# Refinement Orders for Quantitative Information Flow and Differential Privacy +
## Appendix D Proofs of Results about Refinement Comparison  
We call a truncated geometric mechanism'square' if is has the same input and output space (that is, the channel representation is a square matrix). We first show that geometric mechanisms and truncated geometric mechanisms are equivalent to square mechanisms under \(\sqsubseteq^{\rm avg}\). **Lemma A1**. : _Let \(G^{\varepsilon}\) be a geometric mechanism. Then, the reduced (abstract) channel form of \(G^{\varepsilon}\) is the square channel \((\mathcal{X},\mathcal{X},TG^{\varepsilon})\)._  
Proof. : First, note that the square channel is obtained from the (infinite) geometric by summing up all the 'extra' columns (i.e., those columns in \(\mathcal{Y}\setminus\mathcal{X}\)). Now, note that these 'extra' columns are scalar multiples of each other, since each column has the form  
\[ \begin{bmatrix}\frac{(1-\alpha)\cdot\alpha^{k}}{1+\alpha}\\ \\ \frac{(1-\alpha)\cdot\alpha^{k-1}}{1+\alpha}\\ \\ \frac{(1-\alpha)\cdot\alpha^{k-2}}{1+\alpha}\\ \\ \ldots\end{bmatrix} \] (A16)  
for increasing values of \(k\). Thus, the'summing up' operation is a valid reduction operation, and so the infinite geometric is reducible to the corresponding square channel. **Lemma A2**. : _Let \(TG^{\varepsilon}\) be a truncated geometric mechanism. Then, the reduced (abstract) channel form of \(TG^{\varepsilon}\) is the square channel \((\mathcal{X},\mathcal{X},TG^{\varepsilon})\)._  
Proof. : First, note that the truncated geometric is obtained from the infinite geometric by summing up columns at the ends of the matrix. This is exactly the'reduction' step noted above. We can continue, as above, to sum up 'extra' columns until we get a square matrix. **Corollary A1**. : _Any \(\sqsubseteq^{\rm avg}\) refinement that holds for a square geometric mechanism \((\mathcal{X},\mathcal{X},G^{\varepsilon})\) also holds for any truncated geometric mechanism or (the) geometric mechanism \(G^{\varepsilon}\) having domain \(\mathcal{X}\)._  
Note that we only define truncation as far as the square geometric matrix, since at this point the columns of the matrix are linearly independent and can no longer be truncated via matrix reduction operations. We now show that refinement holds for the square geometric mechanisms. **Lemma A3**. : _Let \(TG^{\varepsilon}\) be a square geometric mechanism. Then, decreasing \(\varepsilon\) produces a refinement of it. That is, \(TG^{\varepsilon}\sqsubseteq^{\rm avg}TG^{\varepsilon^{\prime}}\) iff \(\varepsilon\geq\varepsilon^{\prime}\)._Proof. : The square geometric mechanism \(TG^{\varepsilon}\) has the following form:  
\[ \begin{array}{|c|cccc|}\hline TG^{\varepsilon}&x_{1}&x_{2}&\ldots&x_{n}\\ \hline x_{1}&\frac{1}{1+\alpha}&\frac{\alpha\cdot(1-\alpha)}{1+\alpha}&\ldots &\frac{\alpha^{n-1}}{1+\alpha}\\ x_{2}&\frac{\alpha}{1+\alpha}&\frac{1-\alpha}{1+\alpha}&\ldots&\frac{\alpha^{ n-2}}{1+\alpha}\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ x_{n}&\frac{\alpha^{n-1}}{1+\alpha}&\frac{\alpha^{n-2}\cdot(1-\alpha)}{1+ \alpha}&\ldots&\frac{1}{1+\alpha}\\ \hline\end{array}  \tag{120}\]  
where \(\alpha=e^{-\varepsilon}\), and similarly for \(TG^{\varepsilon^{\prime}}\) with \(\alpha=e^{-\varepsilon^{\prime}}\). Now, this matrix is invertible and the inverse has the following form:  
\[ \begin{array}{|c|cccc|}\hline(TG^{\varepsilon})^{-1}&x_{1}&x_{2}&x_{3}&x_{4}& \ldots\\ \hline x_{1}&\frac{1}{1-\alpha}&\frac{-\alpha}{1-\alpha}&0&0&\ldots\\ x_{2}&\frac{-\alpha}{(1-\alpha)^{2}}&\frac{1+\alpha^{2}}{(1-\alpha)^{2}}& \frac{-\alpha}{(1-\alpha)^{2}}&0&\ldots\\ x_{3}&0&\frac{-\alpha}{(1-\alpha)^{2}}&\frac{1+\alpha^{2}}{(1-\alpha)^{2}}& \frac{-\alpha}{(1-\alpha)^{2}}&\ldots\\ x_{4}&0&0&\frac{-\alpha}{(1-\alpha)^{2}}&\frac{1+\alpha^{2}}{(1-\alpha)^{2}}& \ldots\\ \ldots&\ldots&\ldots&\ldots&\ldots\\ \hline\end{array}