the agent gives a grade of 1 when it should be 0 or vice versa). In this case, their grade is reduced by a factor $\beta$ for evaluating incorrectly.

The generalised PeerRank rule continues to satisfy the domain, no discrimination, no dummy, and symmetry properties. For no discrimination, and no dummy, we can just set $\beta=0$ and appeal to the previous results. For the domain property, we need to prove afresh that the additional term cannot take us outside the interval $[0,1]$.

The generalized PeerRank rule returns grades in $[0,1]$.

[[**Proof:  **]{}\
]{}Clearly $X^n_i \geq 0$ for all $n$ as it is the sum of terms which are not negative. We prove that $X^n_i \leq 1$ by induction on $n$ In the step case, suppose $0 \leq X^n_i \leq 1$ for all $i$. Let $X^n_i = 1 - \epsilon$ where $0 \leq \epsilon \leq 1$. Then $$\begin{aligned}
X^{n+1}_i & = & (1-\alpha-\beta)(1-\epsilon) + \frac{\alpha}{\sum_j X^n_j}
\sum_j X^n_j . A_{i,j} + \\
& & \ \ \ \ \frac{\beta}{m}. \sum_j 1 - |A_{j,i} -X^n_j| \\
& \leq & 1 -\alpha - \beta - \epsilon (1-\alpha - \beta) +
\frac{\alpha}{\sum_j X^n_j} \sum_j X^n_j  +  \\
& & \ \ \ \ \ 