Suppose we estimate $x$ from $y_1=ax+z_1$ and get $x_1$. Then suppose we receive $y_2=bx+z_2$ and reestimate $x$ using both $y_1$ and $y_2$ and get $x_2$. Is the means squared error (MSE) of $x_2$ always lower than MSE of $x_1$. In other words is $$E(x-x_1)^2\geq E(x-x_2)^2?$$ Here $a,b$ are know constants and $z_1$ and $z_2$ are Gaussian errors and the estimation is linear minimum mean squared error. The equations can also be vectors in that case error is the trance of the error covariance matrix. By this wiki it seems to be decreasing.
At time $t=0$ a real value $x$ is generated by a random process $X$. Then at time $t=1$, I receive a noise corrupted and scaled version $y_1$ of $x$ denoted by $y_1=ax+z_1$. From $y_1$ I do linear minimum mean squared error (LMMSE) estimation and estimate $x$ and get $x_1$ and the corresponding LMMSE is $E(x-x_1)^2$. Here $a$ is a constant and is known, $z_1$ is Gaussian noise with known mean and variance. LMMSE estimation is straightforward.
Then at time $t=2$, I receive another noise corrupted and scaled version $y_2$ of that same $x$ which is denoted by $y_2=bx+z_2$. Again $b$ is a known constant and $z_2$ is Gaussian noise with known mean and variance. Now I simultaneously use both $y_1$ and $y_2$ to estimate $x$ by LMMSE estimation and I get the value $x_2$ and the corresponding error $E(x-x_2)^2$. My question: is $E(x-x_1)^2\geq E(x-x_2)^2?$. i.e., the time actually does not matter the question is does LMMSE estimation using $n+1$ samples always give a less error than estimating only using $n$ samples out of that $n+1$ samples?