Expected value of mean plus the deviation from the mean Suppose $r$ and $\theta$ are random variables with:
$r = \bar{r} + \tilde{r}$
$\theta = \bar{\theta} + \tilde{\theta}$
where $\bar{r}, \bar{\theta}$ are the means, and $\tilde{r},\tilde{\theta}$ are the deviations of $r$ and $\theta$ from their mean.
I'm trying to verify the following formula:
\begin{align}
\mathbf{E}(r \cos\theta) &= \mathbf{E}[(\bar{r} + \tilde{r}) \cos(\bar{\theta} + \tilde{\theta})] \\
&= \mathbf{E}[(\bar{r} + \tilde{r})(\cos\bar{\theta}\cos\tilde{\theta} - \sin\bar{\theta}\sin\tilde{\theta})] \\
&= \bar{r} \cos\bar{\theta}
\end{align}
\begin{align}
\mathbf{E}(r \sin\theta) &= \mathbf{E}[(\bar{r} + \tilde{r}) \sin(\bar{\theta} + \tilde{\theta})] \\
&= \mathbf{E}[(\bar{r} + \tilde{r})(\sin\bar{\theta}\cos\tilde{\theta} + \cos\bar{\theta}\sin\tilde{\theta})] \\
&= \bar{r} \sin\bar{\theta} \mathbf{E}[{cos \tilde{\theta}]}
\end{align} 
where $\tilde{r}$ and $\tilde{\theta}$ are independent and their probability density functions are symmetric around their means (e.g. Gaussian or uniform).
I really do not understand how to verify these formulas.
Thank you in advance.
 A: I do not think that the statements are true. Consider the example where r is any symmetric distribution (as in symmetric around the expected value) with $E[r]=1$ and $\theta \sim uniform(-\pi,\pi)$. Then
$$E[\cos(\theta)] = \frac{1}{2\pi}\int_{-\pi}^\pi \cos(x) dx = 0 $$
however $\cos(E[\theta]) = \cos(0)=1 $, thus $E[r\cos(\theta)]=0 \neq 1 = E[r]\cos(E[\theta]) $.
A: Your second formula is correct, but the first one is not. 
First we do some expansion:
$$E[r \cos \theta] = E[r] E[\cos \theta] = \bar{r} E[ \cos \bar{\theta} \cos \tilde{\theta} - \sin \bar{\theta} \sin \tilde{\theta}] = \bar{r} \cos \bar{\theta} E[\cos \tilde{\theta}] - \bar{r} \sin \bar{\theta} E[\sin \tilde{\theta}] $$
$$E[r \sin \theta] = E[r] E[\sin \theta] = \bar{r} E[ \sin \bar{\theta} \cos \tilde{\theta} + \cos \bar{\theta} \sin \tilde{\theta}] = \bar{r} \sin \bar{\theta} E[\cos \tilde{\theta}] + \bar{r} \cos \bar{\theta} E[\sin \tilde{\theta}] $$
In the above, the first equal sign (of each line) uses the fact that $r$ and $\theta$ are independent.  The other equal signs rely on all the barred quantities being constants.  So finally the question becomes how to evaluate $E[\cos \tilde{\theta}]$ and $E[\sin \tilde{\theta}]$.
By definition, since $\tilde{\theta}$ is the deviation $\theta - \bar{\theta}$, its own mean is $E[\tilde{\theta}] = 0$.  Moreover, it is given that its distribution is symmetric, i.e. $p(\tilde{\theta}) = p(-\tilde{\theta})$ where $p$ is the PDF.  Since $\sin$ is an odd function, i.e. $\sin x = - \sin (-x)$, this means $E[\sin \tilde{\theta}] = 0$.  The gory details follow:
$$E[\sin \tilde{\theta}] = \int^{+\infty}_{-\infty} \sin \tilde{\theta} p(\tilde{\theta}) d\tilde{\theta} = \int^0_{-\infty} \sin \tilde{\theta} p(\tilde{\theta}) d\tilde{\theta} + \int^{+\infty}_0 \sin \tilde{\theta} p(\tilde{\theta}) d\tilde{\theta}$$
Next we show that the two integrals cancel each other:
$$\int^0_{-\infty} \sin \tilde{\theta} p(\tilde{\theta}) d\tilde{\theta} = \int^0_{+\infty} \sin (-\phi) p(-\phi) (-d\phi) = \int^0_{+\infty} (- \sin \phi) p(\phi) (-d\phi)     $$
$$ = \int^0_{+\infty} \sin \phi \ p(\phi) d\phi = - \int^{+\infty}_0 \sin \phi \ p(\phi) d\phi$$
In the above, the first equality is a change of variable $\phi = -\tilde{\theta}$, the second equality uses both $\sin()$ being odd and $p()$ being even (symmetric), and the rest is just algebraic manipulation.
So $E[\sin \tilde{\theta}] = 0$.  Unfortunately, the same cannot be said of $E[\cos \tilde{\theta}]$: it would depend on the exact distribution.  Indeed, I won't repeat it here but you will find that, because $\cos()$ is even, i.e. $\cos x = \cos (-x)$, the $\int^0_{-\infty}$ integral and the $\int^{+\infty}_0$ integral are equal, as opposed to being opposite signs.  So $E[\cos \tilde{\theta}]$ does not equal $0$ in general.  Thus we are left with:
$$E[r \cos \theta] = \bar{r} \cos \bar{\theta} E[\cos \tilde{\theta}] - \bar{r} \sin \bar{\theta} E[\sin \tilde{\theta}] = \bar{r} \cos \bar{\theta} E[\cos \tilde{\theta}]$$
$$E[r \sin \theta] = \bar{r} \sin \bar{\theta} E[\cos \tilde{\theta}] + \bar{r} \cos \bar{\theta} E[\sin \tilde{\theta}] = \bar{r} \sin \bar{\theta} E[\cos \tilde{\theta}] $$
