# pure death process

A chemical solution contains N molecules of type A and M molecules of type B. An irreversible reaction occurs between type A and type B molecules in which they bond to form a new compound AB. Suppose that in any small time interval of length h, any particular unbounded A molecule will react to any particular unbounded B molecule with probability theta * h + o(h) where theta is a reaction rate. Let X(t) denote the number of unbounded A molecules at time t. Model X(t) as a pure death process by specifying parameters.

The answer is k * ( M - (N-k) ) * theta for k = 0, 1, 2, ... , N

I am unsure about what the expression k * ( M - (N-k) ) represents and would appreciate if someone could explain the rationale behind it to me. The way I approached this question was I took k to represent the number of A molecules remaining. If we want P (X(t) = k), it is equivalent to saying N - k molecules died. If you subtract that from M, that is the remaining number of B molecules remaining to react with. Thus, multiplying that by theta should give you the rate. However, the solution does not match my rationale and adds in a k.

Thanks for the help!

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 The fewer unreacted molecules of A, the more unlikely the next reaction will take place. However, I do not see how this statement leads to the solution? – icobes Mar 24 '11 at 23:30 Perhaps there might be a proportionality to $k$ as there is to $M-(N-k)$? – Henry Mar 24 '11 at 23:56 I am assuming it is a linear death process and that is why you multiply by k. Is that statement correct? How would I go about identifying something as a pure death process where the death rate is just mew vs a linear death process where the death rate is n*mew? I can't make the distinction between the two. Is there anything in the question that might point to it being the case? – icobes Mar 25 '11 at 0:02 In the question you have the phrase ... any particular unbounded A molecule will react to any particular unbounded B molecule ... and you need to deal the particular twice, once for A and once for B, so you need to multiply by the remaining number of As and the remaining number of Bs. – Henry Mar 25 '11 at 6:16