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Proving Markov's Inequality (Probability)

  1. Jan 20, 2013 #1
    1. The problem statement, all variables and given/known data

    If [itex]g(x)\ge 0[/itex], then for any constant ##c>0, r>0##:

    [itex]P(g(X)\ge c)\le \frac{E((g(X))^r)}{c^r}[/itex]

    2. Relevant equations

    I know that [itex]E(g(X))=\int_0^\infty g(x)f(x)dx[/itex] if [itex]g(x)\ge 0[/itex] where ##f(x)## is the pdf of ##X##.

    3. The attempt at a solution

    I tried following a similar proof to what I found here (on page 2: http://www.stat.cmu.edu/~larry/=stat705/Lecture2.pdf).

    I'm really stuck on this. I've tried to look on the Internet for a proof, but the inequalities I find involve X, not g(X) and there isn't any exponent ##r##. I tried doing something similar by replacing X with g(X) but it doesn't quite work out. I still need an exponent ##r## and I don't know how the last step would get me to write something like [itex]c^r P(g(x)\ge c)[/itex]
     
  2. jcsd
  3. Jan 20, 2013 #2

    jbunniii

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    Suppose [itex]g(x)[/itex] has a pdf. Let's call the pdf [itex]f[/itex]. Then
    $$P(g(x) \geq c) = \int_c^\infty f(y) dy$$
    In this integral, we have [itex]y \geq c[/itex]. If [itex]r > 0[/itex] and [itex]c > 0[/itex], then this implies [itex]y^r \geq c^r[/itex]. This in turn is equivalent to [itex]1 \leq y^r/c^r[/itex]. What's the natural next step?

    For the more general case in which [itex]g(x)[/itex] does not necessarily admit a pdf, the proof should be similar, except you would use a Stieltjes integral.
     
    Last edited: Jan 20, 2013
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