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Expectation: Is this proposition true or false?

  1. Jul 17, 2013 #1
    If X is a continuous random variable and E(X) exists, does the limit as x→∞ of x[1 - F(x)] = 0?

    I encountered this, but so far I have neither been able to prove this, nor find a counterexample. I have tried the mathematical definition of the limit, l'Hopital's rule, integration by parts, a double integral (through expectation), and various proof scribbles, but so far, nothing has worked. Can anyone help me with this?

    EDIT: In this case, the function F is the CDF of X.
    Last edited: Jul 17, 2013
  2. jcsd
  3. Jul 17, 2013 #2
    A hint, for the case where [itex]X\geq 0:[/itex]

    [tex]\mathbb E X = \int_0^\infty xf(x)dx \geq \int_0^{\bar x} xf(x)dx + \bar x\int_{\bar x}^\infty f(x)dx.[/tex]

    Think about the pieces of that, and think about limits as [itex] \bar x \to \infty. [/itex]
  4. Jul 17, 2013 #3
    I'm aware of that, and I know how to prove it for the case of [itex]X \geq 0[/itex], but I'm confused about the case of the entire real line.

    Also, what do you mean by [itex]\overline{x}[/itex]?
  5. Jul 17, 2013 #4
    I was just using [itex]\bar x[/itex] as another stand-in variable.

    If you know how to prove it for nonnegative-valued [itex]X[/itex], then you're basically done. The limiting property you care about is the same for [itex]X_+ = \text{max}\{X,0\}[/itex].
  6. Jul 17, 2013 #5
    What do you mean by limiting property?
  7. Jul 17, 2013 #6
    Oh! I get it now! Thank you so much!
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