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given a Poisson process with lambda = 1.4 for 0<=t<2 and lambda = 0.6 for 2<=t<10 if we only focus on the first arrival, what is the distribution of arrival time (X1)? is it still exponential? and is it possible to express the cdf in terms of a closed form function so that we can use the inverse cdf method to simulate it?

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We compute for $0\leqslant t<10$ $$ \mathbb P(X_1\leqslant t, X_1< 2) = \int_0^{t\wedge 2} \frac75 e^{-\frac75 s}\ \mathsf ds = 1 - e^{-\frac75 (t\wedge 2)} $$ where $\wedge$ denotes $\min$, and \begin{align} \mathbb P(X_1\leqslant t, X_1\geqslant 2) &= \mathbb P(X_1\leqslant t\mid X_1\geqslant 2)\mathbb P(X_1\geqslant 2)\\ &= \left(\int_2^t \frac35 e^{-\frac35 s}\ \mathsf ds\right)e^{-\frac{14}5}\\ &= e^{-4} - e^{-\frac15(14-3t)}. \end{align} It follows that for $0\leqslant t<10$, \begin{align} \mathbb P(X_1\leqslant t) &= \mathbb P(X_1\leqslant t, X_1< 2) + \mathbb P(X_1\leqslant t, X_1\geqslant 2)\\ &= 1 - e^{-\frac75 (t\wedge 2)} + e^{-4} - e^{-\frac15(14+3t)} \end{align} Since the arrival rate is zero after $t=10$, we have $$ \mathbb P(X_1=+\infty) = 1 - \mathbb P(X_1\leqslant 10) = e^{-\frac{14}5} - e^{-4} + e^{-\frac{44}5}. $$

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