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Let $X$ be a random variable on $(\Omega, \Sigma, P)$. The expected value of $X$ is defined as $$EX = \int X \,dP.$$
But when we calculate $EX$, we often use $$ EX = \int_{-\infty}^\infty xf(x) dx $$ where $f(x)$ is the density function. How can we prove that these two are equivalent?

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I think you mean $\int_{-\infty}^{\infty} xf(x)dx$. – fgp Nov 13 '12 at 1:09
The first one is measure-theoretic definition, the second one is probabilistic – Alex Nov 13 '12 at 1:44
@Alex: I beg your pardon? – Did Nov 13 '12 at 21:40
@MichaelHardy: I'm pretty sure $\int X\,\mathrm{d}P$ is the correct notation for $E[X]$. – Stefan Hansen Nov 14 '12 at 7:09
Oh: You mean $\int_\Omega X(\omega)\,P(d\omega)$. – Michael Hardy Nov 14 '12 at 14:13
up vote 4 down vote accepted

You have to use two results to obtain the equality. The first being:

If $(X,\mathcal{E},\mu)$ is a measure space, $(Y,\mathcal{F})$ is a measurable space and $\varphi:X\to Y$ be $\mathcal{E}$-$\mathcal{F}$-measurable. Let $\mu_\varphi=\mu\circ\varphi^{-1}$ be the image measure of $\mu$ under $\varphi$. Then $$ \mathcal{L}^1(\mu_\varphi)=\{f:Y\to \mathbb{R},\,f \text{ is }\mathcal{F}\text{-}\mathcal{B}(\mathbb{R})\text{-measurable}\mid f\circ\varphi \in\mathcal{L}^1(\mu)\} $$ and $$ \int_Yf\,\mathrm{d}\mu_{\varphi}=\int_Xf\circ\varphi\,\mathrm{d}\mu,\quad f\in\mathcal{L}^1(\mu_\varphi).\quad (*) $$


1) For $B\in\mathcal{F}$ we have $$ \int_X 1_B\circ\varphi\,\mathrm{d}\mu=\int_X 1_{\varphi^{-1}(B)}\,\mathrm{d}\mu=\mu(\varphi^{-1}(B))=\mu_\varphi(B)=\int_X 1_B\,\mathrm{d}\mu_\varphi. $$

2) If $f,g:Y\to\mathbb{R}$ are non-negative $\mathcal{F}$-measurable functions such that $(*)$ holds, then $$ \int_X(f+g)\circ\varphi\, \mathrm{d}\mu=\int_X (f\circ\varphi+g\circ\varphi)\,\mathrm{d}\mu=\int_X f\circ\varphi\,\mathrm{d}\mu + \int_X g\circ\varphi\,\mathrm{d}\mu\\ =\int_X f\,\mathrm{d}\mu_\varphi+\int_X g\,\mathrm{d}\mu_\varphi, $$ i.e. $(*)$ holds for $f+g$.

3) Let $(f_n)_{n\geq 1}$, $f_n:Y\to\mathbb{R}$, be a sequence of non-negative increasing $\mathcal{F}$-measurable functions obeying $(*)$ such that $f=\lim_{n\to\infty} f_n$ exists pointwise. Then $(f_n\circ \varphi)_{n\geq 1}$ is a sequence of non-negative increasing $\mathcal{E}$-measurable functions and by the monotone convergence theorem (twice), we have $$ \int_X f\circ\varphi\,\mathrm{d}\mu=\lim_{n\to\infty}\int_X f_n\circ\varphi\,\mathrm{d}\mu=\lim_{n\to\infty}\int_X f_n\,\mathrm{d}\mu_\varphi=\int_X f\,\mathrm{d}\mu_\varphi. $$

Then a standard argument yields that $(*)$ holds for all $f\in\mathcal{L}^1(\mu_\varphi)$.

The seconds result:

Let $(X,\mathcal{E},\mu)$ be a measure space and let $g:X\to\mathbb{R}$ be a non-negative $\mathcal{E}$-$\mathcal{B}(\mathbb{R})$-measurable function. Let $\nu$ be the measure with density $g$ with respect to $\mu$, i.e. $\nu(A)=\int_A g\,\mathrm{d}\mu$. Then $$ \mathcal{L}^1(\nu)=\{f:X\to \mathbb{R},\,f \text{ is }\mathcal{E}\text{-}\mathcal{B}(\mathbb{R})\text{-measurable}\mid f\cdot g \in\mathcal{L}^1(\mu)\} $$ and $$ \int_Xf\,\mathrm{d}\nu = \int_X f\cdot g\,\mathrm{d}\mu,\quad f\in\mathcal{L}(\nu). $$

Proof: Recreate the steps 1)-3) from above to this setting.

Try to combine these results with $(X,\mathcal{E},\mu)=(\Omega,\Sigma,P)$, $(Y,\mathcal{F})=(\mathbb{R},\mathcal{B}(\mathbb{R}))$ and $\varphi=X$ in the first result, and $(X,\mathcal{E},\mu)=(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, $\nu=P_X$ and $g=f$ in the second result.

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This kind of takes the result for granted, don't you think? – Did Nov 14 '12 at 7:14
@did: Yes, I guess you're right. Let me add proofs (or an outline at least). I just thought that he may know one or both of the results. – Stefan Hansen Nov 14 '12 at 7:29
+1. $ $ $ $ $ $ – Did Nov 14 '12 at 8:07

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