# Problem 3.6 Conditional Distribution

Show that 
$$P_{C\mid E=2}\sim\mathcal{N}\left(\frac{8}{17},\frac{1}{17}\right)$$
for the SCM
\begin{align}
C:=&\mathcal{N}(0,1)\\
E:=&4C+\mathcal{N}(0,1)
\end{align}

## Proof

The SCM can be seen as a [multivariate normal distribution](https://en.wikipedia.org/wiki/Multivariate_normal_distribution)
$$
P\left(\begin{matrix}C\\E\end{matrix}\right)=
\mathcal{N}\left(\begin{bmatrix}\mu_C\\\mu_E\end{bmatrix},\Sigma\right)
$$
where $\mu_C=\mu_E=0$ because both distributions have an expecation of zero and $\Sigma$ is the covariance matrix which is
\begin{align}
\Sigma:=&\begin{bmatrix}\sigma^2_C&\rho\sigma_C\sigma_E\\\rho\sigma_C\sigma_E&\sigma_C^2\end{bmatrix}\\
=&\begin{bmatrix}1&14\\14&4^2+1\end{bmatrix}
\end{align}

For the bivariate case the conditional distribution is defined as 
$${\displaystyle X_{1}\mid X_{2}=a\ \sim \ {\mathcal {N}}\left(\mu _{1}+{\frac {\sigma _{1}}{\sigma _{2}}}\rho (a-\mu _{2}),\,(1-\rho ^{2})\sigma _{1}^{2}\right).}$$

Applied to $C\mid E=2$ that yields $$\mathcal{N}\left(\frac{8}{17},\frac{1}{17}\right)$$

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2020-02-19T16:30:04.000Z
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1. YES
2. YES