sabri said:
could you please check in this pdf if what i have done is correct or not
hi, I did a whole long write up, hit submit and then got a message that I've been logged out... after I logged back in it to me to a blank page and I lost my whole response.

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back to the matter at hand. I think it is pretty good. One thing in particular is to consider your general case
$q = \sum_{i=0}^n p_I q^i$
you can think about this as $E\big[q^{X_1}\big] $ for $q \in [0,1]$ and you want a fixed point where $E\big[q^{X_1}\big] = q$
or it can be convenient to instead think of the function given by $r(q) = E\big[q^{X_1}\big] - q$.
I'd suggest sketching out $r(q)$. In particular $r(1) = 0$ as you've observed. Note that $r'(1) = E\big[X_1\big]-1$ so this is a positive slope when the mean is $\gt 1$.
When $q \to 0^+$ you have $ r(q) \to p_0 \gt 0$ which suggests the existence of another root $q^* \in (0,1)$. you can differentiate twice to discover that $r''(q) \gt 0$ so it is a convex function and only has those two roots. (Given the slope at $q=1$ in the case of a mean $\leq 1$ you have a linear lower bound of that tangent line saying that there can never be another root $\in (0,1)$)
I'd strongly suggest sketching this out on paper and on the whiteboard -- the curvature makes for a nice picture and discussion with a class...
I had a whole writeup on this fact, but a possible suggestion -- it may be worth trying to prove that $q^*$ is a fixed point of $E\big[q^{X_1}\big]$ implies that it is a fixed point for $E\big[q^{X_n}\big]$ i.e. for all generations... this implies that $q^*$ is an upper bound on total probability of extinction at time $n$ (why? hint: at any generation we have a linear combination involving strictly positive terms that sum to this fixed point value). The fact that $q^* \lt 1$ is an upper bound then gives you an excuse for throwing out the junk root of $1$ in the case of $E\big[X_1\big] \gt 1$. (It also has nice monotone convergence properties in that your cumulative probability of extinction is a monotone non-decreasing sequence bounded above by $q^*$ so the limit $L$ exists and is at most $q^*$, but the limitting value must obey the 'first step analysis' you did and hence it must be the case that $L = q^*$. )
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double check your (c) at the very end -- it seems like you didn't take the complementary probability correctly...