Reading Tanaka 1986, what is the meaning/expansion of $\langle I(\mathbf{r})I(\mathbf{0}) \rangle$ in eq(4)?
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 $I(\mathbf{r})$. $$ F(\mathbf{k}) = \mathscr{F}\{I\} = \int_v I(\mathbf{r}) \exp(-j\mathbf{k}\cdot\mathbf{r})d\mathbf{r} $$
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: $$ \langle I(\mathbf{r})I(\mathbf{0}) \rangle = \int I(\mathbf{r}-\mathbf{s}) \cdot I(\mathbf{s}) d\mathbf{s} $$