원판이나 구처럼 둥근 영역에서 직교좌표로 적분하면 범위가 근호로 복잡해집니다. 좌표를 바꾸면(극좌표, 구면좌표) 범위가 단순해지지만, 그 대가로 “작은 조각의 넓이가 어떻게 바뀌는지”를 곱해 줘야 합니다. 그 배율이 야코비 행렬의 행렬식, 곧 야코비안입니다. 같은 공식이 확률에서 “변수를 바꿀 때 밀도가 어떻게 바뀌는가”를 알려 주고, 이것은 생성 모델(정규화 흐름)의 핵심입니다.
8.1 변수 변환 공식
한 변수의 치환적분(미적분학 9장)은 ∫ f ( x ) d x = ∫ f ( g ( u ) ) ∣ g ′ ( u ) ∣ d u \int f(x)\,dx = \int f(g(u))\,|g'(u)|\,du ∫ f ( x ) d x = ∫ f ( g ( u )) ∣ g ′ ( u ) ∣ d u d x = ∣ g ′ ( u ) ∣ d u dx = |g'(u)|\,du d x = ∣ g ′ ( u ) ∣ d u d u du d u g g g
여러 변수에서는 작은 사각형 d u d v du\,dv d u d v ( x , y ) = T ( u , v ) (x, y) = T(u, v) ( x , y ) = T ( u , v )
정리 8.1 (변수 변환). T T T S S S D D D
∬ D f ( x , y ) d x d y = ∬ S f ( T ( u , v ) ) ∣ det J T ( u , v ) ∣ d u d v . \iint_Df(x, y)\,dx\,dy = \iint_Sf(T(u, v))\,\left|\det J_T(u, v)\right|\,du\,dv. ∬ D f ( x , y ) d x d y = ∬ S f ( T ( u , v )) ∣ det J T ( u , v ) ∣ d u d v . ∣ det J T ∣ |\det J_T| ∣ det J T ∣ 야코비안 (Jacobian)이라 부르고 ∣ ∂ ( x , y ) ∂ ( u , v ) ∣ \left|\dfrac{\partial(x, y)}{\partial(u, v)}\right| ∂ ( u , v ) ∂ ( x , y )
8.2 극좌표
x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ det J = r \det J = r det J = r
d x d y = r d r d θ . dx\,dy = r\,dr\,d\theta. d x d y = r d r d θ . 중심에서 멀수록(r r r d r d θ dr\,d\theta d r d θ
예제 8.2. 반지름 a a a ∫ 0 2 π ∫ 0 a r d r d θ = 2 π ⋅ a 2 2 = π a 2 \int_0^{2\pi}\int_0^ar\,dr\,d\theta = 2\pi\cdot\frac{a^2}2 = \pi a^2 ∫ 0 2 π ∫ 0 a r d r d θ = 2 π ⋅ 2 a 2 = π a 2
예제 8.3. ∬ x 2 + y 2 ≤ 1 ( x 2 + y 2 ) d A = ∫ 0 2 π ∫ 0 1 r 2 ⋅ r d r d θ = 2 π ⋅ 1 4 = π 2 \iint_{x^2 + y^2 \le 1}(x^2 + y^2)\,dA = \int_0^{2\pi}\int_0^1r^2\cdot r\,dr\,d\theta = 2\pi\cdot\frac14 = \frac\pi2 ∬ x 2 + y 2 ≤ 1 ( x 2 + y 2 ) d A = ∫ 0 2 π ∫ 0 1 r 2 ⋅ r d r d θ = 2 π ⋅ 4 1 = 2 π
8.3 가우스 적분
정리 8.4. ∫ − ∞ ∞ e − x 2 d x = π \displaystyle\int_{-\infty}^{\infty}e^{-x^2}\,dx = \sqrt\pi ∫ − ∞ ∞ e − x 2 d x = π
증명. 값을 I I I
I 2 = ∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y = ∬ R 2 e − ( x 2 + y 2 ) d A = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ = 2 π ⋅ 1 2 = π . I^2 = \int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy = \iint_{\mathbb{R}^2}e^{-(x^2 + y^2)}\,dA = \int_0^{2\pi}\int_0^\infty e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\frac12 = \pi. I 2 = ∫ − ∞ ∞ e − x 2 d x ∫ − ∞ ∞ e − y 2 d y = ∬ R 2 e − ( x 2 + y 2 ) d A = ∫ 0 2 π ∫ 0 ∞ e − r 2 r d r d θ = 2 π ⋅ 2 1 = π . e − x 2 e^{-x^2} e − x 2 r r r ∫ e − r 2 r d r \int e^{-r^2}r\,dr ∫ e − r 2 r d r
따름정리 8.5. 정규분포 1 2 π σ e − ( x − μ ) 2 / ( 2 σ 2 ) \frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\mu)^2/(2\sigma^2)} 2 π σ 1 e − ( x − μ ) 2 / ( 2 σ 2 ) x = μ + 2 σ u x = \mu + \sqrt2\sigma u x = μ + 2 σ u
8.4 원기둥좌표와 구면좌표
좌표
변환
부피 조각
원기둥
x = r cos θ x = r\cos\theta x = r cos θ y = r sin θ y = r\sin\theta y = r sin θ z = z z = z z = z d V = r d r d θ d z dV = r\,dr\,d\theta\,dz d V = r d r d θ d z
구면
x = ρ sin ϕ cos θ x = \rho\sin\phi\cos\theta x = ρ sin ϕ cos θ y = ρ sin ϕ sin θ y = \rho\sin\phi\sin\theta y = ρ sin ϕ sin θ z = ρ cos ϕ z = \rho\cos\phi z = ρ cos ϕ d V = ρ 2 sin ϕ d ρ d ϕ d θ dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta d V = ρ 2 sin ϕ d ρ d ϕ d θ
구면좌표에서 ρ \rho ρ ϕ \phi ϕ z z z 0 ≤ ϕ ≤ π 0 \le \phi \le \pi 0 ≤ ϕ ≤ π θ \theta θ 0 ≤ θ < 2 π 0 \le \theta < 2\pi 0 ≤ θ < 2 π
예제 8.6. 반지름 a a a ∫ 0 2 π ∫ 0 π ∫ 0 a ρ 2 sin ϕ d ρ d ϕ d θ = 2 π ⋅ 2 ⋅ a 3 3 = 4 3 π a 3 \int_0^{2\pi}\int_0^\pi\int_0^a\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot2\cdot\frac{a^3}3 = \frac43\pi a^3 ∫ 0 2 π ∫ 0 π ∫ 0 a ρ 2 sin ϕ d ρ d ϕ d θ = 2 π ⋅ 2 ⋅ 3 a 3 = 3 4 π a 3
8.5 확률밀도의 변수 변환
정리 8.7. 확률변수 X \mathbf{X} X p X p_X p X Y = T ( X ) \mathbf{Y} = T(\mathbf{X}) Y = T ( X ) T T T
p Y ( y ) = p X ( x ) ∣ det J T ( x ) ∣ − 1 , x = T − 1 ( y ) . p_Y(\mathbf{y}) = p_X(\mathbf{x})\left|\det J_T(\mathbf{x})\right|^{-1}, \qquad \mathbf{x} = T^{-1}(\mathbf{y}). p Y ( y ) = p X ( x ) ∣ det J T ( x ) ∣ − 1 , x = T − 1 ( y ) . 이유. 확률은 보존되어야 합니다: p Y ( y ) ∣ d y ∣ = p X ( x ) ∣ d x ∣ p_Y(\mathbf{y})\,|d\mathbf{y}| = p_X(\mathbf{x})\,|d\mathbf{x}| p Y ( y ) ∣ d y ∣ = p X ( x ) ∣ d x ∣ ∣ det J ∣ |\det J| ∣ det J ∣
예제 8.8. X X X [ 0 , 1 ] [0, 1] [ 0 , 1 ] Y = 2 X + 3 Y = 2X + 3 Y = 2 X + 3 Y Y Y [ 3 , 5 ] [3, 5] [ 3 , 5 ] 1 2 \frac12 2 1 ∣ J ∣ = 2 |J| = 2 ∣ J ∣ = 2
예제 8.9. X ∼ N ( 0 , 1 ) X \sim N(0, 1) X ∼ N ( 0 , 1 ) Y = e X Y = e^X Y = e X x = ln y x = \ln y x = ln y d y d x = e x = y \frac{dy}{dx} = e^x = y d x d y = e x = y p Y ( y ) = 1 2 π e − ( ln y ) 2 / 2 ⋅ 1 y p_Y(y) = \frac{1}{\sqrt{2\pi}}e^{-(\ln y)^2/2}\cdot\frac1y p Y ( y ) = 2 π 1 e − ( l n y ) 2 /2 ⋅ y 1
8.6 이 장의 요약
∬ f d x d y = ∬ f ( T ) ∣ det J T ∣ d u d v \iint f\,dx\,dy = \iint f(T)\,|\det J_T|\,du\,dv ∬ f d x d y = ∬ f ( T ) ∣ det J T ∣ d u d v 극좌표 r d r d θ r\,dr\,d\theta r d r d θ ρ 2 sin ϕ d ρ d ϕ d θ \rho^2\sin\phi\,d\rho\,d\phi\,d\theta ρ 2 sin ϕ d ρ d ϕ d θ
가우스 적분 π \sqrt\pi π
쓰이는 곳. 정규화 흐름(normalizing flow) 생성 모델은 log p Y ( y ) = log p X ( x ) − log ∣ det J ∣ \log p_Y(\mathbf{y}) = \log p_X(\mathbf{x}) - \log|\det J| log p Y ( y ) = log p X ( x ) − log ∣ det J ∣ z = μ + σ ϵ z = \mu + \sigma\epsilon z = μ + σ ϵ
문제
8.1 [계산] · 확인 문제
극좌표로 ∬ x 2 + y 2 ≤ 4 e − ( x 2 + y 2 ) d A \iint_{x^2 + y^2 \le 4}e^{-(x^2 + y^2)}\,dA ∬ x 2 + y 2 ≤ 4 e − ( x 2 + y 2 ) d A
정답 풀이 ∫ 0 2 π ∫ 0 2 e − r 2 r d r d θ = 2 π ⋅ 1 2 ( 1 − e − 4 ) = π ( 1 − e − 4 ) ≈ 3.0841 \int_0^{2\pi}\int_0^2e^{-r^2}r\,dr\,d\theta = 2\pi\cdot\frac12(1 - e^{-4}) = \pi(1 - e^{-4}) \approx 3.0841 ∫ 0 2 π ∫ 0 2 e − r 2 r d r d θ = 2 π ⋅ 2 1 ( 1 − e − 4 ) = π ( 1 − e − 4 ) ≈ 3.0841
8.2 [계산] · 확인 문제
변환 x = 2 u + v x = 2u + v x = 2 u + v y = u − v y = u - v y = u − v ∣ det J ∣ |\det J| ∣ det J ∣
정답 풀이 J = [ 2 1 1 − 1 ] J = \begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} J = [ 2 1 1 − 1 ] det J = − 2 − 1 = − 3 \det J = -2 - 1 = -3 det J = − 2 − 1 = − 3 ∣ det J ∣ = 3 |\det J| = 3 ∣ det J ∣ = 3
8.3 [계산] · 확인 문제
∫ − ∞ ∞ e − x 2 / 2 d x \displaystyle\int_{-\infty}^{\infty}e^{-x^2/2}\,dx ∫ − ∞ ∞ e − x 2 /2 d x
정답 풀이 x = 2 u x = \sqrt2u x = 2 u 2 ∫ e − u 2 d u = 2 ⋅ π = 2 π ≈ 2.5066 \sqrt2\int e^{-u^2}du = \sqrt2\cdot\sqrt\pi = \sqrt{2\pi} \approx 2.5066 2 ∫ e − u 2 d u = 2 ⋅ π = 2 π ≈ 2.5066
8.4 [계산]
구면좌표로 반지름 1인 공 위에서 ∭ ( x 2 + y 2 + z 2 ) d V \iiint(x^2 + y^2 + z^2)\,dV ∭ ( x 2 + y 2 + z 2 ) d V
정답 풀이 ∫ 0 2 π ∫ 0 π ∫ 0 1 ρ 2 ⋅ ρ 2 sin ϕ d ρ d ϕ d θ = 2 π ⋅ 2 ⋅ 1 5 = 4 π 5 \int_0^{2\pi}\int_0^\pi\int_0^1\rho^2\cdot\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot2\cdot\frac15 = \frac{4\pi}5 ∫ 0 2 π ∫ 0 π ∫ 0 1 ρ 2 ⋅ ρ 2 sin ϕ d ρ d ϕ d θ = 2 π ⋅ 2 ⋅ 5 1 = 5 4 π
8.5 [증명]
X X X p X p_X p X Y = a X + b Y = aX + b Y = a X + b a ≠ 0 a \ne 0 a = 0 p Y ( y ) = 1 ∣ a ∣ p X ( y − b a ) p_Y(y) = \frac{1}{|a|}p_X\!\left(\frac{y - b}{a}\right) p Y ( y ) = ∣ a ∣ 1 p X ( a y − b )
풀이 a > 0 a > 0 a > 0 P ( Y ≤ y ) = P ( X ≤ y − b a ) = F X ( y − b a ) P(Y \le y) = P\!\left(X \le \frac{y-b}{a}\right) = F_X\!\left(\frac{y-b}a\right) P ( Y ≤ y ) = P ( X ≤ a y − b ) = F X ( a y − b ) p Y ( y ) = 1 a p X ( y − b a ) p_Y(y) = \frac1ap_X\!\left(\frac{y-b}a\right) p Y ( y ) = a 1 p X ( a y − b ) a < 0 a < 0 a < 0 P ( Y ≤ y ) = 1 − F X ( y − b a ) P(Y \le y) = 1 - F_X\!\left(\frac{y-b}a\right) P ( Y ≤ y ) = 1 − F X ( a y − b ) − 1 a p X ( ⋅ ) = 1 ∣ a ∣ p X ( ⋅ ) -\frac1ap_X(\cdot) = \frac1{|a|}p_X(\cdot) − a 1 p X ( ⋅ ) = ∣ a ∣ 1 p X ( ⋅ )