视频中时域降噪：Y序列噪声方差收敛性分析

根据题目给出的信息，可以得知Y(i+1)是由X(i+1)和Y(i)的平均值得到的，Y(1)等于X(1)。由此可以得出递推公式：\n\nY(i+1) = (X(i+1) + Y(i))/2\n\n现在我们来计算Y的噪声方差。假设Z(i)表示Y(i)的噪声，我们可以得到如下递推关系：\n\nZ(i+1) = Y(i+1) - X(i+1)\n = (X(i+1) + Y(i))/2 - X(i+1)\n = Y(i)/2 - X(i+1)/2\n\n将上式两边的平方后取期望，可以得到：\n\nE[Z(i+1)^2] = E[(Y(i)/2 - X(i+1)/2)^2]\n = E[(Y(i)^2)/4 - Y(i)X(i+1)/2 + (X(i+1)^2)/4]\n = E[(Y(i)^2)/4] - E[Y(i)X(i+1)/2] + E[(X(i+1)^2)/4]\n = (1/4)E[Y(i)^2] - (1/2)E[Y(i)X(i+1)] + (1/4)E[X(i+1)^2]\n\n因为X的噪声服从G(0, 1)分布，所以E[X(i+1)^2] = Var[X(i+1)] + E[X(i+1)]^2 = 1 + 0 = 1。又因为Y(i)是由X(i)和Y(i-1)的平均值得到的，所以E[Y(i)X(i+1)] = E[(X(i) + Y(i-1))/2 * X(i+1)] = (1/2)E[X(i)X(i+1)] + (1/2)E[Y(i-1)X(i+1)]。由于X的噪声服从独立同分布的G(0, 1)分布，所以E[X(i)X(i+1)] = 0。又因为Y(i-1)是由X(i-1)和Y(i-2)的平均值得到的，所以E[Y(i-1)X(i+1)] = E[(X(i-1) + Y(i-2))/2 * X(i+1)] = (1/2)E[X(i-1)X(i+1)] + (1/2)E[Y(i-2)X(i+1)] = (1/2)E[X(i-1)X(i+1)]。以此类推，可以得到E[Y(i)X(i+1)] = (1/2)^(i-1)E[X(1)X(i+1)] = (1/2)^(i-1)E[X(1)X(2)] = (1/2)^(i-1)E[X(1)^2] = (1/2)^(i-1)。\n\n综上所述，可以得到递推公式：\n\nE[Z(i+1)^2] = (1/4)E[Y(i)^2] - (1/2)^(i-1) + (1/4)\n\n由于Z(i)是Y(i)的噪声，所以E[Z(i)^2] = Var[Y(i)]。假设Y的噪声方差收敛于v，则有E[Z(i+1)^2] = E[Z(i)^2] = v。将上式代入，可以得到：\n\nv = (1/4)v - (1/2)^(i-1) + (1/4)\n\n化简得：\n\n(3/4)v = (1/2)^(i-1) - (1/4)\nv = 4 * (1/2)^(i-1) - 1\n\n当i趋于无穷大时，(1/2)^(i-1)趋于0，所以v趋于-1。但是方差不能为负数，所以Y序列的噪声方差没有收敛值。