Find $E[Y_5\vert Y_1 = y_1]$ where $Y_t = \cos(tW)$ and $W$ is uniform on $[0,2\pi]$

Question

Find $E[Y_5\vert Y_1 = y_1]$ where $Y_t = \cos(tW)$ and $W$ is uniform on $[0,2\pi]$

So I think this is as follows:

Given the definitions of $Y_t$ and $W$ in the title, and Since we know $Y_1=y_1$, we know $W=w$.

That is, $$ w = \cos^{-1}{(1\cdot y_1)} =\cos^{-1}{(y_1)} $$ Thus, we by knowing $Y_1=y_1$ we know $W=w$, and thus $$ E[Y_5\vert Y= y_1] = E[\cos{(5\cdot W)}\vert y_1] = \cos{(5\cdot w)} =\cos{(5\cdot \cos^{-1}{(y_1)})} $$

I am wondering if this simplifies further, or if I made a mistake? It just seems a bit messy to me (although I guess it is easy numerically)

Answer 1

For every real number $w$, $$\cos(5w)=16(\cos w)^5-20(\cos w)^3+5\cos w$$ hence $$Y_5=\cos(5W)=16(\cos W)^5-20(\cos W)^3+5\cos W=16Y_1^5-20Y_1^3+5Y_1$$ This shows that $Y_5$ is $\sigma(Y_1)$-measurable and that $$E(Y_5\mid Y_1)=Y_5=16Y_1^5-20Y_1^3+5Y_1$$ which implies that, for every real number $y_1$, $$E(Y_5\mid Y_1=y_1)=16y_1^5-20y_1^3+5y_1$$ This result holds for every distribution of $W$.