0
$\begingroup$

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 $\{x_t\}_{t = 1}^k$.

$\endgroup$
2
$\begingroup$

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 $C_0 = 1$

$\endgroup$
0
$\begingroup$

Consider $$ \frac{x_{t+1}}{c_{t+1}}-\frac{x_{t}}{c_{t}}=-\alpha x_t $$ Can you take it from here?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.