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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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One more for you: Expected Value = Sum of all the products of the outcomes multiplied by their respective probabilities. –  Emmad Kareem Oct 15 '11 at 21:07
    
Emmad Kareem, your definition holds only for the special case when X is a discrete random variable. I'm looking for a definition for the general case. –  Max Oct 15 '11 at 21:45
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I think most people would probably call that last integral the Riemann-Stieltjes integral. For me, the first is the most logical in an abstract setting, with the latter two resulting as consequences of the formulation, though these last two are important insofar as wanting to be able to actually calculate the desired value. –  cardinal Oct 15 '11 at 22:16
    
Davide Giraudo, can you please show how do you use Fubini's theorem to show what you said? (I'm talking about the general case, where X doesnt have to be continues). –  Max Oct 15 '11 at 22:52
    
@Max: $\mathbb E X = \mathbb E X 1_{X > 0} - \mathbb E (-X) 1_{X < 0}$. Both integrands are nonnegative and can be treated separately. –  cardinal Oct 15 '11 at 23:10
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1 Answer

up vote 0 down vote accepted

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.

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