Let $\{c_t\}_{t = 1}^k$ be a (non-monotone) sequence of real numbers such that $c_t \in (0, 1]$ for all $t = 1, \dots, k$. Consider the recursive sequence $$ \left \{ \begin{array}{ll} x_1 & = c_1 & \\ x_{t + 1} & = c_{k + 1} \left (1 - \alpha \sum_{j = 1}^t x_{j} \right ) & \mbox{for all } 1 \leq t < k\\ \end{array} \right . $$ with $\alpha > 0$ a constant indipendent of $t$. Find a closed-form formula for the sequence $\{x_t\}_{t = 1}^k$.
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2 Answers
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As
$$ \frac{x_{k+1}}{c_{k+1}} = 1-\alpha\sum_{j=1}^{j=k}x_j $$
calling $y_k = \frac{x_k}{c_k}$ we have
$$ y_{k+1}-y_k = -\alpha x_k = -\alpha c_k y_k $$
so we have
$$ y_{k+1}+(\alpha c_k - 1)y_k = 0 $$
this is a recurrence linear equation with solution
$$ y_k = C_0 \prod_{j=1}^{j=k-1}(1-\alpha c_j) $$
and finally
$$ x_k = C_0 c_k\prod_{j=1}^{j=k-1}(1-\alpha c_j) $$
but as $x_{1} = c_1$ follows $C_0 = 1$
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Consider $$ \frac{x_{t+1}}{c_{t+1}}-\frac{x_{t}}{c_{t}}=-\alpha x_t $$ Can you take it from here?
