雙變量高斯的對數似然的負值

Question

我試圖實現一個損失函數，該函數試圖最小化從預測的雙變量高斯分佈參數獲得地面實際值（x，y）的負對數可能性。我在tensorflow實現此 - 這裏是代碼 -

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 

當我做培訓，我可以看到損耗值減少，但它遠遠低於過去0我可以理解，應該是因爲，我們正在將「負面」可能性降至最低。即使損失值正在下降，我也無法獲得準確的結果。如果我爲損失函數編寫的代碼是否正確，是否有人可以幫助驗證？

也是這樣的性質損失理想的訓練神經網絡（特別是RNN）？

Thankss

Answer 1

我看到你發現從品紅sketch-rnn code，我的工作類似的東西。我發現這段代碼本身並不穩定。您需要使用約束來穩定它，因此tf_2d_normal代碼不能單獨使用或解釋。 NaN s和Inf如果您的數據未事先或在丟失功能中未正確標準化，將開始出現在所有地方。

下面是我用Keras建立的更穩定的損失函數版本。這裏可能有一些冗餘，它可能不是完美的滿足你的需求，但我發現它正在工作，你可以測試/適應它。我列出了一些關於如何產生大的負值對數值的內嵌評論：

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 → ∞ 

希望你找到這個值。