Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
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
1  
@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
1  
Oh: You mean $\int_\Omega X(\omega)\,P(d\omega)$. –  Michael Hardy Nov 14 '12 at 14:13

1 Answer 1

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 $$ \int_Yf\,\mathrm{d}\mu_{\varphi}=\int_Xf\circ\varphi\,\mathrm{d}\mu,\quad f\in\mathcal{L}^1(\mu_\varphi).\quad (*) $$

Proof:

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.

share|improve this answer
    
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  
+1. $ $ $ $ $ $ –  Did Nov 14 '12 at 8:07

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.