tensorflow 是否通过 pdf 传播梯度 [英] Does tensorflow propagate gradients through a pdf

问题描述

假设分布函数定义如下:

Lets say, a distribution function is defined as below:

dist = tf.contrib.distributions.Normal(mu, sigma)

从分布中抽取样本

val = dist.pdf(x)

并且该值用于模型中以预测变量

and this value is used in a model to predict a variable

X_hat = f(val)
loss = tf.norm(X_pred-X_hat, ord=2)

如果我想优化变量 mu 和 sigma 以减少我的预测误差，我可以执行以下操作吗?

and if I want to optimize the variables mu and sigma to reduce my prediction error can I do the following?

train = tf.train.AdamOptimizer(1e-03).minimize(loss, var_list=[mu, sigma])

我想知道梯度例程是否通过正态分布传播，或者我应该期待一些问题，因为我在定义分布的参数上采用梯度

I am interested in knowing if the gradient routines are propagated through the normal distribution, or should I expect some issues because I am taking gradients over the parameters defining a distribution

推荐答案

tl;dr: 是的，梯度反向传播将与 tf.distributions.Normal 一起正常工作.

tl;dr: Yes, gradient back propagation will work correctly with tf.distributions.Normal.

dist.pdf(x) 不会从分布中抽取样本，而是返回 x 处的概率密度函数.这可能不是您想要的.

dist.pdf(x) does not draw a sample from the distribution, but rather returns the probability density function at x. This is probably not what you wanted.

要获得随机样本，您真正想要的是调用 dist.sample().对于许多随机分布，随机样本对参数的依赖是重要的，不一定是可反向传播的.

To get a random sample, what you really want is to call dist.sample(). For many random distributions, the dependency of a random sample on the parameters is nontrivial and will not necessarily be backpropable.

然而，正如@Richard_wth 指出的，特别是对于正态分布，可以通过重新参数化来获得对位置和尺度参数(mu 和 sigma>).

However, as @Richard_wth pointed out, specifically for the normal distribution, it is possible through reparametrization to get a simple dependency on the location and scale parameters (mu and sigma).

事实上，在中 of tf.contrib.distributions.Normal(最近迁移到tf.distributions.Normal)，这正是sample实施:

In fact, in the implementation of tf.contrib.distributions.Normal (recently migrated to tf.distributions.Normal), that is exactly how sample is implemented:

def _sample_n(self, n, seed=None):
  ...
  sampled = random_ops.random_normal(shape=shape, mean=0., stddev=1., ...)
  return sampled * self.scale + self.loc

因此，如果您提供尺度和位置参数作为张量，则反向传播将在这些张量上正常工作.

Consequently, if you provide scale and location parameters as tensors, then backpropagation will work correctly on those tensors.

请注意，这种反向传播本质上是随机的:它会根据正常高斯变量的随机抽取而变化.但是，从长远来看(在许多训练示例中)，这可能会如您所愿.

Note that this backpropagation is inherently random: It will vary depending on the random draw of the normal Gaussian variable. However, in the long run (over many training examples), this is likely to work as you expect.

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