determine $\mathbb E(X_1 \mid X_2)$ and $\mathbb V(X_1 \mid X_2)$ Let $(X_1,X_2)$ be a normally distributed vector with $\mathbb E(X_1) = \frac14$, $\mathbb E(X_2) = 0$, $\mathbb V(X_1) = 1$, $\mathbb V(X_2) = \frac14$, $\operatorname{Cov}(X_1,X_2) = -\frac18$.
I want to determine $\mathbb E(X_1 \mid X_2)$ and $\mathbb V(X_1 \mid X_2)$.
I am not quite sure how to apply the definitions of conditional probability.
 A: Using the approach in the comment by @Did,
Let $X_1 = aX_2 + bX_3 + c$ where $X_3 \sim \mathcal{N}(0,1)$ and $X_2$ is independent of $X_3$ we try to find $a, b, c$.
$$\operatorname{Cov}(X_1, X_2) = \frac{-1}{8} \implies  a\operatorname{Var}(X_2) = \frac{-1}{8}$$ 

$$a = \frac{-1}{2}$$

$$E(X_1) = \frac{1}{4} \implies aE(X_2) + c = \frac{1}{4}$$

 $$c = \frac{1}{4}$$

$$\operatorname{Var}(X_1) = 1 \implies a^2\operatorname{Var}(X_2) + b^2 =  1$$

 $$ b = \frac{\sqrt{15}}{4}$$

Therefore, $X_1$ can now be written as,

 $$X_1 = \frac{-X_2}{2} + \frac{\sqrt{15}X_3}{4} + \frac{1}{4}$$

and $E(X_1\mid X_2), V(X_1\mid X_2)$ are,

 $$E(X_1\mid X_2) = \frac{-X_2}{2} + \frac{1}{4}, V(X_1\mid X_2) = \frac{15}{16}$$

A: $\newcommand{\E}{\operatorname{E}}\newcommand{\v}{\operatorname{var}}\newcommand{\c}{\operatorname{cov}}$
I thought of posting an answer involving expectations and covariances, and someone has posted something along those lines, so I'll try a different way of looking at it. You have
$$
\rho = \operatorname{correlation}(X_1,X_2) = \frac{\c(X_1,X_2)}{\sqrt{\v(X_1)\v(X_2)}} = \frac{1/8}{\sqrt{1\cdot1/4}} = \frac 1 4.
$$
Suppose you know that the joint density is
$$
\text{constant} \times
      \exp\left(
        -\frac{1}{2(1-\rho^2)}\left[
          \frac{(x_1-\frac 1 4)^2}{1/2} +
          \frac{(x_2-0)^2}{1} -
          \frac{2\rho(x_1-\frac 1 4)(x_2-0)}{(1/4)\cdot 1}
        \right]
      \right)
$$
Recall that the conditional density of $X_1$ given $X_2=x_2$ is a constant multiple of the expression above as a function of $x_1$ with $x_2$ fixed. And it will be a normal density, having its maximum at its expected value. The expression above has its maximum at the point where the quadratic polynomial inside the $\Big[\text{square brackets}\Big]$ has its minimum.
\begin{align}
& \frac{(x_1-\frac 1 4)^2}{1/2} + \frac{(x_2-0)^2} 1 - \frac{2\rho(x_1-\frac 1 4)(x_2-0)}{(1/4)\cdot 1} \\[10pt]
= {} & 2\left(x_1 - \frac 1 4\right)^2 - 2\left(x_1 - \frac 1 4 \right) x_2 + \cdots\cdots \\
& \quad \text{where “$\cdots\cdots$'' means terms not depending on $x_1$} \\[10pt]
= {} & \frac{(4x_1-1)^2} 8 - 2x_1x_2 + \cdots\cdots \\[10pt]
= {} & \frac 1 8 \left( (16x_1^2 - 8x_1 + 1) - 16 x_1 x_2 \right) + \cdots\cdots \\[10pt]
\propto {} & \underbrace{x_1^2 - \left( \frac 1 2 + x_2 \right)x_1 + \cdots\cdots = x_1^2 -\left( \frac 1 2 + x_2 \right) x_1 + \left( \frac{1+ 2x_2} 4 \right)^2 + \cdots\cdots}_{\Large\text{completing the square}} \\[12pt]
= {} & \left( x_1 - \frac{1+2x_2} 4 \right)^2 + \cdots.
\end{align}
We expect
$$
\left( x_1 - \text{something} \right)^2,
$$
where $\text{“something''}$ does not depend on $x_1,$ and we got $\text{“something''} = \dfrac{1+2x_2} 4.$ Therefore
$$
\E(X_1\mid X_2 = x_2) = \frac{1+2x_2} 4.
$$
PS: I now see that $\rho = \dfrac{-1} 4$ had a minus sign. So just change that in the appropriate places.
A: Since $(X_1,X_2)$ is normally distributed, $$ c_1 X_1+c_2X_2 \overset{d}{=} \mu(c_1,c_2) + \sigma(c_1,c_2) W,\quad \forall c_1,c_2\in \mathbb{R}$$
where $$\mu(c_1,c_2) = \mathbb{E}[c_1 X_1+c_2X_2] = c_1 \mathbb{E}[X_1] + c_2\mathbb{E}[X_2],$$
$$\sigma(c_1,c_2)=\sqrt{\operatorname{Var}[c_1 X_1+c_2X_2]} = \sqrt{c_1^2\operatorname{Var}[ X_1] +c_2^2 \operatorname{Var}[X_2] + 2c_1c_2\operatorname{Cov}[X_1,X_2]},$$ and $W$ is a standard normal random variable.
Now, taking expected value conditioned on $X_2$ of above 
$$ c_1 \operatorname{E}[X_1\mid X_2] + c_2 X_2 = \mu(c_1,c_2) + \sigma(c_1,c_2)\operatorname{E}[W\mid X_2]$$
Similarly,
$$c_1^2\operatorname{Var}[X_1\mid X_2] = \sigma^2(c_1,c_2) \operatorname{Var}[W\mid X_2]$$
Let $c_{1*}, c_{2*}$ be values of $c_1,c_2$ such that $W $ is independent of $X_2$, equivalently, $\operatorname{E}[WX_2] = \operatorname{E}[W] \operatorname{E}[X_2]=0$. This can be done by multiplying the first equation by $X_2$ and taking expected values on both sides. Then, we have
$$\operatorname{E}[X_1\mid X_2] = \frac{\mu(c_{1*}, c_{2*}) - c_{2*}X_2}{c_{1*}}$$
and 
$$ \operatorname{Var}[X_1\mid X_2] = \frac{\sigma^2(c_{1*}, c_{2*})}{c_{1*}^2}$$
