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Suppose I have a probability measure $P([a,b]) = \int_a^b{x}{dx}$ and I want to find the expectation of the random variable $X = w^2$, which is by definition:

$E(X) = \int_{\Omega}{w^2P(dw)}$.

I have no idea what to do here.. I know if the probability measure is the Lebesgue measure then you just turn the $P(dw)$ into $dw$ and integrate "normally" but with such measures I am unsure. Any help appreciated.

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We have $P([-1,0]) =-\frac 12$: is there a typo? –  Davide Giraudo Sep 4 '11 at 16:36
    
Yes, when you say "probability" you will have to restrict to some interval $[A,B]$ where $\int_A^B x\,dx = 1$. And of course your answer for $E(X)$ will depend on which interval you choose for that. –  GEdgar Sep 4 '11 at 16:54
    
To mention the random variable $X=w^2$ is odd, to say the least. Or the OP should explain what is Omega. (That is, after the mysteries of the mass of P and of its nonnegativity are solved.) –  Did Sep 4 '11 at 17:07
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My suppositions in this case has been $\Omega=[0,\sqrt{2}]$ and that $X=w^2$ means $X(w)=w^2$. I think this are the reasonable ones. –  Josué Tonelli-Cueto Sep 4 '11 at 17:14
    
Apologies for the lack of detail. I just made something up from what I had seen recently but couldn't find the exact details. –  Court Sep 4 '11 at 19:21
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In this particular case, you should only apply that $$EX=\int_\Omega\,w^2\,P(dw)=\int_{\Omega}\,w^2\,w\,dw$$ by applying the next theorem:

Theorem. Let $(X,\mathcal{A})$ be a measurable space, and $\mu$ a measure on it and $f:X\rightarrow [0,\infty]$ a measurable function. Then \begin{align} \varphi:\mathcal{A}&\rightarrow [0,\infty]\newline A&\mapsto \int_{A}\,f\,d\mu \end{align} is a measure over $(X,\mathcal{A})$, and for every measurable function $g:X\rightarrow [0,\infty]$ we have $$\int_{\Omega}\,g\,d\varphi=\int_{\Omega}\,g\cdot f\,d\mu$$

Proof: For showing that it is a measure, we only have to show $\sigma$-additivity. So, let $\{E_n\}$ be a denumerable family of disjoint sets of $\mathcal{A}$, then $$\varphi\left(\bigcup E_n\right)=\int_{\bigcup E_n}\,f\,d\mu=\int_{\Omega}\,\chi_{\bigcup E_n}\cdot f\,d\mu$$ $$=\int_{\Omega}\,\sum\chi_{E_n}\cdot f\,d\mu=\sum\int_{\Omega}\chi_{E_n}\cdot f\,d\mu=\sum\int_{E_n}f\,d\mu=\sum\varphi(E_n)$$ thanks to Lebesgue's Monotone Convergence Theorem. Finally, the second part is true by the previous part for characteristic functions, so it will be also true for simple functions. And, so by Lebesgue's Monotone Convergence Theorem for arbitrary measurable functions.

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Thanks for the answer. It seems strange that I haven't found this in any of the books. My teacher solved this problem by differentiating my expression for $P$ and doing possibly something to it and substituting it back in... I think this is different to your answer, any idea on what it might be? –  Court Sep 4 '11 at 19:23
    
@blackcat: I would say he is using the fact that since $f(x)=x$ is a continuous function, so applying more a form of the change of variable theorem for the function $dP=P'([0,x])\,dx$. My approach is more using pure measure theory, with Lebesque integration theory -which I am not sure if your are familair with-. –  Josué Tonelli-Cueto Sep 4 '11 at 21:39
    
Thanks. I have studied Lebesgue measure/integration theory only -- not much about general measures but I guess they're not too different. –  Court Sep 5 '11 at 7:32
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