol{\bar{x}}^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 - \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\bar{x}} + \boldsymbol{\mu}_0^T \boldsymbol{\Lambda} \boldsymbol{\mu}_0 \right) = \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$

নিম্নলিখিত দুটি পদটি স্কেলারগুলি হল: এবং যে কোনও স্কেলার তার ট্রেসের সমান, তাই

\sum_{i = 1}^{N} {(x_{i} - \bar{x})}^{T} Λ (x_{i} - \bar{x}), \frac{N κ_{0}}{κ_{0} + N} {(\bar{x} - μ_{0})}^{T} Λ (\bar{x} - μ_{0}) .

$\sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right), \thinspace \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right).$

t r (W_{0}^{- 1} Λ) + \sum_{i = 1}^{N} {(x_{i} - \bar{x})}^{T} Λ (x_{i} - \bar{x}) + \frac{N κ_{0}}{κ_{0} + N} {(\bar{x} - μ_{0})}^{T} Λ (\bar{x} - μ_{0})

$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)$ হিসাবে আবার লেখা যেতে পারে যেহেতু , উপরের সমষ্টি সমান

t r (W_{0}^{- 1} Λ) + t r (\sum_{i = 1}^{N} {(x_{i} - \bar{x})}^{T} Λ (x_{i} - \bar{x})) + t r (\frac{N κ_{0}}{κ_{0} + N} {(\bar{x} - μ_{0})}^{T} Λ (\bar{x} - μ_{0})) .

$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \right).$

t r (A B C) = t r (C A B)

$tr(\boldsymbol{A} \boldsymbol{B} \boldsymbol{C}) = tr(\boldsymbol{C} \boldsymbol{A} \boldsymbol{B})$

t r (W_{0}^{- 1} Λ) + t r (\sum_{i = 1}^{N} (x_{i} - \bar{x}) {(x_{i} - \bar{x})}^{T} Λ) + t r (\frac{N κ_{0}}{κ_{0} + N} (\bar{x} - μ_{0}) {(\bar{x} - μ_{0})}^{T} Λ) .

$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} \right) + tr \left( \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} \right) + tr \left( \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right).$ এই সত্যটি ব্যবহার করে আমরা সমষ্টিটি আবার লিখতে পারি

t r (A + B) = t r (A) + t r (B)

$tr(\boldsymbol{A} + \boldsymbol{B}) = tr(\boldsymbol{A}) + tr(\boldsymbol{B})$

t r (W_{0}^{- 1} Λ + \sum_{i = 1}^{N} (x_{i} - \bar{x}) {(x_{i} - \bar{x})}^{T} Λ + \frac{N κ_{0}}{κ_{0} + N} (\bar{x} - μ_{0}) {(\bar{x} - μ_{0})}^{T} Λ) = t r ((W_{0}^{- 1} + \sum_{i = 1}^{N} (x_{i} - \bar{x}) {(x_{i} - \bar{x})}^{T} + \frac{N κ_{0}}{κ_{0} + N} (\bar{x} - μ_{0}) {(\bar{x} - μ_{0})}^{T}) Λ) .

$tr \left( \boldsymbol{W}_0^{-1} \boldsymbol{\Lambda} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T \boldsymbol{\Lambda} + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \boldsymbol{\Lambda} \right) = tr \left( \left( \boldsymbol{W}_0^{-1} + \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \right) \boldsymbol{\Lambda} \right).$

এর সবগুলি একসাথে রেখে, যদি আমরা আমরা আছে সম্ভাবনা পূর্বে সমান $\boldsymbol{S} = \sum_{i=1}^N \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right) \left( \boldsymbol{x}_i - \boldsymbol{\bar{x}} \right)^T$ $\times$

| Λ |^{1 / 2} \exp {- \frac{κ_{0} + N}{2} {(μ - \frac{κ_{0} μ + N \bar{x}}{κ_{0} + N})}^{T} Λ (μ - \frac{κ_{0} μ + N \bar{x}}{κ_{0} + N})} \times | Λ |^{(ν_{0} + N - D - 1) / 2} \exp {- \frac{1}{2} t r ((W_{0}^{- 1} + S + \frac{N κ_{0}}{κ_{0} + N} (\bar{x} - μ_{0}) {(\bar{x} - μ_{0})}^{T}) Λ)},

$|\boldsymbol{\Lambda}|^{1/2} \exp \left\{-\frac{\kappa_0 + N}{2} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right)^T \boldsymbol{\Lambda} \left( \boldsymbol{\mu} - \frac{\kappa_0 \boldsymbol{\mu} + N \boldsymbol{\bar{x}}}{\kappa_0 + N} \right) \right\} \\ \times |\boldsymbol{\Lambda}|^{(\nu_0 + N - D - 1)/2} \exp \left\{ - \frac{1}{2} tr \left( \left( \boldsymbol{W}_0^{-1} + \boldsymbol{S} + \frac{N\kappa_0}{\kappa_0 + N}\left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right) \left( \boldsymbol{\bar{x}} - \boldsymbol{\mu}_0 \right)^T \right) \boldsymbol{\Lambda} \right) \right\},$ প্রয়োজন হিসাবে।

— অ্যালেক্স
সূত্র

সাধারণ-উইশার্টের উত্তরোত্তর ব্যয়

স্বাধীনতার ডিগ্রিυ′υ′\upsilon'

স্কেল ফ্যাক্টরκ′κ′\kappa'

গড়μ′μ′\boldsymbol{\mu}'

স্কেল ম্যাট্রিক্সW′W′\boldsymbol{W}'

স্বাধীনতার ডিগ্রি $\upsilon'$

স্কেল ফ্যাক্টর $\kappa'$

গড় $\boldsymbol{\mu}'$

স্কেল ম্যাট্রিক্স $\boldsymbol{W}'$