## 3 From \(R_{10}(t)\) to \(R(t)\)
### Simulation Steps  
Since any conditional PDF is modeled as log-normal within each 10 min interval, we modeled the bivariate probability density function between two successive values \(R_{1,sim}(k)\) and \(R_{1,sim}(k+1)\), in the 10 min interval, as log-normal--we think this modeling can adequately describe at least the central part of the bivariate probability density--therefore, \(R_{1,sim}(k+1)\) can be estimated from \(R_{1,sim}(k)\) by considering the conditional log-normal PDF, as follows. Let \(x\ =\ \log(R_{1,sim}(k))\) and \(y\ =\ \log(R_{1,sim}(k+1))\); then, the conditional PDF has a mean value and standard deviation given by [10; 11; 12]:  
\[m_{y/x}\ =\ \ m_{y}+r\frac{s_{y}}{s_{x}}(x-m_{x}) \tag{2}\]  
\[s_{y/x}\ =\ s_{y}\sqrt{1-r^{2}} \tag{3}\] \begin{table}
\begin{tabular}{c c c c} \hline \hline \(R_{10}\) (mm/h) & Mean & Standard Deviation & Correlation Coefficient \\ \hline
0–2 & –0.60 & 0.75 & 0.94 \\ \hline
2–4 & 0.94 & 0.37 & 0.76 \\ \hline
4–6 & 1.51 & 0.41 & 0.70 \\ \hline
6–8 & 1.83 & 0.49 & 0.72 \\ \hline
8–10 & 2.07 & 0.52 & 0.68 \\ \hline
10–15 & 2.35 & 0.61 & 0.72 \\ \hline
15–20 & 2.62 & 0.76 & 0.71 \\ \hline
20–30 & 3.03 & 0.68 & 0.70 \\ \hline
30–40 & 3.32 & 0.77 & 0.75 \\ \hline \textgreater{}40 & 3.95 & 0.72 & 0.76 \\ \hline \hline \end{tabular}
\end{table} 
Table 2: Mean value, standard deviation of \(\log(R_{1})\), and correlation coefficient between two successive \(\log(R_{1})\) samples within the same 10 min interval for the indicated \(R_{10}\) ranges. The first range starts at 0.2 mm/h. Since within the same 10 min rain rate range (see Table 2), \(m_{x}\ =\ m_{y}\ =\ m\) and \(s_{x}\ =\ s_{y}\ =\ s\), Equations (2) and (3) become:  
\[m_{y/x}\ =\ (1-r)m+rx \tag{4}\]  
\[s_{y/x}\ =\ s\sqrt{1-r^{2}} \tag{5}\]  
Now, we can detail the simulation steps:  
1. Sample 1 of \(R_{1,sim}(t)\). According to sample 1 of \(R_{10}(t)\)--in a real application, this would be the first \(R_{10}\) value of the \(R_{10}(t)\) provided by meteorological institutes--the mean value and the standard deviation of the conditional PDF and the correlation coefficient are selected from Table 1.
2. A standard Gaussian random number \(X\) is generated and denormalized according to the values of step 1, according to the following relationships: \[X\ =\ \frac{x-m}{s}\] (6)  
\[x\ =\ sX+m\ =\ \log(R_{1,sim}) \tag{7}\]  
\[R_{1,sim}\ =\ \exp(sX+m) \tag{8}\]  
For example, let \(R_{10}\ =\ 12\) mm/h and \(X\ =\ -0.40\) (randomly generated). Then, (from Table 2) \(m\ =\ 2.35\), \(s\ =\ 0.61\), \(r\ =\ 0.72\); therefore, \(R_{1,sim}\ =\ \exp(0.61\ \times\ (-0.4)+2.35)\ =\ 8.21\) mm/h. 3. Samples 2 to 10. A standard Gaussian random number is generated for each sample as in step 2, but now it is denormalized according to Equations (4) and (5). Continuing the example of step 2: \(R_{1,sim}(k)\ =\ 8.21\) mm/h. Let \(X\ =\ 0.62\) (randomly generated); then, \(m_{y/x}\ =\ (1-0.72)\ \times\ 2.35+0.72\ \times\ \log(8.21)\ =\ 2.17\) and \(s_{y/x}\ =\ 0.61\ \times\ \sqrt{1-0.72^{2}}\ =\ 0.42\). Therefore, \(R_{1,sim}(k+1)\ =\exp(0.42\ \times\ 0.62+2.17)\ =\ 11.36\) mm/h. Since the next sample depends only on the previous one, the simulation process is a first-order Markov process [13]. 4. The 1 min rain rate sample \(R_{1,sim}(k)\) is scaled to conserve the quantity of water \(Q_{T}\) of the original \(R_{10}\) sample, the latter given by: \[Q_{T}\ =\ \sum_{k\ =\ 1}^{10}R_{1}(k)=\] (9) Now, the quantity of water in the simulation is given by: \[Q_{T,sim}\ =\ \sum_{k\ =\ 1}^{10}R_{1,sim}(k)\] (10)  
Since, in general, \(Q_{T,sim}\
eq Q_{T}\), to conserve the quantity of water, each 1 min sample must be scaled to obtain the rain rate time series \(R_{sim}(t)\):  
\[R_{sim}(t)\ =\ \frac{Q_{T}}{Q_{T,sim}}R_{1,sim}(t) \tag{11}\]
5. The simulation process is repeated by considering the successive value \(R_{10}\) until the last sample of the rain event.