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I've seen three different definitions for expected value of a random variable. The first one is the wikipedia's version: $E[X]=\int_\Omega X\,\mathrm{d}P\,$ (Lebesgue integral).

The second is of my first lecturer: $E[X]=-\int_{-\infty}^0 F_x(t) \,\mathrm{d}t+\int_0^\infty (1-F_x(t))\,\mathrm{d}t \hspace{2 mm}$ (using distribution function).

The third definition (used by my second lecturer) is this: $ E[X]=\int_{-\infty}^\infty \alpha\,\mathrm{d}F_x(\alpha)\,$ where the integral $ \int_{A}^B g(\alpha)\,\mathrm{d}F_x(\alpha)\,$ is defined as $ \int_{A}^B g(\alpha)\, \mathrm{d}F_x(\alpha) \hspace{2 mm} = \lim_{\Delta\alpha\rightarrow 0} \sum_i g(\alpha_i) (F(\alpha_{i+1})-F(\alpha_i)) \qquad \text{(Riemann integral)} . $

Who's right? and are all these three definitions equivalent?

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    I've thought about what you said, and I think I can prove it (tell me if my proof is ok): I define $f(\omega,t):(\Omega\times \mathbb{R}^+)\rightarrow\mathbb{R}^+$ by f(\omega,t)=1_{(X(\omega)>t)}. now the integral \int_0^\infty P(X>t) dt = \int_{\mathbb{R}^+}(\int_\Omega f(\omega,t) dP)dt and by Fubini's theorem it equals $\int_{\Omega}(\int_0^\infty f(\omega,t) dt)dP=\int_{\Omega}(\int_0^{X(\omega)} 1 dt)dP=\int_\Omega X dP$. The question is, could I apply Fubini's theorem in that way?2011-10-16

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We can see that the first and second definition are equivalent by Fubini's theorem. Indeed, we can write $X:=X^+-X^-$ where $X^+,X^-$ are non-negative, then we use the fact that for a non-negative random variable $Y$, we have $E(Y)=\int_0^{+\infty}P(Y\geqslant t)dt.$

The equivalence between the second and the third follows by the definition of Riemann-Stieltjes integral.