This is from MIT's 6.431x.
For the model $X=\Theta+W$, and under the usual independence and normality assumptions for $\Theta$ and $W$, the mean squared error (MSE) of the LMS estimator is $$\frac{1}{(1/\sigma_0^2)+(1/\sigma_1^2)}$$ where $\sigma_0^2$ and $\sigma_1^2$ are the variances of $\Theta$ and $W$, respectively.
Suppose now that we change the observation model to $Y=3\Theta+W$. In some sense the “signal" $\Theta$ has a stronger presence, relative to the noise term $W$, and we should expect to obtain a smaller mean squared error. Suppose $\sigma_0^2=\sigma_1^2=1$. The mean squared error of the original model $X=\Theta+W$ is then $1/2$.
Find the mean squared error of the new model $Y=3\Theta+W$ by thinking of an alternative observation model in which you observe $Y'=\Theta+(W/3)$.
I don't at all understand the solution (see below), in fact I got stuck immediately at the first sentence: Since $Y'$ is just $Y$ scaled by a factor of $1/3$, $Y'$ carries the same information as $Y$, so that $E[\Theta|Y]=E[\Theta|Y']$.
Why does $Y'$ carry the same info as $Y$? Because I would have expected that because it is scaled by a factor of $1/3$, the spread would be smaller, which contributes to a smaller MSE? (And by the same token, a smaller estimate as well)
Solution:
Since $Y'$ is just $Y$ scaled by a factor of $1/3$, $Y'$ carries the same information as $Y$, so that $E[\Theta|Y]=E[\Theta|Y']$. Thus, the alternative observation model $Y'=\Theta+(W/3)$ will lead to the same estimates and will have the same mean squared error as the unscaled model $Y=\Theta+W$. In the equivalent model $Y'$, we have a noise variance of $1/9$ and therefore the mean squared error is $$\frac{1}{\frac{1}{1}+\frac{1}{1/9}}=\frac{1}{10}$$.