2
$\begingroup$

I need to prove that the following constraint is LHC.

$B=\{x \in R^n : px\leqslant pw)$

But Im not capable of finding and sequence $\{x_n\}$ such that $x_n \in B(p_n,w_n) \forall n$ and that $x_n\longrightarrow x$.

I tried with setting $x_n=\frac{x}{1+\beta^n}$, $p_n=\frac{p}{1+\beta^n}$ , $w_n=\frac{w}{1+\beta^n}$ because in that case $x_n\longrightarrow x$. and $x_n \in B(p_n,w_n) \forall n$ but I thing I'm not covering the $\forall p_n, w_n$ part.

Help please and thanks in advance.

$\endgroup$
5
$\begingroup$

I don't believe it is lower semicontinous.

Let $w = (0,\dots,0)$, $p \in \mathbb{R}^n_+$ be any vector such that $p_1 = 0$ (the first coordinate being 0).

The allocation $x=(1,0,\dots,0) \in B(p,w)$.

Define the sequence $p_n = p + (\frac{1}{n},0,\dots,0)$ and $w_n = (\frac{1}{n},0,\dots,0)$. $w_n \rightarrow w$ and $p_n \rightarrow p$.

For any $x^n \in B(p_n,w_n)$, $p_n x^n_1 \leq w_np_n$, so $x_1^n \leq \frac{1}{n}$.

Hence for any sequence such that $x^n \in B(p_n,w_n)$, $x^n \not \rightarrow x$.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Actually, it is. It is proved in De la Fuente's book problem 2.2 Chap 8.I didn't understand what you have done up there, but DLF prove is magical in the sense that I wouldn't have came with that answer. Good name BTW $\endgroup$ – Martin Mendina Jun 21 at 19:26
  • 2
    $\begingroup$ @MartinMendina De la Fuente proves that the Budget correspondence is lhc at points where all prices are positive and the endowment is not zero; so there is no contradiction here. By the way: de la Fuente takes the budget set to include only points in the nonnegative orthant, which is different from your definition. $\endgroup$ – Michael Greinecker Jun 22 at 9:56
  • $\begingroup$ Yes you're right Michael, I was not precise enough. $\endgroup$ – Martin Mendina Jun 22 at 12:09
3
$\begingroup$

One approach could be the following. For a $(p_n,w_n)$ in the sequence and $x \in B(p,w)$ define: $$ \alpha_n = 1 \text{ if } p_n x \le w_n$$ and $$ \alpha_n = \frac{w_n}{p_n x} \text{ if } p_n x > w_n$$ Then define: $$ x_n = \alpha_n x$$ Here $x_n$ equals $x$ if $x$ is in the budget $B(p_n,w_n)$. If not, then $x_n$ is the radial projection of $x$ onto the budget line.

Notice that $$p_n x_n = p_n x \le w_n \text{ if } p_n x \le w_n$$ and $$p_n x_n = p_n \frac{w_n}{p_n x} x = w_n \text{ if } p_n x > w_n$$ which shows that $x_n \in B(p_n, w_n)$.

As such, the only thing left to show is that $x_n \to x$ or equivalently, $\alpha_n \to 1$.

If $p_n \to p \gg 0$ and $w_n \to w > 0$. Then for $n$ big enough one can show that $$ \alpha_n = \min\left\{\frac{w_n}{p_n x}, 1\right\}.$$ As the min function is continuous, it follows that $$ \lim_n \alpha_n = \lim_n \left(\min \left\{\frac{w_n}{p_n x}, 1\right\}\right) = \min\left\{\frac{w}{p x},1\right\} = 1.$$

| improve this answer | |
$\endgroup$

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.