Gumbel 重参数化相关性质证明

Gumbel 的采样过程:

z = a r g m a x i { g i + l o g ( π i ) } , g i = − l o g ( − l o g ( u i ) ) , u i ∼ U ( 0 , 1 ) z=argmax_i \{g_i + log(\pi_i)\}, g_i = -log(-log(u_i)),u_i\sim U(0, 1) z=argmaxi{gi+log(πi)},gi=log(log(ui)),uiU(0,1)

采样得到的随机变量满足一下分布:

g i ∼ G u m b l e ( 0 , 1 ) ( 1 ) g_i \sim Gumble(0, 1) \quad (1) giGumble(0,1)(1)

h i = g i + l o g ( π i ) ∼ G u m b l e ( l o g ( π i ) , 1 ) ( 2 ) h_i = g_i + log(\pi_i)\sim Gumble(log(\pi_i), 1) \quad (2) hi=gi+log(πi)Gumble(log(πi),1)(2)

证明过程:

P ( u ) = P ( U ≤ u ) = u , u ∈ ( 0 , 1 ) P(u) =P(U\le u)= u, u\in(0, 1) P(u)=P(Uu)=u,u(0,1)

G = − l o g ( − l o g ( U ) ) , u ∈ ( 0 , 1 ) G = -log(-log(U)), u\in(0, 1) G=log(log(U)),u(0,1)

P ( g ) = P ( G ≤ g ) = P ( − l o g ( − l o g ( U ) ) ≤ g ) P(g) =P(G\le g) = P(-log(-log(U))\le g) P(g)=P(Gg)=P(log(log(U))g)

= P ( U ≤ e x p ( − e x p ( − g ) ) ) =P(U\le exp(-exp(-g))) =P(Uexp(exp(g)))

= e x p ( − e x p ( − g ) ) = exp(-exp(-g)) =exp(exp(g))

P ( g ) = e x p ( − e x p ( − g ) ) P(g) = exp(-exp(-g)) P(g)=exp(exp(g))

g i ∼ G u m b l e ( 0 , 1 ) g_i\sim Gumble(0, 1) giGumble(0,1)

h i = g i + l o g ( π i ) ∼ G u m b l e ( l o g ( π i ) , 1 ) h_i = g_i + log(\pi_i)\sim Gumble(log(\pi_i), 1) hi=gi+log(πi)Gumble(log(πi),1)

P ( Z = z ) = π i ( 3 ) P(Z=z) = \pi_i \quad(3) P(Z=z)=πi(3)

证明过程:

P ( Z = z ∣ U z = u z ) = ∏ i ≠ z P ( H i < g z + l o g ( π z ) ) P(Z=z | U_z = u_z) = \prod_{i\ne z} P(H_i < g_z + log(\pi_z)) P(Z=zUz=uz)=i=zP(Hi<gz+log(πz))

= ∏ i ≠ z P ( G i + l o g ( π i ) < g z + l o g ( π z ) ) =\prod_{i\ne z} P(G_i + log(\pi_i) < g_z + log(\pi_z)) =i=zP(Gi+log(πi)<gz+log(πz))

= ∏ i ≠ z P ( U i < u z p i / p z ) = \prod_{i\ne z} P(U_i < u_z^{p_i/p_z}) =i=zP(Ui<uzpi/pz)

= ∏ i ≠ z u z p i / p z = u z 1 / p z − 1 = \prod_{i\ne z} u_z^{p_i/p_z} = u_z^{1/p_z - 1} =i=zuzpi/pz=uz1/pz1

P ( Z = z ) = ∫ 0 1 P ( Z = z ∣ U z = u z ) P ( U z = u z ) d u z P(Z = z) = \int_0^1 P(Z=z|U_z = u_z)P(U_z=u_z) du_z P(Z=z)=01P(Z=zUz=uz)P(Uz=uz)duz

= ∫ 0 1 u z 1 / p z − 1 ∗ 1 ∗ d u z = \int_0^1 u_z^{1/p_z - 1} * 1 * du_z =01uz1/pz11duz

= 1 1 / p z u z 1 / p z ∣ 0 1 = \frac{1}{1/p_z}u_z^{1/p_z}|_0^1 =1/pz1uz1/pz01

= p z = p_z =pz

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