How to solve this conditional variance from Kyle 1985?

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Kyle(1985) built a model of an insider, noise trader, and a market maker with an auction model.

This model set $\tilde v \sim N(p_0,\Sigma_0)$,$\tilde u \sim N(0,\sigma_u^2)$ and they are independent. Then the insider set his number of trading as $\tilde x = X(\tilde v)$. The market trader then sets the price as $\tilde p = P(\tilde x + \tilde u)$.

Theorem 1 has proved that if function $X,P$ is linear, the market has an equlibrium with:(1)$X(\tilde v) = \beta(\tilde v - p_0)$;(2)$P(\tilde x + \tilde u) = p_0 + \lambda(\tilde x + \tilde u)$, where $\beta = (\sigma_u^2/\Sigma_0)^{1/2}$, and $\lambda = (\sigma_u^2/\Sigma_0)^{-1/2}/2$.

We define $\Sigma_1 = Var(\tilde v|\tilde p)$. My question is, how to prove that $\Sigma_1 = \Sigma_0/2$?

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The hint here is to think about $Y$ (the sum of informed orders and noise orders) and $v$ (the value of the stock) as bivariate correlated normal random variables.

We know in general that for two normal random variables $X_1\sim N(\mu_1,\sigma_1^2)$ and $X_2\sim N(\mu_2,\sigma_2^2)$, we can write

$X_1 = \mu_1 + \sigma_1 Z_1$

$X_2 = \mu_2 + \sigma_2 (\rho Z_1 + \sqrt{1-\rho^2}Z_2)$

Where $Z_1,Z_2 \sim N(0,1)$ and $\rho$ is the correlation between $X_1$ and $X_2$.

To get the conditional distribution of $X_2$ given $X_1=x$ we let $X_1 = x$ and observe that

$Z_1 = \frac{x-\mu_1}{\sigma_1}$

We can substitute that into the expression for $X_2$ and take the expectation and variance to get the mean and variance of the conditional distribution (we also know that this distribution is normal since it is a linear combination of a standard normal).

If you do all this, the variance turns out to be

$Var(X_2|X_1=x) = (1-\rho^2)\sigma_2^2$

Tying this back to Kyle, let $Y=X_1$ and $v=X_2$. We know how to get the covariance of $Y$ and $v$ (it is $\beta\Sigma_v^2$) And the the variance of $Y=(\beta^2\Sigma_v^2 + \sigma_u^2$) so all we need to compute is the correlation, plug into the above equation (and make all of the cancellations that come from expanding $\beta$ into its equilibrium value) and you'll get the expected result!