MLE of lambda parameter in PowerTransform bijector

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难度
4/5
预计耗时
3-5 天
新手友好度
25/100
Issue 类型
缺陷
描述清晰度
需要澄清
活跃度
停滞
技术栈
python, tensorflow
领域
machine-learning

调研方向

首先复现提供的 PowerTransform 和 TransformedDistribution 训练循环，然后在报告的 lambda 值下，将其对数概率和正向 Jacobian 与 scipy.stats.boxcox 进行比较。确定差异是来自训练过程还是 PowerTransform 的行为；解释清楚该不匹配或识别出一个可复现的缺陷，即视为完成。

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描述

Hi everyone, I'm trying tofind the optimal lambda parameter of the PowerTransform bijector with maximum lilkelihood estimation.
In order to do so I had to modify the constructor of the bijector in order to allow power to be a trainable tf.Variable.
Then code is the following:

import matplotlib.pyplot as plt
import numpy as np
import tensorflow as tf
from tensorflow_probability import bijectors as tfb
from tensorflow_probability import distributions as tfd
from scipy.stats import boxcox

np.random.seed(42)
tf.random.set_seed(42)

# Data exponential generation
data_dist = tfd.Sample(tfd.Exponential(1.), sample_shape=1000)
x_train = data_dist.sample()
plt.hist(x_train.numpy(), bins=120, density=True, alpha=0.6, color='blue', label='Samples')
plt.show()

nf = tfd.TransformedDistribution(
    tfd.Normal(loc=0, scale=1),
    bijector=tfb.PowerTransform(power=tf.Variable(initial_value=1., name='power'))
)

# Training loop
num_steps = 2000
optimizer = tf.keras.optimizers.Adam(learning_rate=0.001)
for step in range(num_steps):
    with tf.GradientTape() as tape:
        loss = -tf.reduce_sum(nf.log_prob(x_train))
        grads = tape.gradient(loss, nf.trainable_variables)
    optimizer.apply_gradients(zip(grads, nf.trainable_variables))

    if step % 100 == 0:
        print(f"Step {step}, Loss: {loss.numpy()}, Grads: {grads}, Power: {nf.trainable_variables[0].numpy()}")

_, llm_lmbda = boxcox(x_train, lmbda=None)
print(f"Scipy MLE power is: {llm_lmbda}") # 0.24647694003577084
print(f"My MLE power is: {nf.trainable_variables[0].numpy()}") # 0.6407918334007263

z_samples = nf.sample(1000)
plt.hist(z_samples.numpy(), bins=120, density=True, alpha=0.6, color='green', label='Samples')
plt.show()

The estimation with the MLE from the scipy library gives lambda = 0.25 but mine gives lambda = 0.64. If I use the bijector with a static value of 0.25 I can recover a distribution that is closer to the original exponential so I believe that there might be a problem with the training procedure or with the computation of the forward Jacobian in the PowerTransform bijector but I can't find it.

Anyone can help with this?

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