静止图像序列噪声方差收敛分析：Y(i+1)=(X(i+1)+Y(i))/2

根据题目给出的假设，Y(i+1)=(X(i+1)+Y(i))/2，可以得到如下递推关系式：\nY(2) = (X(2) + Y(1))/2\nY(3) = (X(3) + Y(2))/2 = (X(3) + (X(2) + Y(1))/2)/2 = X(3)/2 + X(2)/4 + Y(1)/4\nY(4) = (X(4) + Y(3))/2 = (X(4) + (X(3)/2 + X(2)/4 + Y(1)/4))/2 = X(4)/2 + X(3)/4 + X(2)/8 + Y(1)/8\n...\n可以得到递归式：\nY(i) = X(i)/2 + X(i-1)/4 + X(i-2)/8 + ... + X(2)/2^(i-2) + Y(1)/2^(i-1)\n由于X的像素噪声服从G(0, 1)分布，即方差为1，所以每个X(i)的方差也为1。\n因此，Y(i)的方差可以通过对递归式两边同时取方差来计算：\nVar(Y(i)) = Var(X(i)/2 + X(i-1)/4 + X(i-2)/8 + ... + X(2)/2^(i-2) + Y(1)/2^(i-1))\n = Var(X(i)/2) + Var(X(i-1)/4) + Var(X(i-2)/8) + ... + Var(X(2)/2^(i-2)) + Var(Y(1)/2^(i-1))\n = (1/2^2) + (1/4^2) + (1/8^2) + ... + (1/2^(i-2)^2) + (1/2^(i-1)^2)\n = 1/4 + 1/16 + 1/64 + ... + 1/2^(2(i-2)) + 1/2^(2(i-1))\n = 1/4 * (1 + 1/4 + 1/16 + ... + 1/2^(i-2) + 1/2^(i-1))\n = 1/4 * (1 - 1/4^(i-1))/(1 - 1/4)\n = 1/4 * (1 - 1/4^(i-1))/(3/4)\n = 4/3 * (1 - 1/4^(i-1))\n\n当i趋近于无穷大时，噪声方差Var(Y(i))收敛于4/3。