Then, taking into account that l = ∫ τ ( n ) 0 d (Δ t ) v ( n $^{)}$(Δ t ), with the help of Eqs. (2.5) and (2.6) we find Since in the case under consideration κτ ( n ) ≫ 1, the exponential term in Eq. (2.6) can be neglected yielding v ( n ) + = ( f + g ( n $^{)}$) /ν . According to this approximation, the particle velocity at the end of the n th interval is determined by the random force on this interval. Therefore, using the continuity condition for the particle velocity, v ( n ) − = v ( n − 1) + , we obtain v ( n ) − = ( f + g ( n − $^{1)}$) /ν . Substituting these expressions for v ( n ) + and v ( n ) − into Eq. (2.7), we arrive to the following result: It shows that in the weakly underdamped regime the waiting time τ ( n ) depends not only on the random force g ( n $^{)}$, as in the overdamped case, but also on the random force g ( n − $^{1)}$. Since these forces are statistically independent, the probability density of the waiting time can be written in the form if τ ∈ [ τ$_{min}$, τ$_{max}$ ], otherwise it equals zero. It is not difficult to verify that in the overdamped case (when κ = ∞ ) Eq. (2.9) reduces to Eq. (2.3). Next, we express the first two moments of the random variable Y$_{t}$ through the waiting time probability density p ( τ ). Since the moments of N ( t ) are known from the CTRW theory (see, e.g., Ref. [16]), we reproduce here only the main results related to our situation. Introducing the probability P ( n, t ) that N ( t ) = n , we define the k th moment of the particle position Y$_{t}$ in the usual way: ( n ≥ 1), we obtain k l = f + g ( n ) ν τ ( n ) − v ( n ) + − v ( n ) − κ . (2.7) τ ( n ) = κνl + g ( n ) − g ( n − 1) κ ( f + g ( n $^{)}$) . (2.8) p ( τ ) = ∫ g$_{0}$ − g$_{0}$ ∫ g$_{0}$ − g$_{0}$ dgdg $^{′}$u ( g ) u ( g $^{′}$) δ ( τ − κνl + g − g ′ κ ( f + g ) ) (2.9) P ( n, t ) = ∫ t 0 dτ p ( τ ) P ( n − 1 , t − τ ) (2.11) 〈 Y k t 〉 = l k ∞ ∑ n =1 n $^{k}$P ( n, t ) (2.10) ( k = 1 , 2 , . . . ). Then, using the Laplace transform of a function h ( t ), h$_{s}$ = L{ h ( t ) } = ∫ ∞ 0 dt e − $^{st}$h ( t ) (Re s > 0), and taking into account that P (0 , t ) = ∫ ∞ t dτ p ( τ ) and 〈 Y k t 〉 s = l $_{k}$1 − p$_{s}$ s ( p$_{s}$ d dp$_{s}$ ) k 1 1 − p$_{s}$ . (2.12) 6