We have Markov Chain with continuous time, three conditiions and generator: \begin{equation*} Q = \begin{bmatrix} -3 & 1 & 2\\ 1 & -1 & 0\\ 1 & 0 & -1 \end{bmatrix} \end{equation*}
Initial distribution $X_0=1$ almost sure. How to find the distribution of the second moment of the chain jump?
Recall the Kolmogorov forward and backward equations which govern the transient behavior of the process. In particular, we have $$ P'(t) = QP(t), $$ where $P_{ij}(t) = (\mathbb P(X(t) = j\mid X(0)=i)$. The solution to this differential equation is the matrix exponential $$P(t) = \sum_{n=0}^\infty \frac{(tQ)^n}{n!}. $$ In this case, one can use diagonalization to compute powers of $Q$, and we find that $$ P(t) = \left( \begin{array}{ccc} \frac{3 e^{-4 t}}{4}+\frac{1}{4} & \frac{1}{4}-\frac{e^{-4 t}}{4} & \frac{1}{2}-\frac{e^{-4 t}}{2} \\ \frac{1}{4}-\frac{e^{-4 t}}{4} & \frac{1}{12} \left(e^{-4 t}+8 e^{-t}+3\right) & \frac{1}{6} \left(e^{-4 t}-4 e^{-t}+3\right) \\ \frac{1}{4}-\frac{e^{-4 t}}{4} & \frac{1}{12} \left(e^{-4 t}-4 e^{-t}+3\right) & \frac{1}{6} \left(e^{-4 t}+2 e^{-t}+3\right) \\ \end{array} \right). $$ Since $\mathbb P(X(0)=1)=1$, the distribution of the first jump is given by the minimum of two exponentially distributed random variables - namely, it is exponential with rate $3$. To analyze the distribution of the second jump, we must condition on the first jump. Let $\{J_n:n=1,2,\ldots\}$ denote the jump chain. Then $J_2$ is the sum of $J_1$ and a hyperexponential distribution: \begin{align} J_2 = J_1 + Y \end{align} where $$ f_Y(t) = \frac13 e^{-t}+\frac23 e^{-t} = e^{-t}, $$ which in this case happens to be exponential as well. We compute the density of $J_2$ by convolution: \begin{align} f_{J_2}(t) &= f_{J_1}\star f_Y(t)\\ &= \int_0^t f_{J_1}(\tau) f_Y(t-\tau)\ \mathsf d\tau\\ &= \int_0^t 3e^{-3\tau} e^{-(t-\tau)}\ \mathsf d\tau\\ &= \frac32 e^{-3t}(e^{2t}-1). \end{align}