Compute conditional distribution without joint distribution I'm trying to solve problem 3.6 of Elements of causal inference by Peters, Janzing and Schölkopf. And I'm stuck, I'm afraid I'm missing something that might be trivial... the problem is the following:
Given the structual causal model $\mathfrak{C}$ (SCM):
$$\begin{align}C &:= N_C \\ E &:= 4C +N_E \end{align}$$
Where $N_C,N_E\stackrel{i.i.d.}{\sim}\mathcal{N}(0,1)$
Show that $P_{C|E=2}^{\mathfrak{C}}$ is a Gaussian distribution: $$C|E=2 \sim \mathcal{N}\left(\frac{8}{17},\sigma^2=\frac{1}{17}\right)$$
So, I know that the conditional probability mass function of $C$ given $E=e$ is given by 
$$P_{C|E=y}^{\mathfrak{C}}=\frac{P_{CE}^\mathfrak{C}}{P_E^{\mathfrak{C}}}$$
And I can easily compute $P_E^{\mathfrak{C}} = \mathcal{N}(0,17)$, but if I don't have the joint distribution, how can I compute the conditional distribution? 
 A: You can use Lagrange multipliers to solve this, but one approach is related to a Cross Validated question and answer using properties of the sums of independent normal distributed random variables  
Let $D = 4N_C$, so $D \sim \mathcal N(0,16)$ and $E=D+N_E$ with $D$ and  $N_E$ normal and independent so $E \sim \mathcal N(0,17)$
Then, using the linked answer, conditioned on a particular value of $E$ you have the difference in the conditional mean from the unconditional mean for $D$ being the difference between the value of $E$ and its unconditional mean multiplied by its share of the unconditional variance of $E$, so in this case with $E=2$ you get $2\times\frac{16}{16+1}=\frac{32}{17}$.  The conditional variance for $D$ is $\frac{16 \times 1}{16+1} = \frac{16}{17}$ making the conditional distribution of $D \sim \mathcal N\left(\frac{32}{17},\frac{16}{17}\right)$    
But you are actually interested in the conditional distribution of $C=\frac{D}{4}$ given $E=2$, so dividing the mean by $4$ and the variance by $4^2$ you get  $\mathcal N\left(\frac{8}{17},\frac{1}{17}\right)$    
