Link between conditional characteristic function and conditional density Let $X$ and $Y$ be random variables (real-valued). I define
$$E[e^{i\theta X}\mid\sigma(Y)] =: g(Y,\theta)$$
Suppose that $g(Y,\theta) = e^{i\theta Y}e^{-\frac{1}{2}\theta^2}$. Can I then say that the density of $X$ given $Y$, $f_{X\mid Y}(x\mid y)$, exists and is in fact given by
$$f_{X\mid Y}(x\mid y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-y)^2}$$
If yes, can someone refer me to a standard textbook on this result? (I am guessing the proof of such a fact might be quite involved.)
 A: If $f_{X\mid Y}(x\mid y)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-y)^2}$ then
$$\begin{align}\mathsf E(\mathsf e^{\imath\theta X}\mid Y) ~=~& \tfrac 1{\surd (2\pi)}\int_\Bbb R \mathsf e^{\imath\theta x-(x-Y)^2/2}\operatorname d x
\\[1ex] ~=~& \tfrac 1{\surd (2\pi)}\int_\Bbb R \mathsf e^{(-x^2+2(\imath\theta+Y) x-Y^2)/2}  \operatorname d x \\[1ex] ~=~& \tfrac 1{\surd (2\pi)}\int_\Bbb R \mathsf e^{(-x^2+2(\imath\theta+Y) x-(\imath\theta+Y)^2+\theta^2+2\imath\theta Y)/2}  \operatorname d x \\[1ex] ~=~ & \dfrac{\mathsf e^{(\theta^2+\imath\theta Y)/2}}{\surd(2\pi)}\int_\Bbb R \mathsf e^{((\imath x-\imath (\imath\theta+Y))/\surd 2)^2}\operatorname d x
\\[1ex] ~=~ & -\tfrac \imath 2 \mathsf e^{\theta^2/2+\imath\theta Y}\operatorname{erfi}(\dfrac{\imath x-\imath(\imath \theta +Y)}{\surd 2})\Big\vert_{x\to-\infty}^{x\to\infty} \\[1ex] ~=~& \mathsf e^{\theta^2/2+\imath\theta Y}\end{align}$$

Because:  $\dfrac{\partial~\operatorname {erfi}(f(x))}{\partial x}~=~ \dfrac{2f'(x)\mathsf e^{f^2(x)}}{\surd \pi}$
Where $\operatorname {erfi}(\;)$ is the imaginary error function.
