Expectation of random process

Question

I have a random process defined as

$$X(t) = A \sin(\omega t + \phi)$$

where A and $\omega$ are independent. $\phi$ is distributed $U[0,2\pi]$.

I would like to find $E[X(t)]$.

I believe the answer is $0$ because the multiplication of 2 expectation of independent random variables is the product of the expectation of each. i.e. $E[X(t)] = E[A] E[\sin(\omega t + \phi)]$.

Since $\phi$ is distributed uniformly over a cycle, the mean of the $\sin()$ term must be zero thereby making the whole quantity 0.

Is this correct?

Answer 1

Yes, if $(A,\omega,\phi)$ is mutually independent.

Indeed, if $(A,\omega,\phi)$ is mutually independent, then $A$ is independent of $(\omega,\phi)$ so you can write $\mathbb E[X(t)]=\mathbb E[A]\mathbb E[\sin(\omega t+\phi)]$.

Moreover, $\omega$ is independent of $\phi$ so $$ \mathbb E[\sin(\omega t+\phi)]=\mathbb E[\sin(\omega t)\cos(\phi)]+\mathbb E[\sin(\phi)\cos(\omega t)]=\mathbb E[\sin(\omega t)]\mathbb E[\cos(\phi)]+\mathbb E[\sin(\phi)]\mathbb E[\cos(\omega t)]=0 $$

I used the formula $\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$ because it requires the least knowledge in probability theory. If you are more familiar with the probability theory, here is another way to conclude: since $\omega$ and $\phi$ are independent, we have $$ \mathbb E[\sin(\omega t+\phi)]=\mathbb E[f(\omega t)], $$ where $f:\mathbb R\to\mathbb R$ is defined for all $x\in\mathbb R$ by $f(x)=\mathbb E[\sin(x+\phi)]=0$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\int_{0}^{2\pi}A\sin\pars{\omega t + \phi}\,{\dd\phi \over 2\pi}} = \left.\phantom{\Large A} -\,{A \over 2\pi}\,\cos\pars{\omega t + \phi}\,\right\vert_{\ \phi\ =\ 0}^{\ \phi\ =\ 2\pi} \\[5mm] = &\ -\,{A \over 2\pi}\,\cos\pars{\omega t + 2\pi} + {A \over 2\pi}\,\cos\pars{\omega t + 0} = \bbx{0} \end{align}

$\ds{\cos}$ is a periodic function of period $\ds{2\pi}$.