# Help with summation stochastic

I don't understand how to get from the first line to the last. What happens to the summation? Please help! An important piece of information given in the question is that the transition probability from state 0 to state 1 is equal to 1. N.B. o(h) is just an error term that goes to 0 when divided by h and a limit as h->0 is taken.

Thanks for the help!

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This is not my area, but it looks to me like we need to know how $\xi_k$ is defined to show that this is true. –  Alex Becker Mar 19 '11 at 4:30
ξk is a two state markov chain (0, 1) with transition probability matrix 0->0 = 1 , 0->1 = 1, 1->0 = 1 - alpha, and 1->1 = alpha –  icobes Mar 19 '11 at 4:34
you cannot have both 0->0 = 1 and 0->1 = 1. –  Henry Mar 19 '11 at 12:27

I think this is equal to $$(e^{\lambda h}-1)(1- \lambda h+o(h)) \ \ \ (1)$$

$$= e^{\lambda h}- \lambda h e^{\lambda h} + o(h) e^{\lambda h}-1+ \lambda h- o(h) \ \ \ (2)$$

$$= e^{\lambda h}- \lambda h e^{\lambda h} + o(h) e^{\lambda h}-1+ \lambda h- o(h) \ \ \ (3)$$

Also we probably use the fact that the probabilities add up to $1$. And we know that $P(\xi_1 = 1| \xi_0 = 0) = 1$. This simplifies to $\lambda h+o(h)$ probably because we can combine terms to get one $o(h)$ term. Put $x = \lambda h$. So (3) is equal to:

$$= \underbrace{e^{x}- xe^{x} -1}_{o(h) \ \text{term}}+ o(h) e^{x}+ \lambda h- o(h) \ \ \ (3)$$

$$= \lambda h + o(h)$$

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I don't follow how the last line you wrote equals the last line I wrote. Thanks! –  icobes Mar 19 '11 at 5:33