$\theta$ is uniform distributed $[0,\pi]$, $Y=\cos(\theta)$
I need to calculate LMMSE of Y out of $\theta$ $E\hat[Y/\theta]$. To do that I need to calculate cross covariance $COV(Y,\theta)$. Somewhere along the way I have a mistake, but couldn't find where.
This is my calculation:
$$f_Y(y)=\frac{d}{dy}P(Y<y)=\frac{d}{dy}P(\cos(\theta)<y)=\frac{d}{dy}P(\theta>\cos^{-1}(y))=\frac{d}{dy}\int_{\cos^{-1}(y)}^{\pi}\frac{1}{\pi}d\theta=\frac{d}{dy}(\frac{1}{\pi}(\pi-\cos^{-1}(y)))=\frac{1}{\pi}\frac{1}{\sqrt{1-y^2}}$$
Now the mutual PDF calculation is as follow:
$$f(y,\theta)= f(y/\theta)f(\theta)=\frac{1}{\pi^2\sqrt{1-\cos^2(\theta)}}=\frac{1}{\pi^2\sin(\theta)}$$
The covariance calculation:
$$cov(\theta,Y)=\int_{0}^{\pi}\int_{-1}^{\cos(\theta)}y\theta f(y,\theta)dyd\theta=\frac{1}{\pi^2}\int_{0}^{\pi}\frac{\theta}{\sin(\theta)}\int_{-1}^{\cos(\theta)}ydyd\theta=\frac{1}{2\pi^2}\int_{0}^{\pi}\frac{\theta}{\sin(\theta)}(\cos^2(\theta)-1)d\theta=-\frac{1}{2\pi^2}\int_{0}^{\pi}\theta \sin(\theta) d\theta=-\frac{1}{2\pi^2}(\sin(\theta)-\theta \cos(\theta))|_{0}^{\pi}=-\frac{1}{2\pi}$$
The answer should be : $-\frac{2}{\pi}$
The question is very easy!
$$\Theta\sim U[0;\pi]$$
(capital letter is required because it is a rv)
$Y=\cos\Theta$ is a deterministic function thus no double integral is needed:
$$\mathbb{Cov}[Y;\Theta]=\mathbb{E}[\Theta\cdot\cos(\Theta)]-\mathbb{E}[\Theta]\cdot\mathbb{E}[\cos(\Theta)]=$$
$$=\int_0^{\pi}\frac{\theta\cos(\theta)}{\pi}d\theta-\frac{\pi}{2}\cdot\int_0^{\pi}\frac{\cos(\theta)}{\pi}d\theta=-\frac{2}{\pi}$$