Numerical Methods in Computational Finance. Daniel J. Duffy
Чтение книги онлайн.

Читать онлайн книгу Numerical Methods in Computational Finance - Daniel J. Duffy страница 45

Название: Numerical Methods in Computational Finance

Автор: Daniel J. Duffy

Издательство: John Wiley & Sons Limited

Жанр: Ценные бумаги, инвестиции

Серия:

isbn: 9781119719724

isbn:

СКАЧАТЬ Subscript j Baseline right-parenthesis x Superscript j Baseline comma alpha element-of normal double struck upper R period EndLayout"/>

      Then upper P Subscript n is a subspace of P, and it is also a subspace of upper P Subscript m Baseline comma m greater-than-or-equal-to n.

      We say that a subset X of a vector space upper V left-parenthesis upper K right-parenthesis is said to be closed under addition if whenever x 1 comma x 2 element-of upper X, then x 1 plus x 2 element-of upper X. A subset X of a vector space upper V left-parenthesis upper K right-parenthesis is said to be closed under scalar multiplication if whenever normal lamda element-of upper K and x element-of upper X then normal lamda x element-of upper X.

      Theorem 4.1 A subset X of a vector space upper V left-parenthesis upper K right-parenthesis is a subspace if and only if:

      (4.12)normal lamda 1 x 1 plus normal lamda 2 x 2 element-of upper X for normal lamda 1 comma normal lamda 2 element-of upper K comma x 1 comma x 2 element-of upper X period

      An exercise: let x 1 comma ellipsis comma x Subscript r Baseline be any r elements of a vector space upper V left-parenthesis upper K right-parenthesis. Prove that the set U of all elements of upper V left-parenthesis upper K right-parenthesis that can be written in the form sigma-summation Underscript j equals 1 Overscript r Endscripts normal lamda Subscript j Baseline x Subscript j Baseline comma normal lamda Subscript j Baseline element-of upper K comma j equals 1 comma ellipsis comma r forms a subspace of upper V left-parenthesis upper K right-parenthesis.

      We give an example of a subset X of upper K squared defined by:

      (4.13)StartLayout 1st Row 1st Column Blank 2nd Column upper X equals left-brace x equals left-parenthesis t comma t squared right-parenthesis semicolon t element-of upper K right-brace 2nd Row 1st Column Blank 2nd Column x 1 plus x 2 equals left-parenthesis t 1 comma t 1 squared right-parenthesis plus left-parenthesis t 1 comma t 2 squared right-parenthesis equals left-parenthesis t 1 plus t 2 comma left-parenthesis t 1 squared plus t 2 squared right-parenthesis right-parenthesis 3rd Row 1st Column Blank 2nd Column normal lamda x equals left-parenthesis normal lamda t comma normal lamda squared t squared right-parenthesis period EndLayout

      It is easily verified that X is a vector space over K, but X is not a subspace of upper K squared because:

StartLayout 1st Row 1st Column Blank 2nd Column Blank 3rd Column x 1 comma x 2 element-of upper X 2nd Row 1st Column Blank 2nd Column Blank 3rd Column normal lamda left-parenthesis x 1 plus x 2 right-parenthesis factorial equals normal lamda x 1 plus normal lamda x 2 EndLayout

      and these two quantities are thus not the same!

      We are now interested in finding a minimal subspace U of independent vectors containing a set X (U contains X) such that any vector in X can be written as a linear combination of these vectors. In this case we say that X spans U. We are particularly interested in the case upper X equals upper V left-parenthesis upper K right-parenthesis, that is subsets of upper V left-parenthesis upper K right-parenthesis that span upper V left-parenthesis upper K right-parenthesis itself. Such subsets always exist; for example upper X equals upper V left-parenthesis upper K right-parenthesis has this property. We take an example in n-dimensional space. The vectors:

normal lamda 1 e 1 plus ellipsis plus normal lamda Subscript n Baseline e Subscript n Baseline period

      Furthermore, any proper subset of StartSet e 1 comma ellipsis comma e Subscript n Baseline EndSet cannot span upper K Superscript n.