13
Since $a + b=1$ the equations are exactly the same. Substituting in for $a+b$ with $1$ in the third and fourth equations gives the first and second equations.
11
The expression in question is in footnote $11$ of the referenced article. Reading the paper, we see that the decision variable here is "the payout rate", which is the reciprocal of $P$. So equivalently, we can solve the maximization problem with respect to $P$ (and not w.r.t. $Q$). More over, "price elasticity of demand" involves the derivative of $Q$ with ...
9
For continuous-time stochastic dynamic programming, the small, nontechnical Art of Smooth Pasting by Dixit is a wonderful option. It does a very effective job of conveying the basic intuition.
Stokey's more recent The Economics of Inaction is also decent, but for a practical-minded person it probably underperforms Dixit - its much greater length and ...
7
As @user32416 pointed out the first order stationarity conditions are not enough. Specifically it seems that you violate Slater's condition, which states that "the feasible region must have an interior point". There are no $x,y$ for which
$$(x+y-2)^2 < 0.$$
If you rephrase the problem to
$$\max (xy)$$
$$x+y-2 = 0$$
$$x,y \geq 0$$
Slater's condition is ...
6
Dynamic Programming & Optimal Control by Bertsekas
Introduction to Modern Economic Growth by Acemoglu
The Acemoglu book, even though it specializes in growth theory, does a very good job presenting continuous time dynamic programming.
6
They are related and usually fall into the same discussion, but as @Alecos mentions in the comments, the two theorems show different things.
I suppose the connection that you're after is the fact that if
the derivative
$$
\left .\frac{\partial f(x, a)}{\partial a} \right |_{x=x(a)}
$$
exists, then because differentiability implies continuity, you might be ...
6
There is not a single answer, it will depend on the particulars of each problem. Let's look at a standard example.
Consider the benchmark intertemporal optimization problem for the Ramsey model
$$\begin{align}
&\max_u \int^{\infty}_0{e^{-\rho t}u(c)dt}\\
\\
& \text{s.t.}\;\; \dot{k} = i-\delta k\\
& \text{s.t.}\;\; y = f(k)=c+i
\end{align}$$
...
6
This is how you get from your first equation to your second.
your utility function is $u(x_1, x_2)=x_1^a x_2^b$
since $a+b=1$ I'll change it slightly to a and (1-a)
In order to optimise these two choices, you need to maximise utility, wrt your choice variables.
subject to $p_1x_1 + p_2x_2 = w$
using Walras Law. Basically, in order to optimise utility, all ...
6
This is an ill-posed question. Even without going through KKT, your constraint $(x + y - 2)^2 \le 0$, since the left-hand side is a square, means that the only solution that is feasible is the one where the equality binds; i.e. $(x + y - 2)^2 = 0$, or that $|x + y - 2| = 0$ -- of which what you say that $x = y = 1$ is a solution.
5
You are missing a $\min$ operator just before the bracket. The utility maximization problem is as follows, $$\max \ \min [\alpha x_1, ..., \gamma x_3] \\ \ \ \text{such that} \ \ \lambda_1x_1 + ... + \lambda_3x_3 = M$$
Consider the case of two goods with utility $u$ given by $u(x) = \min[\alpha x_1, \beta x_2]$. At the optimum, what do you know about the ...
5
All you need for this particular question is the following. Let $\mathbf{X}$ be a $T \times K$ matrix, $\mathbf{w}$ a K-dimensional vector and $\mathbf{y}$ a T-dimensional vector, then
$$
\begin{eqnarray*}
\frac{\partial \mathbf{w}^{\prime}\mathbf{X}^{\prime}\mathbf{y}}{\partial \mathbf{w}}&=&
\mathbf{X}^{\prime}\mathbf{y}\\
\frac{\partial \mathbf{w}...
5
I think Kamien and Schwartz's Dynamic Optimization: The Calculus of Variations and Optimal Control in Economics and Management is pretty well known.
5
The first order condition of the maximization problem is
\begin{equation}
f'(x)-s=0\iff f'(x)=s
\end{equation}
We can then replace $x$ by $x(s)$ because this is the optimal value given $s$. Since this is true for every $s$, we can differentiate with respect to $s$ which yields
\begin{equation}
f''(x(s))x'(s)=1
\end{equation}
Which can be rewrite as
\...
5
You are unfortunately mistaken. DSGE models are at the heart of monetary policy and the most widely used class of models in this field. To work in monetary there is no real way around learning DSGE.
A very good book to get started. is Walsh (2010) "Monetary Theory and Policy". I can also recommend Gali's book ("Monetary Policy, Inflation, and the Business ...
5
Yes, Bachir, Fabre & Tapia-Garcia extend the Karush-Kuhn-Tucker theorem under mild hypotheses, for an infinite number of variables.
I give hereafter a weaker version of the generalization of Karush-Kuh-Tucker in infinite dimension:
Let $X\subset\mathbb{R}^{\mathbb{N}}$ be a nonempty convex subset of
$\mathbb{R}^{\mathbb{N}}$ and let $x^{*}\in Int\...
5
It is possible to have
$$g(x^*) = b\; {\rm and}\; \lambda^* = 0$$.
When the multiplier is zero and the constraint is equal to zero, then
a) The constraint does not really "bind"
b) That's why the multiplier is zero.
What does it mean "the constraint does not really bind"?
It means that the solution $x^*$, that makes $g(x^*) = b$, would be ...
4
Controlled Markov Processes and Viscosity Solutions by Fleming and Soner includes a number of applications to Finance and Differential Games.
4
A first price standard and reverse auction are formally equivalent to each other, and the same method can be used to solve both:
First Price Auction
In a first price auction, $n$ bidders choose their bid, $b_i$, as a function of their value $v_i$ (distributed according to $F$. They seek to maximise their expected payoff:
$$[v_i-b_i(v_i)]\Pr(b_i\geq\max_j ...
4
Quoting the OP from a comment
What sort of conditions on utility function and constraint enables us
to apply envelope theorem only after we established the continuity of
value function by Berge's theorem?
people.hss.caltech.edu/~pbs/expfinance/Readings/Lucas1978.pdf
In the referenced Lucas (1978) paper, Proposition 1 establishes that
where $v(z,y;...
4
These are proper Marshallian demand functions, even though Income does not appear in them. This is due to specific form of the utility function (and the candidate solution of all goods being purchased at strictly positive quantities). It emerges that there is no income effect for goods $x_1$ and $x_2$ - optimal uncompensated demand does not depend after all ...
4
Some confusion and incorrect statements in the answers already given, including the "accepted" answer. (E.g. obvious distinctions between necessity and sufficiency of different conditions for KKT are mixed up.)
There are different formulations of KKT theorem.
The KKT Theorems one might consider for your example are the following.
For notational simplicity,...
4
As mentioned in the other answer, the Lagrange multiplier is the marginal effect on the value (optimized) function, when the constrained is "relaxed" marginally. In your case then it should be interpreted "how much studying time changes as the required average expected grade..." increases?
Well, this is a good example to show that how we set up the ...
4
Non-negativity constraints have nothing special and are not critical for the general validity of the Karush-Kuhn-Tucker approach. First, realize that we could have $x \geq a >0$ and then we could write $x-a \geq 0$ and view this "non-negativity" constraint as just one more inequality constraint on the solution.
When a decision variable is weakly bounded, ...
4
Questions with numbers are usually not as good as questions without numbers. If you had written down the formula for CV and EV you would probably have noticed that your premise is false.
CV and EV are not supposed to have opposing signs. You can see this from their definitions where
$$
CV = e(p_1,u_1) - e(p_1,u_0), \hskip 20pt EV = e(p_0,u_1) - e(p_0,u_0).
$...
4
You are right. The problem here is a corner solution.
Let us define the optimal combination $X^* = \beta_1e_1^* + \beta_2e_2^*$.
The first derivative gives you
$$X^* = \dfrac{l_1}{2\beta_1} $$
whereas the second gives you
$$X^* = \dfrac{l_2}{2\beta_2}$$
They can only be true "by chance". This is, the existence of an interior solution cannot be ...
4
Here are two methods. First method: the substitution can be made by inverting $f$. Since $f$ is strictly increasing and continuous, $f^{-1}$ is well-defined. The constraint can therefore be written
\begin{equation*}
x = f^{-1}\Big(\dfrac{c-(1-P)f(y)}{P}\Big)
\end{equation*}
The objective becomes
\begin{equation*}
\max_{y \geq 0}{P \Big[a-f^{-1}\Big(\dfrac{c-...
4
Let $g$ be the inverse function of $f$ defined over range of $f$. Notice that $g$ is increasing and strictly convex. We can rewrite the maximization problem as:
\begin{eqnarray*} \max\limits_{u\geq 0, \ v \geq 0} & P(a - g(u)) + (1-P)(b-g(v)) \\ \text{s.t.} & Pu + (1-P)v = c\end{eqnarray*}
where $u=f(x)$ and $v = f(y)$.
Solving above is equivalent ...
4
What the book calls the "Weierstrass Theorem" is more commonly known as the Extreme Value Theorem, which states that a continuous function defined on a compact domain attains a maximum and a minimum.
A common proof of this theorem involves the use of the Bolzano–Weierstrass theorem, which you learned in your math course, and which says every bounded ...
4
Your intuition is correct. Say you know that $Z=X\cdot Y=0$ You don't know if $X=0$ or $Y=0$ or both are zero. Even if you know that $X=0$ you have no idea if $Y=0$, $Y<0$ , or $Y>0$.
Consider the potentially satiated utility function:
$$ \max_{X,Y} U(X,Y) = min(X+Y, 5)$$
$$ S.T. \:p_x X + p_y Y + p_z Z\leq M$$
Assume, for simplicity, that $p_x = p_y =...
4
Because you are talking about constraints, it appears you do not consider the case of inserting such "access costs" (because this is what they are) in the utility function. It implies that they do not create disutility directly, only direct or indirect monetary costs.
Let $d$ be the distance in some units. Let $C_d = c_dd$ be a linear (for ...
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