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$Y$ is a Gaussian distributed random variable with zero mean and known variance: $Y\sim N(0,\sigma_y)$. We measure $Y$ with a sensor, which is corrupted by white Gaussian noise: $Z=Y+V$; $V\sim N(0,\sigma_v)$.

How can I design a random variable $X$ depending on $Z$, $\sigma_y$, $\sigma_v$, and $\sigma_x$ which results in $X$ having the distribution $X\sim N(0,\sigma_x)$ and having a maximal correlation with $Y$?

Initial idea: We can take $\hat{Y}$, the minimum mean square estimation (MMSE) of $Y$, and let $X=t\hat{Y}$ with $t$ chosen to ensure that $X$ has the desired variance. But later I found this intuitive idea is not correct. I wonder if someone can answer this question.

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    $\begingroup$ What's wrong with the intuitive idea? $\endgroup$
    – Matt F.
    Jan 18 at 3:52
  • $\begingroup$ @MattF. Thanks for help editing the question. Please see my answer for the intuitive idea. $\endgroup$
    – Jing Zhou
    Jan 18 at 19:46
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It can be obtained that the MMSE of $Y$ is given by $\hat{Y}=\frac{\sigma_y}{\sigma_y+\sigma_v}Z=kZ$, and the covariance of $\hat{Y}$ is $k^2(\sigma_y+\sigma_v)$. Now suppose $X=t\hat{Y}$, to ensure $X$ has covariance $\sigma_x$, we get $$t^2k^2(\sigma_y+\sigma_v)=\sigma_x$$ Solving $t$ and substituting into $X$ leads to $$X=\sqrt{\frac{\sigma_x}{k^2(\sigma_y+\sigma_v)}}\hat{Y}=\sqrt{\frac{\sigma_x}{k^2(\sigma_y+\sigma_v)}}kZ=\sqrt{\frac{\sigma_x}{\sigma_y+\sigma_v}}Z$$ It seems that $X$ has nothing to do with the MMSE of $Y$. We just define a new variable $$\bar{t}=\sqrt{\frac{\sigma_x}{\sigma_y+\sigma_v}}$$ Then $X$ can be derived directly by $$X=\bar{t}Z$$ We can also calculate the expectation of $XY$, i.e., $$\mathbb{E}[XY]=\bar{t}\mathbb{E}[ZY]=\bar{t}\sigma_y$$ My intuitive idea is that: If we know the exact value of $Y$, we can simply set $X=\frac{\sigma_x}{\sigma_y}Y$, which ensure maximal correlation between $X$ and $Y$. But now, since we cannot know the exact value of $Y$, we use its MMSE estimation. Now the above derivation states that we can simple use the corrupted measurement $Z$, to design $X$. I wonder if there is any in deep explanation of this problem.

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  • $\begingroup$ I think it’s right that a rescaled version of $Z$ has the highest correlation. $\endgroup$
    – Matt F.
    Jan 18 at 22:59
  • $\begingroup$ I made a mistake in the above derivation. The covariance of $\hat{Y}$ (MMSE of $Y$), is given by $$Cov[\hat{Y}]=\frac{\sigma_y\sigma_v}{\sigma_y+\sigma_v}$$ $\endgroup$
    – Jing Zhou
    Jan 20 at 16:01

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