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The Statement of the Problem:

For fixed $r \gt 0$ and $m$, find the maximum value of $1-rx^2-y^2$ on the constraint set $x+y=m$. Find the value function $f^*(r,m)$ and compute $\frac{\partial f^*}{\partial r}$ and $\frac{\partial f^*}{\partial m}$ and confirm the envelope theorem in this case. How does the maximum value change if we replace $m$ with $m+\epsilon$ for small $\epsilon$?

Where I Am:

I set up my first order Lagrange multiplier conditions and found that $x=\frac{m}{1+r}$ and $y= \frac{m}{\frac{1}{r}+1}$ giving me my maximum value:

$$ f\left(\frac{m}{1+r},\frac{m}{\frac{1}{r}+1}\right)=1-r\left(\frac{m}{1+r}\right)^2-\left(\frac{m}{\frac{1}{r}+1} \right)^2 $$

and the value function

$$ f^*(r,m)=1-r\left(\frac{m}{1+r}\right)^2-\left(\frac{m}{\frac{1}{r}+1} \right)^2 $$

Taking partial derivatives, gives me:

$$ \frac{\partial f^*}{\partial r}=\frac{-m^2}{(1+r)^2} \text{ and } \frac{\partial f^*}{\partial m}=\frac{-2mr}{1+r}$$

Now, the problem asks me to "confirm the envelope theorem in this case." In my book, I have the following equality for the statement of the envelope theorem:

$$ \frac{\partial f^*(\mathbf{\bar r})}{\partial r_j}=\left[\frac{\partial f(\mathbf x , \mathbf r)}{\partial r_j} \right]_{(\mathbf x = \mathbf x^*(\mathbf{\bar r}), \mathbf r = \mathbf{\bar r})} $$

which I played with, but didn't get anywhere with it. I assume that that is what the question is asking me, though, i.e. to show the two sides are equal. I'm just not too sure what sides those are.

And, for the last part, regarding $m+\epsilon$, I assume I'm just supposed to make a qualitative statement about the maximum value, i.e. whether it increases or decreases, but perhaps there's more to it? Any ideas? Thanks in advance.

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1 Answer 1

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The optimization problem is

$$\max_{x,y} 1-rx^2-y^2$$

subject to $$ x+y=m$$.

So the decision varibles are $\mathbf{x}=(x,y)$ and the parameters are $\mathbf{r}=(r,m)$. That's how we connect the symbols in the definition to the symbols in the problem.

It will be useful to define the objective $f(\mathbf{x},\mathbf{r})=1-rx^2-y^2$ and constraint function $g(\mathbf{x},\mathbf{r})=x+y-m$.

As you discovered, the optimal solution is

$\mathbf{x^*}=\left(\frac{m}{r+1}, \frac{mr}{r+1}\right)$

The optimal value depends on the parameterization of the problem. For a specific parameterization $\mathbf{\bar{r}}$, the optimal value is $\mathbf{x^*}\left(\mathbf{\bar{r}}\right)$.

It's also important to understand that $f^*\left(\mathbf{r}\right)=f\left(\mathbf{x^{*}(r)},\mathbf{r}\right)$.

With that understood, there doesn't seem to be much depth to the Envelope Theorem as you described it.

Sections 1.2 and 1.3 of this document: http://www.sfu.ca/~wainwrig/Econ331/env-theorem2.pdf might help. From there, it seems that the Envelope Theorem is about showing that $$f(\mathbf{x^*(r_1)},\mathbf{r_1})= f(\mathbf{x^*(r_1)},\mathbf{r_2})$$.

The left-hand-side of eq. 25 from that document is the quantity $$\frac{df^*(\mathbf{x},\mathbf{r})}{d\mathbf{r}}$$

which you computed as $$\left[ -\left( \frac{m}{r+1} \right)^2, \frac{-2mr}{r+1} \right]$$.

The point is that in eq. 25, the terms that look like $f_x + \lambda g_x$ are zero, since those are exactly the first-order optimality conditions. (Notice that you should have already computed a value for $\lambda$ since you computed a value for $x^*$.) And all that you are left with is

$$\frac{df^*(\mathbf{x},\mathbf{r})}{d\mathbf{r}} = \left. \frac{df(\mathbf{x},\mathbf{r})}{d\mathbf{r}}\right|_{\mathbf{x}=\mathbf{x^*},\mathbf{r}=\mathbf{r}} + \lambda \left. \frac{dg(\mathbf{x},\mathbf{r})}{d\mathbf{r}}\right|_{\mathbf{x}=\mathbf{x^*},\mathbf{r}=\mathbf{r}}$$

The first term, before you substitute $\mathbf{x^*}$, is $[-x^2,0]$. The second term is $\lambda [0,-1]$.

(I'm using $[\cdots,\cdots]$ to denote a vector. Since $\mathbf{r}$ is a two-dimensional vector, a gradient of a scalar-valued function with respect to \mathbf{r} is also a two-dimensional vector (in the dual space)).

If you plug in all that you know into $[-x^2,0] + \lambda [0,-1]$, you will get $[\frac{df^*}{dr}, \frac{df^*}{dm}]$, and the expressions will agree with what you computed directly (perhaps with a sign error in the second component, in which case just negate the definition of $g$ (it doesn't matter because the constraint is $g=0$).

As for the second part of the question, you should internalize that the meaning of a Lagrange multiplier is the cost associated with a constraint being active. In this case, if you make $m$ smaller, then the optimal value increases (and if you make it bigger, then the optimal value decreases). You can already see this because you computed $\frac{df^*}{dm}$. You should also draw a picture of the level sets of $f$ (they're ellipses) and the constraint set $g=0$ (which is a line) to geometrically confirm.

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