No. In the large $n$ limit, this is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$.

We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute $$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ Putting these together gives $E[(a_{ij}-b_{ij})^2]=\frac{2-3\pi}{5}$.

No. In the large $n$ limit, this is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$.

We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute $$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ Putting these together gives $E[(a_{ij}-b_{ij})^2]=2-\frac{3\pi}{5}$.

No. This is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$.

We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute $$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ Putting these together gives $E[(a_{ij}-b_{ij})^2]=\frac{2-3\pi}{5}$.

No. In the large $n$ limit, this is equivalent to asking whether the expected value of $(a_{ij}-b_{ij})^2$ is $(1/3)^2$, but in fact the expected value is $2-\frac{3\pi}{5} \approx .115$.

We can compute the expectation as follows: $a_{ij}^2 = r_i^2 + r_j^2 + 2 r_i r_j \cos \theta_{ij}$, where $\theta_{ij}$ is the angle between the vectors from the origin to points $i$ and $j$. The variable $\theta_{ij}$ is independent of $r_i,r_j$ and $\cos \theta_{ij}$ has expected value zero. So $$E[a_{ij}^2]=2E[r_i^2]=2\frac{\int_0^1 r^2 \cdot r^2\,dr}{\int_0^1 r^2 \,dr} = \frac{6}{5}$$ We have $b_{ij}=a_{ij} \sin \phi_{ij}$, where $\phi_{ij}$ is the angle between the vector from $i$ to $j$ and the vertical. Furthermore, $\phi_{ij}$ is independent of $a_{ij}$. So $E[(a_{ij}-b_{ij})^2]=E[a_{ij}^2] E[(1-\sin \phi_{ij})^2]$. We can compute $$ E[(1-\sin \phi_{ij})^2]=\frac{\int_0^{\pi} (1-\sin \phi)^2 \sin \phi\, d\phi}{\int_0^{\pi} \sin \phi\, d\phi}=\frac{5}{3}-\frac{\pi}{2}$$ Putting these together gives $E[(a_{ij}-b_{ij})^2]=\frac{2-3\pi}{5}$.