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\section   We start by applying the \textbf{P(E)-Theory} to the case of a quantum dot which is unequally coupled to a resistive enviroment. For such calculation we utilize the following expression  \begin{equation}  \Gamma_{i\rightarrow j}^{l}= 2\pi \int |t_{l}|^2 \rho(\epsilon_i-\mu_i) \bar\rho(\epsilon_j-\mu_j)P_{l}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j\,  \end{equation}  Let us consider a processes in which we fill up a quantum dot with one electron. In this case we need to compute $\Gamma_{0\right $\Gamma_{0\rightarrow  1}^{1}$. We denote the leads with labels $l=\{1,2\}$ for left and right contacts. The dot occupation in this case change from $n=0$ to $n=1$. Therefore $\rho(\epsilon_i-\epsilon_j)$ $\rho(\epsilon_i-\mu_i)=f(\epsilon_i-\mu_1)$ and $\bar\rho(\epsilon_j-\mu_j)=\delta(\epsilon_j-\mu_d)$  \begin{equation}  \Gamma_{0\rightarrow 1}^{1}= 2\pi \int |t_{1}|^2 f(\epsilon_i-\mu_1) \delta(\epsilon_j-\mu_d)P_{1}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j  \end{equation}  On the other hand when the dot becomes empty, an electron in the dot leave it through barrier $2$, the transition rate is   \begin{equation}  \Gamma_{2\rightarrow 0}^{1}= 2\pi \int |t_{2}|^2 \rho(\epsilon_i-\mu_i) \bar\rho(\epsilon_j-\mu_j) P_{2}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j  \end{equation}  where now $\rho(\epsilon_i-\mu_i)=\delta(\epsilon_i-\mu_d)$, and $\bar\rho(\epsilon_j-\mu_j)=[1-f(\epsilon_j-\mu_2)]$. The transition rate read  \begin{equation}  \Gamma_{2\rightarrow 0}^{1}= 2\pi \int |t_{2}|^2 [1-f(\epsilon_j-\mu_2)] \delta(\epsilon_i-\mu_d) P_{2}(\epsilon_i-\epsilon_j)d\epsilon_i d\epsilon_j  \end{equation}