Products and Convolutions of Gaussian Probability Density Functions
Products and Convolutions of Gaussian Probability Density Functions
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发表时间：
2013
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通讯作者：
P. Bromiley
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作者：
P. Bromiley
众所周知，两个高斯概率密度函数（PDF）的乘积和卷积也是高斯分布的。本备忘录提供了两种情况下所得高斯函数的平均值和标准差的推导。这些结果可用于计算在各种算法中作为中间步骤应用的平滑效果。 1 两个高斯 PDF 的乘积 我们希望在最一般的情况下，即不相同的均值下，找到两个高斯 PDF 的乘积 f(x) = 1 √ 2πσf e − (x−μf ) 2 2σ2 f 和 g(x) = 1 √ 2πσg e − (x−μg) 2 2σ2 g (1)。乘积得出 f(x)g(x) = 1 2πσfσg e − ( (x−μf ) 2 2σ2 f + (x−μg) 2 2σ2 g ) (2) 检查指数中的项 α = (x− μf ) 2σ2 f + (x− μg) 2σ2 g (3) 我们可以展开两个二次方程并收集 x 的幂项，得到 α = (σ f + σ 2 g)x 2 − 2(μfσ g + μgσ 2 f )x+ μ 2 fσ 2 g + μ 2 gσ 2 f 2σ2 fσ 2 g (4) 除以 x 的系数，得到 α = x − 2 μfσ 2 g+μgσ 2 f σ2 f +σ2 g x+ μ2fσ 2 g+μ 2 gσ 2 f σ2 f +σ2 g 2σ2 f σ2 g σ2 f +σ2 g (5) 这又是 x 的二次方程，所以等式： 2 是高斯函数。比较等式中的项。 5 化为通常的高斯形式 P (x) = 1 √ 2πσ e (x−μ)2 2σ2 = 1 √ 2πσ e (x2−2μx+μ2) 2σ2 (6) 由于我们可以添加与 x 无关的项 γ 来完成 α 的平方，因此在可以忽略归一化的情况下，这足以完成证明。两个高斯 PDF 的乘积与平均值为等式 1 中 x 系数一半的高斯 PDF 成正比。 5 和标准差，即分母一半的平方根，即 σfg = √ σ2 fσ 2 g σ2 f + σ 2 g 和 μfg = μfσ 2 g + μgσ 2 f σ2 f + σ 2 g (7) 一般来说，乘积本身并不是 PDF，因为由于存在比例因子，它不会具有正确的归一化。现在，我们可以直接用未知的缩放常数以通常的高斯形式写出乘积 f(x)g(x)，或者从等式 1 开始。 5 明确获得缩放常数。采用后一种方法，假设 γ 是完成 α 平方所需的项，即 γ = ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 − ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) = 0 (8) 将此项添加到 α 得到 α = x − 2x μfσ 2 g+μgσ 2 f σ2 f +σ2 g + ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) + μ2fσ 2 g+μ 2 gσ 2 f σ2 f +σ2 g − ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) (9) 经过一些处理后，可简化为 α = ( x− μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2 σ2 f σ2 g σ2 f +σ2 g + (μf − μg) 2(σ2 f + σ 2 g) = (x− μfg) 2 2σ2 fg + (μf − μg) 2(σ2 f + σ 2 g) 代回方程： 2 给出 f(x)g(x) = 1 2πσfσg exp [
It is well known that the product and the convolution of two Gaussian probability density functions (PDFs) are also Gaussian. This memo provides derivations for the mean and standard deviation of the resulting Gaussians in both cases. These results are useful in calculating the effects of smoothing applied as an intermediate step in various algorithms. 1 The Product of Two Gaussian PDFs We wish to find the product of two Gaussian PDFs f(x) = 1 √ 2πσf e − (x−μf ) 2 2σ2 f and g(x) = 1 √ 2πσg e − (x−μg) 2 2σ2 g (1) in the most general case i.e. non-identical means. The product gives f(x)g(x) = 1 2πσfσg e − ( (x−μf ) 2 2σ2 f + (x−μg) 2 2σ2 g ) (2) Examining the term in the exponent α = (x− μf ) 2σ2 f + (x− μg) 2σ2 g (3) we can expand the two quadratics and collect terms in powers of x to give α = (σ f + σ 2 g)x 2 − 2(μfσ g + μgσ 2 f )x+ μ 2 fσ 2 g + μ 2 gσ 2 f 2σ2 fσ 2 g (4) Dividing through by the coefficient of x gives α = x − 2 μfσ 2 g+μgσ 2 f σ2 f +σ2 g x+ μ2fσ 2 g+μ 2 gσ 2 f σ2 f +σ2 g 2 σ2 f σ2 g σ2 f +σ2 g (5) This is again a quadratic in x, and so Eq. 2 is a Gaussian function. Compare the terms in Eq. 5 to a the usual Gaussian form P (x) = 1 √ 2πσ e (x−μ)2 2σ2 = 1 √ 2πσ e (x2−2μx+μ2) 2σ2 (6) Since we can add a term γ that is independent of x to complete the square in α, this is sufficent to complete the proof in cases where the normalisation can be ignored. The product of two Gaussian PDFs is proportional to a Gaussian PDF with a mean that is half the coefficient of x in Eq. 5 and a standard deviation that is the square root of half of the denominator i.e. σfg = √ σ2 fσ 2 g σ2 f + σ 2 g and μfg = μfσ 2 g + μgσ 2 f σ2 f + σ 2 g (7) In general, the product is not itself a PDF as, due to the presence of the scaling factor, it will not have the correct normalisation. We can now either write down the product f(x)g(x) in the usual Gaussian form directly, with an unknown scaling constant, or proceed from Eq. 5 to obtain the scaling constant explicitly. Taking the latter route, suppose that γ is the term required to complete the square in α i.e. γ = ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 − ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) = 0 (8) Adding this term to α gives α = x − 2x μfσ 2 g+μgσ 2 f σ2 f +σ2 g + ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) + μ2fσ 2 g+μ 2 gσ 2 f σ2 f +σ2 g − ( μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2σ2 f σ2 g (σ2 f +σ2 g) (9) After some manipulation, this reduces to α = ( x− μfσ 2 g+μgσ 2 f σ2 f +σ2 g )2 2 σ2 f σ2 g σ2 f +σ2 g + (μf − μg) 2(σ2 f + σ 2 g) = (x− μfg) 2 2σ2 fg + (μf − μg) 2(σ2 f + σ 2 g) Substituting back into Eq. 2 gives f(x)g(x) = 1 2πσfσg exp [