Dynamic programming, optimal consumption-savings (finite horizon) problem

Question

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 $0 < b < 1$, 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.

Answer 1

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 $c^*_{T−1}(w_{𝑇−1})=\frac{\alpha\beta}{1 + \alpha \beta}w_{T-1}$. Plugging back into $V_{T-1}$, we have solved for the value of $V_{T-1}$, \begin{align} V_{T-1}[w] &= \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})) \\ &=\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.

Answer 2

I still can't comment, so this is not another answer it is adding some math to the given answer Since

$ Log X*Y = log X + log Y $

$ Log(X ^ i) = i * log X $

So substituting

$ Ln Z +1/2 Ln X + 1/2 Ln Y = Ln (z * Srqt(X*Y)) $

will simplify the expression a lot, but I have to reduce it in handwritten form because I'm writing from a phone now

You will reach

$ max [ w*(w-c)] $

which clearly happens at

$ c=w/2 $

(try any values)

2- Also, are you sure of ur use of the term "dynamic programming"???

3- And I don't understand from where came the term

$ [ Alpha*Beta / (Alpha*Beta +1)] $ , that's why I substituted twice (in the original formula before this reduction too)

4- Finally, I don't see why didn't u substitute $ V(t) = u(t) $ from the beginning to simplify things? I simplified ur formulas mathematically, but I don't get how u reached the EQ (I admit I'm not specialized in ur field I came for an interdisciplinary problems). I get that

$ Saving(t) = Return_of_Saving(t-1) + w(t)- c(t) $ Then u say her expected wealth "w" at time t+1 is an average of two values. So
$ W(t+1)= (alpha + beta) /2 * W(t) + Income(t+1) - C(t+1) $ - that u have to add a factor for the decrease of money value & the increase of prices with time, that's ur "b" in the maximization formula right?

Anyways, if u don't care to explain just be informed of these logarithmic relations & use them to simplify