解：将概率分布代入对数似然函数，

l(ψ,μ0,μ1,∑)=∑i=1mlogpX∣Y(x(i)∣y(i);μ0,μ1,∑)+∑i=1mlogpY(y(i);ψ)l(\psi,\mu_0,\mu_1,\sum)=\sum^m_{i=1}{log{p_{X|Y}(x^{(i)}|y^{(i)};\mu_0,\mu_1,\sum)}}+\sum^m_{i=1}log{p_Y}(y^{(i)};\psi)l(ψ,μ0​,μ1​,∑)=∑i=1m​logpX∣Y​(x(i)∣y(i);μ0​,μ1​,∑)+∑i=1m​logpY​(y(i);ψ)

=∑i=1m(1−y(i))log1(2π)n/2∣∑∣1/2exp(12(x(i)−μ0)T∑−1(x(i)−μ0))=\sum^m_{i=1}(1-y^{(i)}){log \frac{1}{(2\pi)^{n/2}|\sum|^{1/2}}exp(\frac{1}{2}(x^{(i)}-\mu_0)^T\sum^{-1}(x^{(i)}-\mu_0))}=∑i=1m​(1−y(i))log(2π)n/2∣∑∣1/21​exp(21​(x(i)−μ0​)T∑−1(x(i)−μ0​))

+∑i=1my(i)log1(2π)n/2∣∑∣1/2exp(12(x(i)−μ1)T∑−1(x(i)−μ1))+\sum^m_{i=1}y^{(i)}{log \frac{1}{(2\pi)^{n/2}|\sum|^{1/2}}exp(\frac{1}{2}(x^{(i)}-\mu_1)^T\sum^{-1}(x^{(i)}-\mu_1))}+∑i=1m​y(i)log(2π)n/2∣∑∣1/21​exp(21​(x(i)−μ1​)T∑−1(x(i)−μ1​))

+∑i=1mlogψy(i)(1−ψ)1−y(i)+\sum^m_{i=1}{log\psi^{y^{(i)}}(1-\psi)^{1-y^{(i)}}}+∑i=1m​logψy(i)(1−ψ)1−y(i)

求取l(ψ,μ0,μ1,∑)l(\psi,\mu_0,\mu_1,\sum)l(ψ,μ0​,μ1​,∑)的最大值，令

∂∂ψl(ψ,μ0,μ1,∑)=0\frac{\partial}{\partial\psi}l(\psi,\mu_0,\mu_1,\sum)=0∂ψ∂​l(ψ,μ0​,μ1​,∑)=0 （1）

∇μ0l(ψ,μ0,μ1,∑)=0\nabla_{\mu_0}l(\psi,\mu_0,\mu_1,\sum)=0∇μ0​​l(ψ,μ0​,μ1​,∑)=0 （2）

∇μ1l(ψ,μ0,μ1,∑)=0\nabla_{\mu_1}l(\psi,\mu_0,\mu_1,\sum)=0∇μ1​​l(ψ,μ0​,μ1​,∑)=0 （3）

∇∑l(ψ,μ0,μ1,∑)=0\nabla_{\sum}l(\psi,\mu_0,\mu_1,\sum)=0∇∑​l(ψ,μ0​,μ1​,∑)=0 （4）

对于（1）式：

∂∂ψ∑i=1my(i)logψ+(1−y(i))log(1−ψ)=0\frac{\partial}{\partial\psi}{\sum^m_{i=1}y^{(i)}log\psi+(1-y^{(i)})log(1-\psi)}=0∂ψ∂​∑i=1m​y(i)logψ+(1−y(i))log(1−ψ)=0

∑i=1my(i)ψ+1−y(i)1−ψ=0{\sum^m_{i=1}\frac{y^{(i)}}{\psi}+\frac{1-y^{(i)}}{1-\psi}}=0∑i=1m​ψy(i)​+1−ψ1−y(i)​=0

∑i=1my(i)(1−ψ)+(1−y(i))ψ=0{\sum^m_{i=1}y^{(i)}{(1-\psi)}+(1-y^{(i)}){\psi}}=0∑i=1m​y(i)(1−ψ)+(1−y(i))ψ=0

∑i=1my(i)=mψ{\sum^m_{i=1}y^{(i)}}=m\psi∑i=1m​y(i)=mψ

ψ=∑i=1m1{y(i)=1}m\psi=\frac{\sum^m_{i=1}1\{y^{(i)}=1\}}{m}ψ=m∑i=1m​1{y(i)=1}​

对于（2）式：

∇μ0∑i=1m(1−y(i))(x(i)−μ0)T∑−1(x(i)−μ0)=0\nabla_{\mu_0}\sum^m_{i=1}(1-y^{(i)})(x^{(i)}-\mu_0)^T\sum^{-1}(x^{(i)}-\mu_0)=0∇μ0​​∑i=1m​(1−y(i))(x(i)−μ0​)T∑−1(x(i)−μ0​)=0

∑i=1m(1−y(i))(x(i)−μ0)T∑−1(x(i)−μ0)=0\sum^m_{i=1}(1-y^{(i)})(x^{(i)}-\mu_0)^T\sum^{-1}(x^{(i)}-\mu_0)=0∑i=1m​(1−y(i))(x(i)−μ0​)T∑−1(x(i)−μ0​)=0

∑i=1m(1−y(i))[∑−1(x(i)−μ0)d(x(i)−μ0)T+(x(i)−μ0)T∑−1d(x(i)−μ0)]=0\sum^m_{i=1}(1-y^{(i)})[\sum^{-1}(x^{(i)}-\mu_0)d(x^{(i)}-\mu_0)^T+(x^{(i)}-\mu_0)^T\sum^{-1}d(x^{(i)}-\mu_0)]=0∑i=1m​(1−y(i))[∑−1(x(i)−μ0​)d(x(i)−μ0​)T+(x(i)−μ0​)T∑−1d(x(i)−μ0​)]=0

∑i=1m(1−y(i))∑−1(x(i)−μ0)=0\sum^m_{i=1}(1-y^{(i)})\sum^{-1}(x^{(i)}-\mu_0)=0∑i=1m​(1−y(i))∑−1(x(i)−μ0​)=0

∑i=1m(1−y(i))(x(i)−μ0)=0\sum^m_{i=1}(1-y^{(i)})(x^{(i)}-\mu_0)=0∑i=1m​(1−y(i))(x(i)−μ0​)=0

∑i=1m(1−y(i))x(i)=∑i=1m(1−y(i))μ0\sum^m_{i=1}(1-y^{(i)})x^{(i)}=\sum^m_{i=1}(1-y^{(i)})\mu_0∑i=1m​(1−y(i))x(i)=∑i=1m​(1−y(i))μ0​

μ0=∑i=1m1{y(i)=0}x(i)/∑i=1m1{y(i)=0}\mu_0=\sum^m_{i=1}1\{y^{(i)}=0\}x^{(i)}/\sum^m_{i=1}1\{y^{(i)}=0\}μ0​=∑i=1m​1{y(i)=0}x(i)/∑i=1m​1{y(i)=0}

对于（3）式，类同（2）式：

μ0=∑i=1m1{y(i)=1}x(i)/∑i=1m1{y(i)=1}\mu_0=\sum^m_{i=1}1\{y^{(i)}=1\}x^{(i)}/\sum^m_{i=1}1\{y^{(i)}=1\}μ0​=∑i=1m​1{y(i)=1}x(i)/∑i=1m​1{y(i)=1}

对于（4）式：

∇∑(−m2log∣∑∣)−12∑i=1m(1−y(i))(x(i)−μ0)T∑−1(x(i)−μ0)−12∑i=1my(i)(x(i)−μ1)T∑−1(x(i)−μ1)=0\nabla_{\sum}(-\frac{m}{2}log|\sum|)-\frac{1}{2}\sum^m_{i=1}(1-y^{(i)})(x^{(i)}-\mu_0)^T\sum^{-1}(x^{(i)}-\mu_0)-\frac{1}{2}\sum^m_{i=1}y^{(i)}(x^{(i)}-\mu_1)^T\sum^{-1}(x^{(i)}-\mu_1)=0∇∑​(−2m​log∣∑∣)−21​∑i=1m​(1−y(i))(x(i)−μ0​)T∑−1(x(i)−μ0​)−21​∑i=1m​y(i)(x(i)−μ1​)T∑−1(x(i)−μ1​)=0

∇∑(mlog∣∑∣)+∇∑∑i=1m(1−y(i))(x(i)−μ0)T∑−1(x(i)−μ0)+∇∑∑i=1my(i)(x(i)−μ1)T∑−1(x(i)−μ1)=0\nabla_{\sum}(mlog|\sum|)+\nabla_{\sum}\sum^m_{i=1}(1-y^{(i)})(x^{(i)}-\mu_0)^T\sum^{-1}(x^{(i)}-\mu_0)+\nabla_{\sum}\sum^m_{i=1}y^{(i)}(x^{(i)}-\mu_1)^T\sum^{-1}(x^{(i)}-\mu_1)=0∇∑​(mlog∣∑∣)+∇∑​∑i=1m​(1−y(i))(x(i)−μ0​)T∑−1(x(i)−μ0​)+∇∑​∑i=1m​y(i)(x(i)−μ1​)T∑−1(x(i)−μ1​)=0

已知协方差矩阵Si=1m∑i=1m(x(i)−μi)(x(i)−μi)TS_i=\frac{1}{m}\sum^m_{i=1}(x^{(i)}-\mu_i)(x^{(i)}-\mu_i)^TSi​=m1​∑i=1m​(x(i)−μi​)(x(i)−μi​)T，将通过SiS_iSi​简化表达上式

∇∑∑i=1m(x(i)−μi)T∑−1(x(i)−μi)\nabla_{\sum}\sum^m_{i=1}(x^{(i)}-\mu_i)^T\sum^{-1}(x^{(i)}-\mu_i)∇∑​∑i=1m​(x(i)−μi​)T∑−1(x(i)−μi​)

=∇∑tr(∑i=1m(x(i)−μi)T∑−1(x(i)−μi))=\nabla_{\sum}tr(\sum^m_{i=1}(x^{(i)}-\mu_i)^T\sum^{-1}(x^{(i)}-\mu_i))=∇∑​tr(∑i=1m​(x(i)−μi​)T∑−1(x(i)−μi​))

=∇∑tr(∑i=1m(x(i)−μi)(x(i)−μi)T∑−1)=\nabla_{\sum}tr(\sum^m_{i=1}(x^{(i)}-\mu_i)(x^{(i)}-\mu_i)^T\sum^{-1})=∇∑​tr(∑i=1m​(x(i)−μi​)(x(i)−μi​)T∑−1)

=∇∑tr(miSi∑−1)=\nabla_{\sum}tr(m_iS_i\sum^{-1})=∇∑​tr(mi​Si​∑−1)

其中mi=∑k=1m1{y(k)=i}m_i=\sum^m_{k=1}1\{y^{(k)}=i\}mi​=∑k=1m​1{y(k)=i}，

∇∑tr(miSi∑−1)=−miSiT∑−2\nabla_{\sum}tr(m_iS_i\sum^{-1})=-m_iS_i^T\sum^{-2}∇∑​tr(mi​Si​∑−1)=−mi​SiT​∑−2，

而∇∑(mlog∣∑∣)=m1∣∑∣∣∑∣∑−1=m∑−1\nabla_{\sum}(mlog|\sum|)=m\frac{1}{|\sum|}|\sum|\sum^{-1}=m\sum^{-1}∇∑​(mlog∣∑∣)=m∣∑∣1​∣∑∣∑−1=m∑−1，

因此，（4）式可简化为

m∑−1−∑i2miSiT∑−2=0m\sum^{-1}-\sum_i^{2}m_iS_i^T\sum^{-2}=0m∑−1−∑i2​mi​SiT​∑−2=0

∑=1m∑i2miSiT\sum=\frac{1}{m}\sum_i^{2}m_iS_i^T∑=m1​∑i2​mi​SiT​

∑=1m∑i=1m(x(i)−μy(i))T(x(i)−μy(i))\sum=\frac{1}{m}\sum_{i=1}^{m}(x^{(i)}-\mu_{y^{(i)}})^T(x^{(i)}-\mu_{y^{(i)}})∑=m1​∑i=1m​(x(i)−μy(i)​)T(x(i)−μy(i)​)