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ofer zeitouni
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The LHS of the expression you wrote is the minimum mean square error (i.e., the expectation $\inf E(X-\hat X)^2$ where $\hat X$ is measurable on $Y$). On the other hand, the expression you wrote ($\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2$$\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2)$) is the error of the optimal linear estimator, so it always bounds from above the optimal error, that is $\hat \sigma^2\geq E Var(X+Y)$$\hat \sigma^2\geq E Var(X|Y)$; equality is achieved in the Gaussian case.

The LHS of the expression you wrote is the minimum mean square error (i.e., the expectation $\inf E(X-\hat X)^2$ where $\hat X$ is measurable on $Y$). On the other hand, the expression you wrote ($\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2$) is the error of the optimal linear estimator, so it always bounds from above the optimal error, that is $\hat \sigma^2\geq E Var(X+Y)$; equality is achieved in the Gaussian case.

The LHS of the expression you wrote is the minimum mean square error (i.e., the expectation $\inf E(X-\hat X)^2$ where $\hat X$ is measurable on $Y$). On the other hand, the expression you wrote ($\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2)$) is the error of the optimal linear estimator, so it always bounds from above the optimal error, that is $\hat \sigma^2\geq E Var(X|Y)$; equality is achieved in the Gaussian case.

Source Link
ofer zeitouni
  • 7.6k
  • 1
  • 22
  • 38

The LHS of the expression you wrote is the minimum mean square error (i.e., the expectation $\inf E(X-\hat X)^2$ where $\hat X$ is measurable on $Y$). On the other hand, the expression you wrote ($\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2$) is the error of the optimal linear estimator, so it always bounds from above the optimal error, that is $\hat \sigma^2\geq E Var(X+Y)$; equality is achieved in the Gaussian case.