# Iterative Learning for K-Approval Votes in Crowdsourcing Systems
## Appendix A Proof of Sub-Gaussianity  
We now prove that for all \(d\in[2:D]\), \(\mathbf{\hat{x}}_{d}^{(k)}\) is _sub-Gaussian_ with some \(m_{k}\) and \(\tilde{\sigma}_{k}^{2}\). Recurrence relations of the evolution of the MGFs (moment generating functions) on \(\mathbf{\hat{x}}_{d}\) and \(\mathbf{y_{p}}\) are stated as:  
\[\mathbb{E}[e^{\lambda\mathbf{\hat{x}}_{d}^{(k)}}]=\left(\mathbb{E}_{\mathbf{p }}\left[\mathbf{p^{+}}\mathbb{E}\left[e^{\lambda\mathbf{y}_{p}^{(k-1)}}| \mathbf{p}\right]+\mathbf{p^{-}}\mathbb{E}\left[e^{-\lambda\mathbf{y}_{p}^{(k- 1)}}|\mathbf{p}\right]+(1-\mathbf{p^{+}}-\mathbf{p^{-}})\right]\right)^{\hat{l}}, \tag{11}\]  
\[\mathbb{E}[e^{\lambda\mathbf{y}_{p}^{(k)}}]=\left(p\cdot\mathbb{E}\left[e^{ \lambda\left(\frac{D-K}{D}\mathbf{\hat{x}}_{d}^{(k)}\right)}|(1-p)\cdot \mathbb{E}\left[e^{-\lambda\left(\hat{x}\right)}\mathbf{\hat{x}}_{d}^{(k)} \right]\right)^{\hat{l}}, \tag{12}\]  
where \(\mathbf{p^{+}}=p\cdot\frac{D-K}{D-1}\) and \(\mathbf{p^{-}}=(1-p)\cdot\frac{K}{D-1}\). Using the above MGFs and _mathematical induction_, we can prove that \(\mathbf{\hat{x}}_{d}^{(k)}\) are _sub-Gaussian_, for all \(d\in[2:D]\). First, for \(k=1\), we prove that all of \(\mathbf{\hat{x}}_{d}^{(1)}\) are _sub-Gaussian_ random variables with mean \(m_{1}=\mu\hat{l}\) and variance \(\tilde{\sigma}_{1}^{2}=2\hat{l}\), where \(\mu\equiv\mathbb{E}[\frac{D}{D-1}(p_{j}-\frac{1}{D})]\). Using _Gaussian_ initialization of \(\mathbf{y}_{p}\sim\mathcal{N}(1,1)\), we obtain \(\mathbb{E}[e^{\lambda\mathbf{y}_{p}^{(0)}}]=e^{\lambda+(1/2)\lambda^{2}}\) regardless of \(p\). Substituting this into Equation (13), we have:  
\[\mathbb{E}[e^{\lambda\mathbf{\hat{x}}_{d}^{(1)}}] =\left(\mathbb{E}_{\mathbf{p}}\left[\mathbf{p^{+}}e^{\lambda+(1 /2)\lambda^{2}}+\mathbf{p^{-}}e^{-\lambda+(1/2)\lambda^{2}}(1-\mathbf{p^{+} }-\mathbf{p^{-}})\right]\right)^{\hat{l}} \tag{13}\] \[\leqslant\left(\mathbb{E}[a]e^{\lambda}+\left(\mathbb{E}[\hat{a} ]e^{-\lambda}\right)^{\hat{l}}e^{(1/2)\lambda^{2}}\] \[\leqslant e^{(\lambda\lambda+\lambda^{2})\hat{l}},\]  
where \(a=\frac{1+\mathbf{p^{+}}-\mathbf{p^{-}}}{2}\), \(\quad\hat{a}=1-a=\frac{1-\mathbf{p^{+}}+\mathbf{p^{-}}}{2}\). The first inequality follows from the fact that \(2\leqslant e^{\lambda}+e^{-\lambda}\) for any \(\lambda\in\mathbb{R}\), and the second inequality follows from the fact that:  
\[be^{z}+(1-b)e^{-z}\leqslant e^{(2b-1)z+(1/2)z^{2}}, \tag{14}\]for any \(z\in\mathbb{R}\) and \(b\in[0,1]\) (cf. _[PERSON]_ 2008, Lemma A.1.5) [39]. From the \(k^{th}\) inductive hypothesis, we can assume that \(\mathbb{E}[e^{\lambda\mathbf{k}_{d}^{(k)}}]\leqslant e^{m_{k}\lambda+(1/2) \tilde{\sigma}_{k}^{2}\lambda^{2}}\) for \(|\lambda|\leqslant 1/(2m_{k-1}\hat{r})\). Now, we will show that \(\mathbb{E}[e^{\lambda\mathbf{k}_{d}^{(k+1)}}]\leqslant e^{m_{k+1}\lambda+(1/2 )\tilde{\sigma}_{k+1}^{2}\lambda^{2}}\) for \(|\lambda|\leqslant 1/(2m_{k}\hat{r})\). For any \(|\lambda|\leqslant 1/(2m_{k}\hat{r})\) and \(p\in[0,1]\), we get:  
\[\mathbb{E}[e^{\lambda\mathbf{y}_{F}^{(k)}}]\leqslant\Big{[}\Big{(}pe^{(1/2)m _{k}\lambda}+\tilde{p}e^{-(1/2)m_{k}\lambda}\Big{)}\Big{]}^{p}\cdot e^{(\frac{ D-2K}{2D})m_{k}\lambda+\frac{1}{2}(\frac{D-K}{D})^{2}\tilde{\sigma}_{k}^{2} \lambda^{2}}.\]  
Similar to (A3)'s process, from (A1), we get:  
\[\mathbb{E}[e^{\lambda\mathbf{k}_{d}^{(k+1)}}]\leqslant\mathbb{E}_{\mathbf{p} }\Big{(}a\mathbb{E}\Big{[}e^{\lambda\mathbf{y}_{F}^{(k)}}\Big{]}+a\mathbb{E} \Big{[}e^{-\lambda\mathbf{y}_{F}^{(k)}}\Big{]}\Big{)}^{\hat{l}}.\]  
with \(2\leqslant e^{\lambda}+e^{-\lambda}\) for any \(\lambda\in\mathbb{R}\).