Tensorflow 二元高斯分布对数似然的负值

我试图实现一个损失函数，它试图最小化从预测的二元高斯分布参数中获得地面真值（x，y）的负对数可能性。我在tensorflow中实现这个- 这是代码-

def tf_2d_normal(self, x, y, mux, muy, sx, sy, rho):
    '''
    Function that implements the PDF of a 2D normal distribution
    params:
    x : input x points
    y : input y points
    mux : mean of the distribution in x
    muy : mean of the distribution in y
    sx : std dev of the distribution in x
    sy : std dev of the distribution in y
    rho : Correlation factor of the distribution
    '''
    # eq 3 in the paper
    # and eq 24 & 25 in Graves (2013)
    # Calculate (x - mux) and (y-muy)
    normx = tf.sub(x, mux)
    normy = tf.sub(y, muy)
    # Calculate sx*sy
    sxsy = tf.mul(sx, sy)
    # Calculate the exponential factor
    z = tf.square(tf.div(normx, sx)) + tf.square(tf.div(normy, sy)) - 2*tf.div(tf.mul(rho, tf.mul(normx, normy)), sxsy)
    negRho = 1 - tf.square(rho)
    # Numerator
    result = tf.exp(tf.div(-z, 2*negRho))
    # Normalization constant
    denom = 2 * np.pi * tf.mul(sxsy, tf.sqrt(negRho))
    # Final PDF calculation
    result = -tf.log(tf.div(result, denom))
    return result

当我进行训练时，我可以看到损失值在下降，但远远低于0。我能理解这应该是因为，我们正在最小化“负面”可能性。即使损失值在减少，我也无法得到准确的结果。有人能帮我验证一下我为loss函数编写的代码是否正确吗

对于训练神经网络（特别是RNN），这种损失的性质也是可取的吗

谢谢

我知道你找到了洋红的，我正在做类似的事情。我发现这段代码本身并不稳定。您需要使用约束来稳定它，因此不能单独使用或解释

tf\u 2d\u normal

代码<如果您的数据事先未正确规范化或丢失函数中未正确规范化，则code>NaNs和

Inf

s将开始到处出现

下面是我用Keras构建的一个更稳定的损失函数版本。这里可能有一些冗余，它可能不适合您的需要，但我发现它可以工作，您可以测试/调整它。我包含了一些关于负日志值可能产生多大的内联注释：

def r3_bivariate_gaussian_loss(true, pred):
    """
    Rank 3 bivariate gaussian loss function
    Returns results of eq # 24 of http://arxiv.org/abs/1308.0850
    :param true: truth values with at least [mu1, mu2, sigma1, sigma2, rho]
    :param pred: values predicted from a model with the same shape requirements as truth values
    :return: the log of the summed max likelihood
    """
    x_coord = true[:, :, 0]
    y_coord = true[:, :, 1]
    mu_x = pred[:, :, 0]
    mu_y = pred[:, :, 1]

    # exponentiate the sigmas and also make correlative rho between -1 and 1.
    # eq. # 21 and 22 of http://arxiv.org/abs/1308.0850
    # analogous to https://github.com/tensorflow/magenta/blob/master/magenta/models/sketch_rnn/model.py#L326
    sigma_x = K.exp(K.abs(pred[:, :, 2]))
    sigma_y = K.exp(K.abs(pred[:, :, 3]))
    rho = K.tanh(pred[:, :, 4]) * 0.1  # avoid drifting to -1 or 1 to prevent NaN, you will have to tweak this multiplier value to suit the shape of your data

    norm1 = K.log(1 + K.abs(x_coord - mu_x))
    norm2 = K.log(1 + K.abs(y_coord - mu_y))

    variance_x = K.softplus(K.square(sigma_x))
    variance_y = K.softplus(K.square(sigma_y))
    s1s2 = K.softplus(sigma_x * sigma_y)  # very large if sigma_x and/or sigma_y are very large

    # eq 25 of http://arxiv.org/abs/1308.0850
    z = ((K.square(norm1) / variance_x) +
         (K.square(norm2) / variance_y) -
         (2 * rho * norm1 * norm2 / s1s2))  # z → -∞ if rho * norm1 * norm2 → ∞ and/or s1s2 → 0
    neg_rho = 1 - K.square(rho)  # → 0 if rho → {1, -1}
    numerator = K.exp(-z / (2 * neg_rho))  # → ∞ if z → -∞ and/or neg_rho → 0
    denominator = (2 * np.pi * s1s2 * K.sqrt(neg_rho)) + epsilon()  # → 0 if s1s2 → 0 and/or neg_rho → 0
    pdf = numerator / denominator  # → ∞ if denominator → 0 and/or if numerator → ∞
    return K.log(K.sum(-K.log(pdf + epsilon())))  # → -∞ if pdf → ∞

希望你觉得这很有价值