What is the meaning of these angle brackets $\langle I(\mathbf{r})I(\mathbf{0}) \rangle$ in the convolution theorem? Reading Tanaka 1986, what is the meaning/expansion of
$\langle I(\mathbf{r})I(\mathbf{0}) \rangle$  in eq(4)?
Power Spectrum of 2D-FT
Equation (4):
$$P(\mathbf{k}) = |F(\mathbf{k})|^2 =F\cdot F^* = \int_v\langle I(\mathbf{r})I(\mathbf{0}) \rangle \exp(-j\mathbf{k}\cdot\mathbf{r})d\mathbf{r} $$
where $F(\mathbf{k})$ is the Fourier Transform of $I(\mathbf{r})$.
$$ F(\mathbf{k}) = \mathscr{F}\{I\} = \int_v I(\mathbf{r}) \exp(-j\mathbf{k}\cdot\mathbf{r})d\mathbf{r} $$
 A: My try to the answer:
From the convolution theorem:
 $$\mathscr{F}\{f\}\cdot\mathscr{F}\{g\} = \mathscr{F}\{f*g\}$$
where:
$$\{f*g\}(\mathbf{r}) = \int f(\mathbf{s})\cdot g(\mathbf{r}-\mathbf{s}) d\mathbf{s}$$
$$\mathscr{F\{f*g\}(\mathbf{k})} = \int_v\biggr(\int f(\mathbf{s})\cdot g(\mathbf{r}-\mathbf{s}) d\mathbf{s}\biggl) \exp(-j\mathbf{k}\cdot\mathbf{r})d\mathbf{r}$$
In our case, $f = g = I(\mathbf{r})$ and assuming the FT is real, so $F^* = F$:
$$ \mathscr{F}\{I\}\cdot \mathscr{F}\{I\} = \int_v\biggr(\int I(\mathbf{s})\cdot I(\mathbf{r}-\mathbf{s}) d\mathbf{s}\biggl) \exp(-j\mathbf{k}\cdot\mathbf{r})d\mathbf{r}$$
so:
$$ \langle I(\mathbf{r})I(\mathbf{0}) \rangle  = \int I(\mathbf{s})\cdot I(\mathbf{r}-\mathbf{s}) d\mathbf{s} $$
But not familiar with this notation, what does the $\mathbf{0}$ represent?
Edit: As @AnonSubmitter85 suggested, the angles refer to the inner product, and the arguments might reffer to the offset, reordering the integral:
$$ \langle I(\mathbf{r})I(\mathbf{0}) \rangle  = \int  I(\mathbf{r}-\mathbf{s}) \cdot I(\mathbf{s}) d\mathbf{s} $$
