Mara Cook

2022-06-29

statement of the proof was that $S$ was closed from below, and bounded from below. It is fine when they say that $B$ is bounded below which is also obvious from the definition. So, there should be a g.l.b. for the set $B$, which is denoted by ${b}_{0}$.

Now why there should exist a sequence of points $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$ in the risk set $S$ such that $\sum {p}_{j}{y}_{j}\to {b}_{0}$? Is ${b}_{0}$ a limit point of $B$? Is $B$ closed ? Even if this happen why ${b}_{0}$ which is greatest lower bound of $B$ should belong to $B$?

Next, I guess they apply Bolzano Weierstrass theorem to say that ${\mathbf{y}}^{\mathbf{0}}$ is a limit point of the sequence $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$. But why the last step $\sum {p}_{j}{y}_{j}^{0}={b}_{0}$?

Now why there should exist a sequence of points $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$ in the risk set $S$ such that $\sum {p}_{j}{y}_{j}\to {b}_{0}$? Is ${b}_{0}$ a limit point of $B$? Is $B$ closed ? Even if this happen why ${b}_{0}$ which is greatest lower bound of $B$ should belong to $B$?

Next, I guess they apply Bolzano Weierstrass theorem to say that ${\mathbf{y}}^{\mathbf{0}}$ is a limit point of the sequence $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$. But why the last step $\sum {p}_{j}{y}_{j}^{0}={b}_{0}$?

trajeronls

Beginner2022-06-30Added 21 answers

Since ${b}_{0}=infB$, there exists a sequence $({\beta}_{n}{)}_{n\ge 1}$ of elements of $B$ such that $\underset{n}{lim}{\beta}_{n}={b}_{0}$. As in the book, write ${\beta}_{n}=\sum _{j=1}^{k}{p}_{j}{y}_{j}^{(n)}$ where $({y}_{1}^{(n)},\dots ,{y}_{k}^{(n)})\in S$.

Since $({\beta}_{n}{)}_{n\ge 1}$ converges, it is bounded above by some $B$, hence for fixed $j$, we have ${p}_{j}{y}_{j}^{(n)}\le \sum _{j=1}^{k}{p}_{j}{y}_{j}^{(n)}\le B$. Since ${p}_{j}>0$, this yields $\mathrm{\forall}n\ge 1,\phantom{\rule{thickmathspace}{0ex}}{y}_{j}^{(n)}\le \frac{B}{{p}_{j}}$. Consequently, $({\mathbf{y}}^{(n)}{)}_{n\ge 1}$ is bounded coordinate-wise, hence bounded in ${\mathbb{R}}^{k}$. Applying Bolzano-Weierstrass gives some limit point ${\mathbf{y}}^{0}$ and a subsequence $({\mathbf{y}}^{({n}_{m})}{)}_{m\ge 1}$ such that $\underset{m}{lim}{\mathbf{y}}^{({n}_{m})}={\mathbf{y}}^{0}$. This implies convergence w.r.t each coordinate: $\underset{m}{lim}{y}_{j}^{({n}_{m})}={y}_{j}^{0}$.

Letting $m\to \mathrm{\infty}$ in $\sum _{j=1}^{k}{p}_{j}{y}_{j}^{({n}_{m})}={\beta}_{{n}_{m}}$ yields

$\sum _{j=1}^{k}{p}_{j}{y}_{j}^{0}={b}_{0}$

Since $({\beta}_{n}{)}_{n\ge 1}$ converges, it is bounded above by some $B$, hence for fixed $j$, we have ${p}_{j}{y}_{j}^{(n)}\le \sum _{j=1}^{k}{p}_{j}{y}_{j}^{(n)}\le B$. Since ${p}_{j}>0$, this yields $\mathrm{\forall}n\ge 1,\phantom{\rule{thickmathspace}{0ex}}{y}_{j}^{(n)}\le \frac{B}{{p}_{j}}$. Consequently, $({\mathbf{y}}^{(n)}{)}_{n\ge 1}$ is bounded coordinate-wise, hence bounded in ${\mathbb{R}}^{k}$. Applying Bolzano-Weierstrass gives some limit point ${\mathbf{y}}^{0}$ and a subsequence $({\mathbf{y}}^{({n}_{m})}{)}_{m\ge 1}$ such that $\underset{m}{lim}{\mathbf{y}}^{({n}_{m})}={\mathbf{y}}^{0}$. This implies convergence w.r.t each coordinate: $\underset{m}{lim}{y}_{j}^{({n}_{m})}={y}_{j}^{0}$.

Letting $m\to \mathrm{\infty}$ in $\sum _{j=1}^{k}{p}_{j}{y}_{j}^{({n}_{m})}={\beta}_{{n}_{m}}$ yields

$\sum _{j=1}^{k}{p}_{j}{y}_{j}^{0}={b}_{0}$

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