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Thread: Simple Random Variable

  1. #1
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    Simple Random Variable

    Hi everyone,sorry for my poor english (i'm italian),my name is Stefano ; For a week i 'm trying to solve the following probability question:
    If $\displaystyle P $ is a probability measure on $\displaystyle A '$ respect to $\displaystyle A $ ( i have supposed $\displaystyle ( A$$\displaystyle A ')$ is a measurable space) and if $\displaystyle f $ is a function from $\displaystyle A $ to $\displaystyle R $ (real numbers set ,sorry for my poor Latex too) , if the set $\displaystyle \{ f (b) \mid b \in A\} $ is finite , if $\displaystyle f $ is measurable respect to $\displaystyle ( A$$\displaystyle A ')$ and $\displaystyle ( R$$\displaystyle Borel R )$ ,if $\displaystyle n $ is a natural $\displaystyle \neq 0 $ , $\displaystyle \forall i \in \{ 1,...,n\} B_{i}\in A' $ such that $\displaystyle \forall b\in A $ $\displaystyle f(b)= \sum_ {i=1}^{n} d_{i}*I(B_ {i}(b))$ where $\displaystyle d_{i} $ are real numbers and $\displaystyle I(B_ {i}) $ denotes the indicator function of $\displaystyle B_{i}$ and ,analogously $\displaystyle n' $ is a natural $\displaystyle \neq 0 $ , $\displaystyle \forall i \in \{ 1,...,n'\} C_{i}\in A' $ such that $\displaystyle \forall b\in A $ $\displaystyle f(b)= \sum_ {i=1}^{n'} e_{i}*I(C_ {i}(b))$ where $\displaystyle e_{i} $ are real numbers ,how can i prove that $\displaystyle \sum_{i=1}^{n} d_{i}*P(B_{i}) $ $\displaystyle = \sum_{i=1}^{n'} e_{i}*P(C_{i})$ ? . My idea was to account the standard representation of f setting $\displaystyle \{ f (b) \mid b \in A\} =\{k_{1},...,k_{r}\}$ where $\displaystyle r $ ,natural $\displaystyle \neq 0 $, is the cardinality of $\displaystyle \{ f (b) \mid b \in A\}$ , $\displaystyle \forall i\in\{1,..,r \} $ let $\displaystyle F_{i} =\{b \in A \mid f(b)=k_ {i}\} $ so there is the standard representation of $\displaystyle f $ $\displaystyle f(b)= \sum_ {i=1}^{r} k_{i}*I(F_ {i}(b))$ ,and i tried to show $\displaystyle \sum_ {i=1}^{r} k_{i}*P(F_ {i})$ is equal to one between $\displaystyle \sum_ {i=1}^{n} d_{i}*P(B_ {i})$ and $\displaystyle \sum_ {i=1}^{n'} e_{i}*P(C_ {i})$ .To make this i observed : $\displaystyle k{1}*P(F_{1}) +...+k_{r}*P(F_{r}) = (d_{1}*IB_{1}(p_{1}) +...+d{n}*IB_{n}(p_{1})) * P(F_{1}) +...+(d_{1}*IB_{1}(p_{r}) +...+d{n}*IB_{n}(p_{r}) * P(F_{r})= d_{1}*[(P(F_{1})*IB_{1}(p_{1})+...+(P(F_{r}*IB_{1}(p_{r})] +...+d_{n}*[(P(F_{1}*IB_{n}(p_{1})+...+(P(F_{r}*IB_{n}(p_{r})]$ where $\displaystyle p_{i} $ are fixed in $\displaystyle F_{i} $ .Intuitively it results $\displaystyle [(P(F_{1)}*IB_{i}(p_{1})+...+(P(F_{r})*IB_{i}(p_{r} )] = P(B_{i}) $ ma i can't prove it.I hope someone help me ,thank you so much.
    Last edited by TheDifferentialProability; Nov 29th 2012 at 05:40 AM.
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