I've had my test for this course and I think I failed it again. The hardest part for me is findig the correct distributions. This is a test exercise I couldn't figure out or at least, I probably failed the question. I hope I can get your help, so next time I won't fail my test.
Given: $$ \begin{align} X_{k+1} &= X_k\\ Y_k &= X_k^2 - V_k^2 \end{align} $$ Also $X_0$ and $V_k$ are $\sim U(0,1)$. With $V_k$ white noise, in this case means always independent of $X_k$. I did the test and I came up with several approaches, but the reason I think they failed is since the outcome of the different ways were not the same.
First: Determine $E[X_0|Y_0]$, for simplicity $X_0 = X$ and $Y_0 = Y$. The approach I did that I thought was most promising: $$ \begin{align} E[X|Y] &= \int x f_{X|Y} dx = \int x \left(\frac{d}{dx}P(X\leq x|Y =y)\right)dx\\ &= \int x \left(\frac{d}{dx}P(V_0^2\leq x^2 - y|Y =y)\right)dx\\ &= \int x \left(\frac{x}{\sqrt{x^2 - y}} \mathbb{1}_{x^2-1<y<x^2}\mathbb{1}_{0<x<1} \right)dx\\ &= \left(\mathbb{1}_{-1<y<0}\int_{0}^{\sqrt{y+1}} + \mathbb{1}_{0<y<1}\int_{\sqrt{y}}^{1}\right)\frac{x^2}{\sqrt{x^2 - y}}dx \end{align} $$ I couldn't get any further than this. What should I have done? Are those steps done right? I also thought of computing ${f_{X,Y}(x,y)}{f_Y(y)}$ and using a parametrization $X_0 = r \cosh(\theta)$ and $V_0 = r \sinh(\theta)$. But how I should do that in this case is beyond me at the moment--although I guess this can't be done either since $\cosh(\theta) \geq 1$.
Second: coming up...
There are two ways you could tackle the problem:
(a) Use "change of variables" from $X$ and $V$ to $X$ and $Y$ and then find the required conditional distribution and expectation.
(b) Use the "direct" method, which is what you've done.
For this particular problem, I think (a) is a slightly simpler but I will use (b) here since it is in line with your approach.
I'll assume for $E(X\mid Y=y)$ that $y\geq 0$. The working is similar in the $y\lt 0$ case.
\begin{eqnarray*} E(X\mid Y=y) &=& \int_x xP(X=x\mid X^2-V^2 = y)\;dx \\ &=& \int_x xP(X=x\cap V^2 = x^2-y)\;dx \;/\; P(X^2-V^2 = y) \\ &=& \int_x xP(X=x)P(V^2 = x^2-y)\;dx \;/\; P(X^2-V^2 = y) \\ &=& \int_{x=\sqrt{y}}^{1} x\cdot 1\cdot \dfrac{1}{2\sqrt{x^2-y}}\;dx \;/\; P(X^2-V^2 = y) \\ && \qquad\qquad\qquad\qquad\text{(if $U=V^2,\;$ then $\frac{dv}{du}=\frac{1}{2\sqrt{u}}\;$ so $f_U(u)=\frac{1}{2\sqrt{u}}$)} \\ &=& \left[\frac{1}{2}\sqrt{x^2-y}\right]_{x=\sqrt{y}}^{1} \bigg/ P(X^2-V^2 = y) \\ &=& \frac{1}{2}\sqrt{1-y} \bigg/ P(X^2-V^2 = y). \qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{(1)} \end{eqnarray*}
We now find $P(X^2-V^2 = y)$ by way of $P(X^2-V^2 \leq y)$.
\begin{eqnarray*} P(X^2-V^2 \leq y) &=& 1 - \int_{x=\sqrt{y}}^{1} \sqrt{x^2-y}\;dx \\ &=& 1 - \frac{1}{2} \left[ x\sqrt{x^2-y} - y\ln \left(\sqrt{x^2-y} + x\right) \right]_{x=\sqrt{y}}^{1} \\ &=& 1 - \frac{1}{2} \left[ \sqrt{1-y} - y\ln \left(\sqrt{1-y} + 1\right) +y\ln\left(\sqrt{y}\right) \right]. \\ \end{eqnarray*}
Differenting wrt $y$, we have
\begin{eqnarray*} P(X^2-V^2 = y) &=& - \frac{1}{2} \left[ \frac{-1}{2\sqrt{1-y}} - \ln \left(\sqrt{1-y} + 1\right) +\frac{y}{\sqrt{1-y}+1}\cdot \frac{1}{2\sqrt{1-y}} + \ln\left(\sqrt{y}\right) + \frac{1}{2} \right] \\ &=& \frac{1}{2}\ln \left(\sqrt{1-y} + 1\right) - \frac{1}{2}\ln\left(\sqrt{y}\right) \qquad\qquad\qquad\qquad\qquad\text{after simplification.} \end{eqnarray*}
Substituting into $(1)$ we have, for $y\geq 0$
\begin{eqnarray*} E(X\mid Y=y) &=& \frac{\sqrt{1-y}}{\ln \left(\sqrt{1-y} + 1\right) - \ln\left(\sqrt{y}\right)}. \end{eqnarray*}