1
$\begingroup$

Given income $y$ and a vector of commodity prices $p\in R_+^L$, the set of feasible consumption bundles is described by the budget correspondence, $B(p,y)=\{x\in R_+^L:px\leq y\}$. $B(p,y)$ is both upper and lower hemicontinuous, as proved in Chapter 8 Problem 2.2 in Mathematical Methods and Models for Economists by de la Fuente.

Here is the proof: To establish that $B$ is an lhc correspondence, we need to show that given any price-income seqwuence $\{(p_n,y_n)\}$ converging to $(p,y)>>0$ and an arbitrary point $x\in B(p,y)$, there exists a companion sequence of consumption bundles $\{x_n\}$ with $x_n\in B(p_n,y_n)$ for all $n$ that converges to $x$.

Let $x_n=x$ if $x\in B(p_n,y_n)$ and $x_n=\frac{y_n}{p_nx}x$ if otherwise. Notice that $x_n$ is feasible for $(p_n,y_n)$ by construction, because $x_n$ is defined as the largest fraction of the bundle $x$ that the consumer can afford with income $y_n$ and prices $p_n$. It's also clear that $\{x_n\}\to x$. If $x$ lies in the interior of the budget set, then we have $x_n=x$ for $n$ sufficiently large. Otherwise, $y=px$ and $\lim x_n=\lim\frac{y_n}{p_nx}x=\frac{y}{px}x=x$.

You can also see here for the proof.

But I'm having trouble understanding this proof. Here is the definition of lhc: A correspondence is lhc at $a$ if $\forall b\in F(a)$, $\exists a_n$ and $b_n$ such that $a_n\to a$ and $b_n\to b$.

But in this proof, $p_n$ is taken as an arbitrary sequence. $p_n$ should correspond to the $a_n$ sequence in the definition of lhc. In the definition, $a_n$ should be "there exists", not "for all". Can someone explain why?

$\endgroup$
2
  • 1
    $\begingroup$ This is the definition of lhc: $F$ is lower hemicontinuous at $a$ if for every sequence $(a_n)$ converging to $a$ and every $b \in F(a)$ there exists a sequence $(b_n)$ converging to $b$ with $b_n \in F (a_n)$. $\endgroup$
    – Amit
    Nov 15, 2023 at 8:24
  • $\begingroup$ And the budget correspondence will not be lhc on all of $\mathbb{R}^l_+$. Goods with a price of zero will cause trouble. $\endgroup$ Nov 15, 2023 at 23:21

1 Answer 1

4
$\begingroup$

I think you need to recheck the definition you provided for lhc. This is because the definition you provided is always true for any correspondence at any point $(a,b)$ satisfying $b\in F(a)$. To see this, consider constant sequences $a_n=a$, and $b_n=b$. Clearly, $a_n\rightarrow a$, and $b_n\rightarrow b$. Additionally, $b_n\in F(a_n)$ for all $n\in \mathbb{N}$ also holds because $a_n=a$, $b_n=b$ and $b\in F(a)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.