AICS Modeling

Brief, sometimes rough, data stories and tutorials from the modeling group at the Allen Institute for Cell Science

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Currently an unfinished rough draft

toy 2D → 3D problem

import numpy as np
import numba

exact solution

Our map is a simple one, from 2D $\boldsymbol{z}$ space to 3D $\boldsymbol{x}$ space. We endow the 2D latent space with a gaussian density, i.e \begin{align} \rho(\boldsymbol{z}) &= \frac{1}{2\pi} e^{-\tfrac{1}{2} |\boldsymbol{z}|^2 } \end{align} As a toy map, take \begin{align} \boldsymbol{z} &= (z_1, z_2)
\boldsymbol{x}(\boldsymbol{z}) &= \tfrac{1}{\sqrt[4]{2}} \left(\tfrac{1}{\sqrt{2}}z_1^2, \tfrac{1}{\sqrt{2}}z_2^2, z_1 z_2 \right) \end{align} The Jacobian of this transformation is \begin{align} J(\boldsymbol{x}) &= \frac{d \boldsymbol{x}}{d \boldsymbol{z}}
&= \sqrt[4]{2} \pmatrix{ z_1 & 0 \cr 0 & z_2 \cr \tfrac{1}{\sqrt{2}}z_2 & \tfrac{1}{\sqrt{2}}z_1 } \end{align} A scalar measure Jacobian we can use is $D(\boldsymbol{z}) = \sqrt{ \det{ \left( J(\boldsymbol{z})^T J(\boldsymbol{z}) \right)} }$ (see

\begin{align} J(\boldsymbol{x})^T J(\boldsymbol{x}) &= \sqrt{2} \pmatrix{z_1 & 0 & \tfrac{1}{\sqrt{2}}z_2 \cr 0 & z_2 & \tfrac{1}{\sqrt{2}}z_1} \pmatrix{z_1 & 0 \cr 0 & z_2 \cr \tfrac{1}{\sqrt{2}}z_2 & \tfrac{1}{\sqrt{2}}z_1}
&= \pmatrix{\sqrt{2}z_1^2 + \tfrac{1}{\sqrt{2}}z_2^2 & \tfrac{1}{\sqrt{2}} z_1 z_2 \cr \tfrac{1}{\sqrt{2}} z_1 z_2 & \tfrac{1}{\sqrt{2}}z_1^2 + \sqrt{2} z_2^2 } \end{align} Taking the determinant we get \begin{align} \det{ \left( J(\boldsymbol{z})^T J(\boldsymbol{z}) \right)} &= z_1^4 + 2 z_1^2 z_2^2 + z_2^4
&= \left( z_1^2 + z_2^2 \right)^2 \end{align} Our measure is then simply \begin{align} D(\boldsymbol{z}) &= \sqrt{ \det{ \left( J(\boldsymbol{z})^T J(\boldsymbol{z}) \right)} }
&= \sqrt{\left( z_1^2 + z_2^2 \right)^2}
&= z_1^2 + z_2^2 \end{align}

To find the expectation of this measure over the entire $\boldsymbol{z}$ space, we integrate over the space, weighting by the density of $\boldsymbol{z}$ in that space: \begin{align} \Bbb E [ D(\boldsymbol{z}) ] &= \int D(\boldsymbol{z}) \rho(\boldsymbol{z}) d\boldsymbol{z}
&= \int_{-\infty}^\infty \int_{-\infty}^\infty \left( z_1^2 + z_2^2 \right) \frac{1}{2\pi} e^{-\tfrac{1}{2} (z_1^2 + z_2^2)^2 } d z_1 d z_2
&= \int_{0}^{2\pi} \int_{0}^\infty \frac{1}{2\pi} e^{ -\tfrac{1}{2} r^2 } r^2 r dr d\theta
&= \int_{0}^\infty e^{ -\tfrac{1}{2} r^2 } r^3 r dr
&= 2 \end{align}

numerical solution

def x(z):
    z1,z2 = z
    return np.array( [z1**2/np.sqrt(2.0), z2**2/np.sqrt(2.0), z1*z2] ) / np.sqrt(np.sqrt(2.0))
def sqrt_det_jacobian(z, delta=1e-6):
    J = np.zeros((len(x(z)),len(z)))
    for i,z_i in enumerate(z):
        z_delta = z.copy()
        z_delta[i] += delta
        J[:,i] = (x(z_delta)-x(z))/delta
    det_JtJ = np.linalg.det(, J) )
    return np.sqrt(det_JtJ)
z_points = np.random.randn(1024,2)
jac_samples = np.zeros(len(z_points))
for i,zi in enumerate(z_points):
    jac_samples[i] = sqrt_det_jacobian(zi,delta=1e-9)

notes re pytorch