PRML一书中关于贝叶斯曲线拟合结论的推导细节
我们令训练数据集为
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(X,T)
(X,T), 对于一个新的点
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p(t|x, X, T)
p(t∣x,X,T)
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p(t|x,X,T) =int p(t|x,w,X,T)dw= int p(t|x,w)p(w|X,T)dw \
p(t∣x,X,T)=∫p(t∣x,w,X,T)dw=∫p(t∣x,w)p(w∣X,T)dw
其中，
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w=[w_0,w_1,...,w_M]^T
w=[w0,w1,...,wM]T 是
M
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M阶多项式的参数
在PRML一书中，直接给出了这么一个结论
预测分布由高斯分布 N ( t ∣ m ( x ) , s 2 ( x ) ) mathcal N(t|m(x),s^2(x)) N(t∣m(x),s2(x))给出
其中均值 m ( x ) = β ϕ ( x ) T S ∑ n = 1 N ϕ ( x n ) t n m(x) = betaphi(x)^TSsum_{n=1}^{N}phi(x_n)t_n m(x)=βϕ(x)TS∑n=1Nϕ(xn)tn
方差 s 2 ( x ) = β − 1 + ϕ ( x ) T S ϕ ( x ) s^2(x) = beta^{-1} + phi(x)^TSphi(x) s2(x)=β−1+ϕ(x)TSϕ(x)
其中 S − 1 = α I + β ∑ n = 1 N ϕ ( x n ) ϕ ( x n ) T S^{-1} = alpha I +betasum_{n=1}^{N}phi(x_n)phi(x_n)^T S−1=αI+β∑n=1Nϕ(xn)ϕ(xn)T, ϕ ( x ) = [ 1 , x , x 2 , . . . , x M ] T phi(x)=[1,x,x^2,...,x^M]^T ϕ(x)=[1,x,x2,...,xM]T
这个结论给的非常突然，让人无所适从，我决定花点时间分析一下，并记录下来，以供大家参考！
如上式所述，其可以被写成积分的形式，我们利用一些结论来进行分析
- 首先对于t的分布应该是一个高斯分布，
p ( t ∣ x , w ) = N ( t ∣ y ( x , w ) , β − 1 ) p(t|x,w) = mathcal N(t|y(x,w), beta^{-1}) p(t∣x,w)=N(t∣y(x,w),β−1)
- 对于分布 p ( w ∣ X , T ) p(w|X,T) p(w∣X,T), 其正比于先验分布和似然的乘积
p ( w ∣ X , T , α , β ) ∝ p ( T ∣ X , w , β ) p ( w ∣ α ) p(w|X,T, alpha, beta) varpropto p(T|X, w, beta)p(w|alpha) p(w∣X,T,α,β)∝p(T∣X,w,β)p(w∣α)
如果
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alpha, beta
α,β 已知， 可以写成：
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p(w|X,T) = frac{p(T|X,w)p(X,w)}{p(X,T)} = frac{p(T|X,w)p(w|X)p(X)}{p(T|X)p(X)} = frac{p(T|X,w)p(w|X)}{p(T|X)}
p(w∣X,T)=p(X,T)p(T∣X,w)p(X,w)=p(T∣X)p(X)p(T∣X,w)p(w∣X)p(X)=p(T∣X)p(T∣X,w)p(w∣X)
所以我们可以继续将积分式子改写：
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p(t|x,X,T) = int mathcal N(t|y(x,w), beta^{-1}) prod_{n=1}^Nmathcal N(t_n|y(x_n, w), beta^{-1})frac{p(w|X)}{p(T|X)}dw
p(t∣x,X,T)=∫N(t∣y(x,w),β−1)n=1∏NN(tn∣y(xn,w),β−1)p(T∣X)p(w∣X)dw
y ( x , w ) = w 0 + w 1 x + w 2 x 2 + ⋯ + w M x M = ϕ ( x ) T [ w 0 , w 1 , . . . , w M ] T = ϕ ( x ) T w ϕ ( x ) = [ 1 , x , x 2 , . . . , x M ] T y(x,w) = w_0 + w_1x + w_2x^2 + cdots + w_Mx^M = phi(x)^T[w_0, w_1, ..., w_M]^T = phi(x)^Tw\ phi(x) = [1,x,x^2,...,x^M]^T y(x,w)=w0+w1x+w2x2+⋯+wMxM=ϕ(x)T[w0,w1,...,wM]T=ϕ(x)Twϕ(x)=[1,x,x2,...,xM]T
从而，对于高斯分布
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mathcal N(t|y(x,w), beta^{-1})
N(t∣y(x,w),β−1), 可以写成:
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mathcal N(t|y(x,w), beta^{-1}) = frac{1}{sqrt{2pibeta^{-1}}}exp(-frac{(t-y(x,w))^2}{2beta^{-1}}) = frac{1}{sqrt{2pibeta^{-1}}}exp(-frac{(t-phi(x)^Tw)^2}{2beta^{-1}})
N(t∣y(x,w),β−1)=2πβ−11exp(−2β−1(t−y(x,w))2)=2πβ−11exp(−2β−1(t−ϕ(x)Tw)2)
同样的，我们可以写出N个高斯分布乘积的形式
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prod_{n=1}^Nmathcal N(t_n|y(x_n, w), beta^{-1})=(frac{1}{2pibeta^{-1}})^{N/2}exp(-frac{beta}{2}sum_{n=1}^N(t_n-phi(x_n)^Tw)^2)
n=1∏NN(tn∣y(xn,w),β−1)=(2πβ−11)N/2exp(−2βn=1∑N(tn−ϕ(xn)Tw)2)
于是，如下：
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mathcal N(t|y(x,w), beta^{-1}) prod_{n=1}^Nmathcal N(t_n|y(x_n, w), beta^{-1}) = (frac{1}{2pibeta^{-1}})^{(N+1)/2}exp(-frac{beta}{2}((t-phi(x)^Tw)^2 + sum_{n=1}^N(t_n-phi(x_n)^Tw)^2))
N(t∣y(x,w),β−1)n=1∏NN(tn∣y(xn,w),β−1)=(2πβ−11)(N+1)/2exp(−2β((t−ϕ(x)Tw)2+n=1∑N(tn−ϕ(xn)Tw)2))
如果我们将
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exp
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p(w|X) = p(w|alpha) = (frac{alpha}{2pi})^{(M+1)/2}exp(-frac{alpha}{2}w^Tw)
p(w∣X)=p(w∣α)=(2πα)(M+1)/2exp(−2αwTw)， 且
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p(T|X)=1
p(T∣X)=1
则有
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p(t|x,X,T) = (frac{beta}{2pi })^{frac{N+1}{2}} (frac{alpha}{2pi})^{ frac{M+1}{2}}int expBig(-frac{alpha}{2}w^Tw -frac{beta}{2}((t-phi(x)^Tw)^2 + sum_{n=1}^N(t_n-phi(x_n)^Tw)^2) Big ) dw
p(t∣x,X,T)=(2πβ)2N+1(2πα)2M+1∫exp(−2αwTw−2β((t−ϕ(x)Tw)2+n=1∑N(tn−ϕ(xn)Tw)2))dw
注意到，高斯积分的形式
∫
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int _{-infin}^{+infin}e^{-a(x+b)^2} dx = sqrt{frac{pi}{a}}
∫−∞+∞e−a(x+b)2dx=aπ
故,
∫
exp
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int expBig(-frac{alpha}{2}w^Tw -frac{beta}{2}((t-phi(x)^Tw)^2 + sum_{n=1}^N(t_n-phi(x_n)^Tw)^2) Big ) dw = intexp(-k(w+b)^2+u)dw\ k = frac{alpha}{2} + frac{beta}{2}(phi(x)^Tphi(x) + sum_{n=1}^{N}phi(x_n)^Tphi(x_n))
∫exp(−2αwTw−2β((t−ϕ(x)Tw)2+n=1∑N(tn−ϕ(xn)Tw)2))dw=∫exp(−k(w+b)2+u)dwk=2α+2β(ϕ(x)Tϕ(x)+n=1∑Nϕ(xn)Tϕ(xn))
相当于对于一个二次式进行配方，我们简单记作：
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-m_1w^2-m_2(m_3w^2+m_4w+m_5)\ m_1 = frac{alpha}{2}, m_2=frac{beta}{2}\ m_3 = phi(x)^Tphi(x) + sum_{n=1}^Nphi(x_n)^Tphi(x_n)\ m_4 =-2(tphi(x)^T + sum_{n=1}^Nt_nphi(x_n)^T)\ m_5 = t^2 + sum_{n=1}^Nt_n^2
−m1w2−m2(m3w2+m4w+m5)m1=2α,m2=2βm3=ϕ(x)Tϕ(x)+n=1∑Nϕ(xn)Tϕ(xn)m4=−2(tϕ(x)T+n=1∑Ntnϕ(xn)T)m5=t2+n=1∑Ntn2
从而，
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-(m_1+m_2m_3)w^2 - m_2m_4w - m_2m_5=-(m_1+m_2m_3)(w^2+frac{m_2m_4}{m_1+m_2m_3}w + frac{m_2m_5}{m_1+m_2m_3})\ =-(m_1+m_2m_3)[(w+frac{m_2m_4}{2(m_1+m_2m_3)})^2 + frac{m_2m_5}{m_1+m_2m_3} - frac{m_2^2m_4^2}{4(m_1+m_2m_3)^2}]\ =-(m_1+m_2m_3)[(w+frac{m_2m_4}{2(m_1+m_2m_3)})^2 + frac{4(m_1+m_2m_3)m_2m_5 -m_2^2m_4^2}{4(m_1+m_2m_3)^2}]\ =-(m_1+m_2m_3)(w+frac{m_2m_4}{2(m_1+m_2m_3)})^2 + frac{4(m_1+m_2m_3)m_2m_5 -m_2^2m_4^2}{4(m_1+m_2m_3)}
−(m1+m2m3)w2−m2m4w−m2m5=−(m1+m2m3)(w2+m1+m2m3m2m4w+m1+m2m3m2m5)=−(m1+m2m3)[(w+2(m1+m2m3)m2m4)2+m1+m2m3m2m5−4(m1+m2m3)2m22m42]=−(m1+m2m3)[(w+2(m1+m2m3)m2m4)2+4(m1+m2m3)24(m1+m2m3)m2m5−m22m42]=−(m1+m2m3)(w+2(m1+m2m3)m2m4)2+4(m1+m2m3)4(m1+m2m3)m2m5−m22m42
存在一个常数项，即
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frac{4(m_1+m_2m_3)m_2m_5 -m_2^2m_4^2}{4(m_1+m_2m_3)} = frac{beta(alpha + beta m_3)m_5 - 4^{-1}beta^2m_4^2}{2(alpha + beta m_3)}
4(m1+m2m3)4(m1+m2m3)m2m5−m22m42=2(α+βm3)β(α+βm3)m5−4−1β2m42
从而,
∫
exp
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exp
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int expBig(-frac{alpha}{2}w^Tw -frac{beta}{2}((t-phi(x)^Tw)^2 + sum_{n=1}^N(t_n-phi(x_n)^Tw)^2) Big ) dw = sqrt{frac{pi}{k}}exp( frac{beta(alpha + beta m_3)m_5 - 4^{-1}beta^2m_4^2}{2(alpha + beta m_3)})
∫exp(−2αwTw−2β((t−ϕ(x)Tw)2+n=1∑N(tn−ϕ(xn)Tw)2))dw=kπexp(2(α+βm3)β(α+βm3)m5−4−1β2m42)
代入到原式得
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=
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exp
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m
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p(t|x,X,T)= (frac{beta}{2pi })^{frac{N+1}{2}} (frac{alpha}{2pi})^{ frac{M+1}{2}}sqrt{frac{pi}{k}}exp( frac{beta(alpha + beta m_3)m_5 - 4^{-1}beta^2m_4^2}{2(alpha + beta m_3)})\
p(t∣x,X,T)=(2πβ)2N+1(2πα)2M+1kπexp(2(α+βm3)β(α+βm3)m5−4−1β2m42)
对于指数部分的系数：
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(frac{beta}{2pi })^{frac{N+1}{2}} (frac{alpha}{2pi})^{ frac{M+1}{2}}sqrt{frac{pi}{k}}=Big((frac{beta^{N+1}alpha^{M+1}}{(2pi)^{N+M+2}} (frac{alpha}{2}+frac{beta}{2}(phi(x)^Tphi(x) + sum_{n=1}^{N}phi(x_n)^Tphi(x_n))))Big)^{1/2}
(2πβ)2N+1(2πα)2M+1kπ=(((2π)N+M+2βN+1αM+1(2α+2β(ϕ(x)Tϕ(x)+n=1∑Nϕ(xn)Tϕ(xn)))))1/2
而指数部分为:
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frac{beta(alpha + beta m_3)m_5 - 4^{-1}beta^2m_4^2}{2(alpha + beta m_3)} = frac{beta(alpha + beta m_3)(t^2 + t_{sum}) - frac{1}{4}beta^2(-2(tphi(x)^T + sum_{n=1}^Nt_nphi(x_n)^T))^2}{2(alpha +beta m_3)}\ =frac{beta(alpha + beta m_3)(t^2 + t_{sum}) - beta^2(tphi(x)^T + q)^2}{2(alpha +beta m_3)} =frac{[alpha beta + beta^2 m_3 - beta^2phi(x)^Tphi(x)]cdot t^2 -2beta^2 phi(x)^Tq^T t + (alphabeta+beta^2 m_3)t_{sum}-beta^2qq^T}{2(alpha +beta m_3)} \ =frac{[alpha beta + beta^2 v]cdot t^2 -2beta^2 phi(x)^Tq^T t + (alphabeta+beta^2 m_3)t_{sum}-beta^2qq^T}{2(alpha +beta m_3)}\ =(alpha beta + beta^2 v)frac{t^2 -2frac{beta^2 phi(x)^Tq^T}{alpha beta + beta^2 v} t + frac{(alphabeta+beta^2 m_3)t_{sum}-beta^2qq^T}{alpha beta + beta^2 v}}{2(alpha +beta m_3)} =(alpha beta + beta^2 v)frac{ (t -frac{beta^2 phi(x)^Tq^T}{alpha beta + beta^2 v})^2 -frac{(beta^2 phi(x)^Tq^T)^2}{(alpha beta + beta^2 v)^2} + frac{(alphabeta+beta^2 m_3)t_{sum}-beta^2qq^T}{alpha beta + beta^2 v}}{2(alpha +beta m_3)} \ t_{sum} = sum_{n=1}^Nt_n^2\ q = sum_{n=1}^Nt_nphi(x_n)^T\ v = sum_{n=1}^Nphi(x_n)^Tphi(x_n)
2(α+βm3)β(α+βm3)m5−4−1β2m42=2(α+βm3)β(α+βm3)(t2+tsum)−41β2(−2(tϕ(x)T+∑n=1Ntnϕ(xn)T))2=2(α+βm3)β(α+βm3)(t2+tsum)−β2(tϕ(x)T+q)2=2(α+βm3)[αβ+β2m3−β2ϕ(x)Tϕ(x)]⋅t2−2β2ϕ(x)TqTt+(αβ+β2m3)tsum−β2qqT=2(α+βm3)[αβ+β2v]⋅t2−2β2ϕ(x)TqTt+(αβ+β2m3)tsum−β2qqT=(αβ+β2v)2(α+βm3)t2−2αβ+β2vβ2ϕ(x)TqTt+αβ+β2v(αβ+β2m3)tsum−β2qqT=(αβ+β2v)2(α+βm3)(t−αβ+β2vβ2ϕ(x)TqT)2−(αβ+β2v)2(β2ϕ(x)TqT)2+αβ+β2v(αβ+β2m3)tsum−β2qqTtsum=n=1∑Ntn2q=n=1∑Ntnϕ(xn)Tv=n=1∑Nϕ(xn)Tϕ(xn)
故，我们可以从上式中，直接推出均值
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m(x)=frac{beta^2 phi(x)^Tq^T}{alpha beta + beta^2 v} = frac{beta phi(x)^Tq^T}{alpha + beta v} = (alpha+beta sum_{n=1}^Nphi(x_n)^Tphi(x_n))^{-1}(betaphi(x)^Tsum_{n=1}^{N}phi(x_n)t_n) = beta phi(x)^TSsum_{n=1}^{N}phi(x_n)t_n\ S^{-1} = alpha+beta sum_{n=1}^Nphi(x_n)^Tphi(x_n)
m(x)=αβ+β2vβ2ϕ(x)TqT=α+βvβϕ(x)TqT=(α+βn=1∑Nϕ(xn)Tϕ(xn))−1(βϕ(x)Tn=1∑Nϕ(xn)tn)=βϕ(x)TSn=1∑Nϕ(xn)tnS−1=α+βn=1∑Nϕ(xn)Tϕ(xn)
倘若上述配方成功，方差为
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s^2(x)=frac{alpha + beta m_3}{alpha beta + beta^2 v} = frac{alpha+beta(phi(x)^Tphi(x) + sum_{n=1}^Nphi(x_n)^Tphi(x_n))}{alphabeta + beta^2sum_{n=1}^Nphi(x_n)^Tphi(x_n) } =frac{alpha+betasum_{n=1}^Nphi(x_n)^Tphi(x_n) + beta phi(x)^Tphi(x)}{alphabeta + beta^2sum_{n=1}^Nphi(x_n)^Tphi(x_n) }\ =frac{1}{beta} + frac{phi(x)^Tphi(x)}{alpha + beta sum_{n=1}^Nphi(x_n)^Tphi(x_n) }\ = beta^{-1} + Sphi(x)^Tphi(x)
s2(x)=αβ+β2vα+βm3=αβ+β2∑n=1Nϕ(xn)Tϕ(xn)α+β(ϕ(x)Tϕ(x)+∑n=1Nϕ(xn)Tϕ(xn))=αβ+β2∑n=1Nϕ(xn)Tϕ(xn)α+β∑n=1Nϕ(xn)Tϕ(xn)+βϕ(x)Tϕ(x)=β1+α+β∑n=1Nϕ(xn)Tϕ(xn)ϕ(x)Tϕ(x)=β−1+Sϕ(x)Tϕ(x)
最后
以上就是苗条缘分最近收集整理的关于PRML一书中关于贝叶斯曲线拟合结论的推导细节的全部内容，更多相关PRML一书中关于贝叶斯曲线拟合结论内容请搜索靠谱客的其他文章。
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