I have a plan of the open lecture in which I explain the probability theory for Finance. The contents are similar to the one which has been held in last summer.
1.Basic concepts of probabilities
2.Elementary Lebesgue theory
3.Normal distributions and Brownian motions
4.Conditional expectations
5.Ito's calculus
The lecture will be given over a period of 12 hours. Because the time of the lecture is extended one hour, I will give some changes in the explanations.
Especially, in introducing two methods for solving Black-Scholes formula, I will explain how to use definitions and propositions of the probability theory.
If you are interested in, please join the seminar (← link) in Aug-Sept.
2014/06/09
2014/06/01
functions 2
In preceding post, we defined a function as follows.
A relation $f$ from a member of a set $X$ to a set $Y$ which is many to one or one to one is called a function from $X$ to $Y$. Then, we write
$f:X\rightarrow Y$
There are usually two metric spaces $S,T$ , and $X\subset S$ and $Y\subset T$ are assumed.
$Y=f(X)=\left\{ f(x) | x\in X \right\}$ is the range or the image of the function.
Conversely, the inverse image of the function is
$f^{-1}(Y)=\left\{ x | f(x)\in Y \right\}$
As it is the domain or a part of the domain of the function, $f^{-1}(Y)\subset X$
By using the distance function $d$ on $S$ and $d'$ on$T$ , we shall define again continuous functions by epsilon-delta proofs.
A function $f$ is continuous at a point c if, for any $\epsilon>0$, there is a $\delta>0$
such that if $d(x,c)<\delta$ , $d'(f(x),f(c))<\epsilon$
Next definition is derived from above. But no distance fuctions are used.
A function is continuous if and only if the inverse image of a open set is open.
Similarly, the definition by a closed set is available.
A function is continuous if and only if the inverse image of a closed set is closed.
Two definitions are equivalent each other. However these are not true on the range or the image.
A relation $f$ from a member of a set $X$ to a set $Y$ which is many to one or one to one is called a function from $X$ to $Y$. Then, we write
$f:X\rightarrow Y$
There are usually two metric spaces $S,T$ , and $X\subset S$ and $Y\subset T$ are assumed.
$Y=f(X)=\left\{ f(x) | x\in X \right\}$ is the range or the image of the function.
Conversely, the inverse image of the function is
$f^{-1}(Y)=\left\{ x | f(x)\in Y \right\}$
As it is the domain or a part of the domain of the function, $f^{-1}(Y)\subset X$
By using the distance function $d$ on $S$ and $d'$ on$T$ , we shall define again continuous functions by epsilon-delta proofs.
A function $f$ is continuous at a point c if, for any $\epsilon>0$, there is a $\delta>0$
such that if $d(x,c)<\delta$ , $d'(f(x),f(c))<\epsilon$
Next definition is derived from above. But no distance fuctions are used.
A function is continuous if and only if the inverse image of a open set is open.
Similarly, the definition by a closed set is available.
A function is continuous if and only if the inverse image of a closed set is closed.
Two definitions are equivalent each other. However these are not true on the range or the image.