Problem is about getting the covariance of two random variables that are not independent: $\operatorname{cov}(\tilde{x}\mid(\tilde{y}=y),\tilde{x})= \text{ ?}$
$$\tilde{x}\sim N(\mu_1,\sigma^2_1)$$
$$\tilde{x}\mid (\tilde{y}=y) \sim N(\mu_2,\sigma^2_2)$$
where $\tilde{y}=\tilde{x}+\tilde{n}$ where $n$ is also normal distributed. So there is a change in both the mean and variance for the conditional distribution.
Now my question is to get the covariance between $\tilde{x} \mid (\tilde{y}=y)$ and $\tilde{x}$.
I tried to deduct it as following but I am stuck:
\begin{align}
& \operatorname{cov}(\tilde{x}\mid(\tilde{y}=y),\tilde{x})=E((\tilde{x} \mid (\tilde{y} = y) \times \tilde{x})-E(\tilde{x} \mid (\tilde{y} = y) E(\tilde{x}) \\[10pt]
= {} & E(\tilde{x}\mid (\tilde{y}=y)\times \tilde{x})-\mu_1\mu_2=\text{ ?}
\end{align}
Best Answer
This question says
$$\tilde{x}\sim N(\mu_1,\sigma^2_1),$$
$$\tilde{x}\mid (\tilde{y}=y) \sim N(\mu_2,\sigma^2_2).$$
This means that given the event that that $\bar y=y$, the conditional distribution of $\bar x$ is $N(\mu_2,\sigma_2^2)$. This is a perfectly correct use of the notation $\text{“ } \tilde{x}\mid (\tilde{y}=y) \sim N(\mu_2,\sigma^2_2)\text{ ''}$.
However, the request for a covariance between $\bar x$ and something called $\text{“ }\bar x\mid (\bar y=y)\text{ ''}$ makes it appear that the poster thinks there is some random variable involved called $\bar x\mid (\bar y=y)$. The question seems to contemplate a joint distribution between that random variable and the random variable $\bar x$, so that there would be a covariance between them. That would be the case if $y$ were a random variable, but nothing to that effect appears in the question. That is a misunderstanding of the meaning of the notation. The notation does not refer to something called $\text{“ }\bar x\mid (\bar y=y)\text{ ''}$ Rather, the meaning of the notation is as explained in the paragraph above.
One could speak of a random variable $\operatorname{E}(\bar x\mid \bar y)$, which would be a function of $\bar y$, and ask for the covariance between that and $\bar x$. However, without knowing anything about the joint distribution of $\bar x$ and $\bar y$, one could not specify which distribution that is.