# Pure Birth Process Probabilities

I have no clue how to proceed with this question. Please help me in deriving the differential equations! Thanks very much... I really appreciate it :)

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$$P_0 (t + h) = {\rm P}(N(t + h)\;{\rm is} \; {\rm even}|N(t)\;{\rm is} \; {\rm even})P_0 (t) + {\rm P}(N(t + h)\;{\rm is} \; {\rm even}|N(t)\;{\rm is} \; {\rm odd})P_1 (t).$$ Put $\Delta_t(h) = N(t+h)-N(t)$. Then, as $h \downarrow 0$, $$P_0 (t + h) = {\rm P}(\Delta_t(h)=0|N(t)\;{\rm is} \; {\rm even})P_0 (t) + {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm odd})P_1 (t) + o(h).$$ Hence, $$P_0 (t + h) = [1 - {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm even})]P_0 (t) + {\rm P}(\Delta_t(h)=1|N(t)\;{\rm is} \; {\rm odd})P_1 (t) + o(h),$$ leading to $$P_0 (t + h) = (1-\beta h)P_0 (t) + \alpha h P_1 (t) + o(h).$$ Rearranging gives $$\frac{{P_0 (t + h) - P_0 (t)}}{h} = \alpha P_1 (t) - \beta P_0 (t) + \frac{{o(h)}}{h}.$$ Letting $h \downarrow 0$ thus gives $$P'_0 (t) = \alpha P_1 (t) - \beta P_0 (t) .$$ Similarly, $$P'_1 (t) = \beta P_0 (t) - \alpha P_1 (t) ,$$ which also follows trivially from $P_0 (t) + P_1 (t) = 1$.