20190824

20190824

2つの互いに独立な確率変数

(Ω,F,P)(\Omega, \mathcal{F}, P)
(X,MX)(\mathcal{X}, \mathcal{M}_{\mathcal{X}})
(Y,MY)(\mathcal{Y}, \mathcal{M}_{\mathcal{Y}})

X:Ω→XX: \Omega \to \mathcal{X}
Y:Ω→YY: \Omega \to \mathcal{Y}
(X,Y):ω↦(X(ω),Y(ω))(X, Y): \omega \mapsto (X(\omega), Y(\omega))

PX,Y=P∘(X,Y)−1P_{X, Y} = P \circ (X, Y)^{-1}
PX=PX,Y(⋅×Y)=P∘X−1P_X = P_{X, Y}(\cdot \times \mathcal{Y}) = P \circ X^{-1}
PY=PX,Y(X×⋅)=P∘Y−1P_Y = P_{X, Y}(\mathcal{X} \times \cdot) = P \circ Y^{-1}

PX,YP_{X, Y}も直積測度PX×PYP_X \times P_Yも,(X×Y,MX×MY)(\mathcal{X} \times \mathcal{Y}, \mathcal{M}_{\mathcal{X}} \times \mathcal{M}_{\mathcal{Y}})上の測度.

XXとYYが独立なら,PX,Y=PX×PYP_{X, Y} = P_X \times P_Y.

f:X×Y→[0,∞)f: \mathcal{X} \times \mathcal{Y} \to [0, \infty)

フビニ:

∫X×Yf(x,y)d(PX×PY)=∫X{∫Yf(x,y)dPY}dPX=∫Y{∫Xf(x,y)dPX}dPY \begin{aligned} \int_{\mathcal{X} \times \mathcal{Y}} f(x, y) d(P_X \times P_Y) &= \int_{\mathcal{X}} \left\{\int_{\mathcal{Y}} f(x, y) dP_Y\right\}dP_X\\ &= \int_{\mathcal{Y}} \left\{\int_{\mathcal{X}} f(x, y) dP_X\right\}dP_Y \end{aligned}

XXとYYが独立なら,EPX,Y[f(x,y)]=EPX[EPY[f(x,y)]]=EPY[EPX[f(x,y)]]\mathbb{E}_{P_{X, Y}} \left[f(x, y)\right] = \mathbb{E}_{P_X} \left[ \mathbb{E}_{P_Y} \left[f(x, y)\right]\right] = \mathbb{E}_{P_Y} \left[ \mathbb{E}_{P_X} \left[f(x, y)\right]\right]

empirical distributionとsample

f:X→[0,∞)f: \mathcal{X} \to [0, \infty)
DX(m)=1m∑i=1mδxiD^{(m)}_X = \frac{1}{m} \sum_{i=1}^m \delta_{x_i}
EPX[f]≈EDX(m)[f]=1m∑i=1mf(xi):=g(x)\mathbb{E}_{P_X} \left[f\right] \approx \mathbb{E}_{D^{(m)}_X} [f] = \frac{1}{m} \sum_{i=1}^m f(x_i) := g(\bm{x})
S:ω↦(X1(ω),…,Xm(ω))S: \omega \mapsto (X_1(\omega), \ldots, X_m(\omega))
PS=P∘S−1P_S = P \circ S^{-1}
EPS[g]=EP[g(S)]=1m∑i=1mEP[f(Xi)]=1m∑i=1mEPXi[f]=1m∑i=1mEPX[f]=EPX[f]\mathbb{E}_{P_S} \left[g\right] = \mathbb{E}_P \left[g(S)\right] = \frac{1}{m}\sum_{i=1}^m \mathbb{E}_P \left[f(X_i)\right] = \frac{1}{m}\sum_{i=1}^m \mathbb{E}_{P_{X_i}} \left[f\right] = \frac{1}{m}\sum_{i=1}^m \mathbb{E}_{P_{X}} \left[f\right] = \mathbb{E}_{P_X} \left[f\right]

Complexity

XXとYYは独立とする.

sup⁡g∈GYg(X)\sup_{g \in G} Y g(X)が大きい方が複雑度が高い.

EP[sup⁡g∈GYg(X)]=EPX,Y[sup⁡g∈Gyg(x)]≈EDX,Y(m)[sup⁡g∈Gyg(x)]=1m∑i=1msup⁡g∈Gyig(xi)=:f(x,y) \begin{aligned} \mathbb{E}_P \left[\sup_{g \in G} Y g(X)\right] &= \mathbb{E}_{P_{X, Y}} \left[\sup_{g \in G} y g(x)\right] \\ &\approx \mathbb{E}_{D^{(m)}_{X, Y}}\left[\sup_{g \in G} y g(x)\right] \\ &= \frac{1}{m} \sum_{i=1}^m \sup_{g \in G} y_i g(x_i) =: f(\bm{x}, \bm{y}) \end{aligned}

SY:ω↦(Y1(ω),…,Ym(ω))S_Y: \omega \mapsto (Y_1(\omega), \ldots, Y_m(\omega))

for x∈Xm\bm{x} \in \mathcal{X}^m, C^(G)=EPSY[f(x,⋅)]\hat{\mathscr{C}}(G) = \mathbb{E}_{P_{S_Y}} \left[f(\bm{x}, \cdot)\right]を,empirical complexityと呼ぶことにする.

SX:ω↦(X1(ω),…,Xm(ω))S_X: \omega \mapsto (X_1(\omega), \ldots, X_m(\omega))

EPSX[C^(G)]\mathbb{E}_{P_{S_X}} \left[\hat{\mathscr{C}}(G)\right]を,complexityと呼ぶことにする.

complexityのイメージは多分こんな感じ.Y∈{−1,1}\mathcal{Y} \in \left\{-1, 1\right\}にして,PYP_Yをuniformだとすれば,Rademacher complexityになるはず.

うーんでも,ちょっとまだきになるところがあるので,明日やろうと思う.

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