Expected value of $\sin^2(0.01n)$ for discrete $n$ Is it possible to calculate the expected value of $\sin^2(0.01n)$, with $n$ taking non-negative integer values?

Normally, when we wish to find the expected value of the sine function, we integrate (or add) its corresponding values within a single period (I think?), and then take its mean. Is such a method possible for the function under question?
Disclaimer: This is, so far, one of my most downvoted answers on the site. Since it is perfectly correct, and it answers the question, the downvotes might be due to some extra-mathematical reasons. In any case... happy reading!


The phrase "expected value" has no meaning when applied to a sequence $\mathbf x=(x_n)_{n\geqslant1}$, but one can look at the Cesàro means $(c_n(\mathbf x))_{n\geqslant1}$ and at their limit $\mathcal C(\mathbf x)$ defined, if it exists, as $$\mathcal C(\mathbf x)=\lim_{n\to\infty}c_n(\mathbf x)$$ where, for every $n\geqslant1$, $$c_n(\mathbf x)=\frac1n\sum_{k=1}^nx_k$$
In the present case, $x_n=\sin^2(an)$ with $a=0.01$ hence $2x_n=1-\cos(2an)=1-\Re(\mathrm e^{\mathrm ian})$ and $$2c_n(\mathbf x)=1-\frac1n\Re\left(\sum_{k=1}^n\mathrm e^{\mathrm iak}\right)$$ 
The sum in the RHS is geometric with ratio $z=\mathrm e^{\mathrm ia}$. Thus, the sums in the RHS are bounded since $|z|=1$ and $z\ne1$, hence the sequence $(c_n(\mathbf x))_{n\geqslant1}$ converges and its limit is $$\mathcal C(\mathbf x)=\frac12$$
Disclaimer: This is, so far, one of my most downvoted answers on the site. Since it is perfectly correct, and it answers the question, the downvotes might be due to some extra-mathematical reasons. In any case... happy reading!


The phrase "expected value" has no meaning when applied to a sequence $\mathbf x=(x_n)_{n\geqslant1}$, but one can look at the Cesàro means $(c_n(\mathbf x))_{n\geqslant1}$ and at their limit $\mathcal C(\mathbf x)$ defined, if it exists, as $$\mathcal C(\mathbf x)=\lim_{n\to\infty}c_n(\mathbf x)$$ where, for every $n\geqslant1$, $$c_n(\mathbf x)=\frac1n\sum_{k=1}^nx_k$$
In the present case, $x_n=\sin^2(an)$ with $a=0.01$ hence $2x_n=1-\cos(2an)=1-\Re(\mathrm e^{\mathrm ian})$ and $$2c_n(\mathbf x)=1-\frac1n\Re\left(\sum_{k=1}^n\mathrm e^{\mathrm iak}\right)$$ 
The sum in the RHS is geometric with ratio $z=\mathrm e^{\mathrm ia}$. Thus, the sums in the RHS are bounded since $|z|=1$ and $z\ne1$, hence the sequence $(c_n(\mathbf x))_{n\geqslant1}$ converges and its limit is $$\mathcal C(\mathbf x)=\frac12$$