Dynamic programming, optimal consumption-savings (finite horizon) problem Let $w_t$ denote a consumer's wealth at time $t$ and $c_t$,
the amount she chooses to consume, so her savings exiting this time period are $w_t-c_t$. Given this savings decision, her savings $w_{t+1}$ at time $t+1$ are determined by a random process in which $w_{t+1}=\alpha w_t$ or $w_{t+1}=\beta w_t$, where $\alpha,\beta$ are postitive constants, each with probability $\dfrac{1}{2}$ and independent of past rates of return on her savings. At time $t$, when she is choosing $c_t$, she knows $w_t$ but she doesn’t know
anything about future returns except the probabilistic law just given; she does know her past consumption decisions, of course, and past rates of return. Considering a log utility, i.e. $u(t)=ln(c_t)$ and assuming that the consumer is impatient, so that he discounts future utility by a factor $b$ each period, where $0 &lt; b &lt; 1$, then the consumer's decision problem can be written as follows:
$$v(c_t)=\max_{c_t \geqslant 0}\sum_{t=0}^{T}b^{t}u(c_t)$$

If the consumer can not borrow, then what is the optimal consumption level for the consumer?

$\underline{Note:}$ The problem is based on David M. Kreps' microeconomic theory book, but it is adjusted to be a finite horizon problem. Kreps, in his book, solves this problem in a fuzzy way for $T=3$, that is not obvious to me. However, due to the fact that I know little of dynamic programming, since I am in the begining of examining this topic, I would appreciate it, if someone could provide a solution in the finite horizon. If the problem is not set in a rigorous way, I would be glad too to see somene making any appropriate change. I beleive that it is a classic problem in the field of economics.
Your value function is as follows:
$$
V_t[w] = \max_{c_t \in[0,w]} \left\{u(c_t) + \frac{1}{2}V_{t+1}[\alpha(w_t - c_t)] + \frac{1}{2}V_{t+1}[\beta(w_t-c_t)] \right\}
$$
with the terminal condition
$$
V_{T}[w_T] = \max_{c_T \in [0,w_T]} u(c_T)
$$

So, we can solve this via backward induction. Clearly, at the final period $T$, since $u$ is monotonic, we consume everything, so $V_T[w_T] = u(w_T) = \ln(w_T)$. 

Let us now consider one period previous, so period $T-1$. The value function is:
$$
V_{T-1}[w] = \max_{c_{T-1} \in [0,w_{T-1}]} \left\{u(c_{T-1}) + \frac{1}{2}V_{T}[\alpha(w_{T-1} - c_{T-1})] + \frac{1}{2}V_{T}[\beta(w_{T-1}-c_{T-1})] \right\}
$$
We already know what $V_T[\cdot]$ is, so substituting,
$$
V_{T-1}[w] = \max_{c_{T-1} \in [0,w_{T-1}]} \left\{u(c_{T-1}) + \frac{1}{2}u(\alpha(w_{T-1} - c_{T-1})) + \frac{1}{2}u(\beta(w_{T-1}-c_{T-1})) \right\}
$$
considering the case $u(\cdot) = \ln(\cdot)$,
$$
V_{T-1}[w] = \max_{c_{T-1} \in [0,w_{T-1}]} \left\{\ln(c_{T-1}) + \frac{1}{2}\ln(\alpha(w_{T-1} - c_{T-1})) + \frac{1}{2}\ln(\beta(w_{T-1}-c_{T-1})) \right\}
$$
Take first order conditions, and we can get optimal $c_{T-1}^*(w_{T-1})$.

Then we have solved what $V_{T-1}[\cdot]$ is! We can then do the same procedure by considering $V_{T-2}$. Repeat until we get to $V_0$.

EDIT to reflect comment.
You now know that $c^*_{T−1}(w_{−1})=\frac{\alpha\beta}{1 + \alpha \beta}w_{T-1}$.
Plugging back into $V_{T-1}$, we have solved for the value of $V_{T-1}$,
\begin{align}
V_{T-1}[w] &amp;= \ln(\frac{\alpha\beta}{1 + \alpha \beta}w_{T-1}) + \frac{1}{2}\ln(\alpha(w_{T-1} - \frac{\alpha\beta}{1 + \alpha \beta}w_{T-1})) + \frac{1}{2}\ln(\beta(w_{T-1}-\frac{\alpha\beta}{1 + \alpha \beta}w_{T-1})) \\
&amp;=\ln(\frac{\alpha\beta}{1 + \alpha \beta}w_{T-1}) + \frac{1}{2}\ln(\alpha(\frac{w_{T-1}}{1 + \alpha \beta})) + \frac{1}{2}\ln(\beta(\frac{w_{T-1}}{1 + \alpha \beta}))
\end{align}

Finally, lets now go to $T-2$. The value function is
$$
V_{T-2}[w] = \max_{c_{T-2} \in[0,w_{T-2}]} \left\{\ln(c_{T-2}) + \frac{1}{2}V_{T-1}[\alpha(w_{T-2} - c_{T-2})] + \frac{1}{2}V_{T-1}[\beta(w_{T-2}-c_{T-2})] \right\}
$$
We just solved what $V_{T-1}$ is! Plug it in, and repeat.

This expression will probably blow up quite quickly, so it will be a pain to solve analytically.