This is an example of a tutorial but I think has not been solved properly. Please help me!
$X(T) = A \sin(\omega_0 t + \Phi)$
$A$ and $\phi$ are independent
$A$ is uniformly distributed over $(0,1)$
$\Phi$ is uniformly distributed over $(0, 2\pi)$
Is it mean-ergodic?
It is solved as follows:
$E[X(t)]=E[A]E[\sin(\omega_0t+\Phi)]=0$
$R_X(\tau)=E[X(t)X(t+\tau)]=E[A]E[\sin(\omega_0t+\Phi)\sin(\omega_0t+\omega_0\tau+\Phi)]=\frac{1}{2}E[A^2]\cos(\omega_0\tau)$
so it is stationary. Since its mean value is zero
$C_X(\tau)=\frac{1}{2}E[A^2]\cos(\omega_0\tau)$$\sigma^2_{\hat\mu_X}=\frac{E[A^2]}{2T}\{\int_{-2T}^{2T}[1-\frac{|\tau|}{2T}]\frac{\cos(\omega_0\tau)}{2}d\tau\}$
The next step is where I have problem with. I don't know whether we can substitute the fourier transform of $f$ when computing $\int_a^b f$ or are the fourier transforms computed here correct? Even I don't know the integration has been done correctly? The solution continues as follows:
We can use the result from Fourier transform that time multiplication is frequency convolution and evaluate the integral shown above.
$$[1-\frac{|\tau|}{2T}]\Longleftrightarrow 2T[\frac{\sin(\omega T)}{\omega T}]^2\quad and\quad\frac{\cos(\omega_0\tau)}{2}\Longleftrightarrow \frac{\pi}{2}[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]$$
Substituting the Fourier transforms of the triangular and the cosine functions in the equation for vairance, we obtain
$$\begin{align} \sigma^2_{\hat\mu_X}&=\frac{1}{2\pi}\frac{E[A^2]}{2T}\int_{-\infty}^{\infty}2T[\frac{\sin(\omega T)}{\omega T}]^2\frac{\pi}{2}[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]d\omega\\ &=\frac{E[A^2]}{2\pi}[\frac{\sin(\omega_0T)}{\omega_0T}]^2 \end{align}$$ $T\to\infty\,,\sigma^2_{\hat\mu_X}\to 0\,,for\;\omega_0\neq 0$ Thus $X(t)$ is mean-ergodic.
1- I don't know if the fourier transform calculation $[1-\frac{|\tau|}{2T}]\Longleftrightarrow 2T[\frac{\sin(\omega T)}{\omega T}]^2$ is true?
2- I don't know whether we can compute the integral of the fourier transform of a function instead of the original function?
3- Regarding that multiplication in time is equivalent to the convolution in frequency space I don't know whether substituting $[1-\frac{|\tau|}{2T}]\frac{\cos(\omega_0\tau)}{2}$ in $\sigma^2_{\hat\mu_X}=\frac{E[A^2]}{2T}\{\int_{-2T}^{2T}[1-\frac{|\tau|}{2T}]\frac{\cos(\omega_0\tau)}{2}d\tau\}$ by $2T[\frac{\sin(\omega T)}{\omega T}]^2\frac{\pi}{2}[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]$ in $\sigma^2_{\hat\mu_X}=\frac{1}{2\pi}\frac{E[A^2]}{2T}\int_{-\infty}^{\infty}2T[\frac{\sin(\omega T)}{\omega T}]^2\frac{\pi}{2}[\delta(\omega+\omega_0)+\delta(\omega-\omega_0)]d\omega$ is done correctly?