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Thread: Markov's inequality

  1. #1
    Senior Member dokrbb's Avatar
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    Markov's inequality

    Hi there,

    I am working on a proof here. So assume $E[Y^{n}] \leq k^n$ for any $k \geq 0 $ and $n \geq 1$, where $Y $ is either a continuous or discrete non-negative RV.

    I need to show that for some $\delta > 0, P(Y > k + \delta) = 0$

    My first attempt is as follows:
    $P(Y > k + \delta) = P(Y - k > \delta)$, and since $Y - k > \delta \rightarrow |Y-k| > \delta$
    $P(|Y -k| > \delta) \leq \frac{E[|Y - k|]}{\delta}$ by Marcov's inequality,
    $ \frac{E[|Y - k|]}{\delta} = \frac{E[Y] - k}{\delta}$ and by the given condition $E[Y^{n}] \leq k^n$
    $\frac{E[Y] - E[Y]}{\delta} = 0$

    Is this oversimplified, or should I use induction somehow, or should I approach it differently...

    Thanks in advance for your help
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  2. #2
    MHF Contributor
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    Re: Markov's inequality

    Hey dokrbb.

    I would add a few things - firstly I would make sure you put Y^n and k^n for general n and make sure that the answer can't be negative since it is a probability therefore making the numerator zero as a consequence.
    Thanks from dokrbb
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  3. #3
    Senior Member dokrbb's Avatar
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    Re: Markov's inequality

    Hey chiro,

    Thanks, but I am not sure about your second suggestion, making sure the answer is not negative - should I define strictly $k^n = E[Y]^n$?
    Can't see any other way of doing it
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  4. #4
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    Re: Markov's inequality

    Ok, I think you're a bit off base here.

    Reading the Wiki on moment generating functions I see the following nifty property.

    $P[Y\geq t] \leq \dfrac{E[Y^n]}{t^n}$

    let $t = k+\delta$

    $P[Y \geq k + \delta] = \dfrac{E[Y^n]}{(k+\delta)^n} \leq \dfrac{k^n}{(k+\delta)^n} =

    \left(\dfrac{k}{k+\delta}\right)^n,~\forall n \geq 1,~k>0$

    $\delta>0 \Rightarrow \displaystyle{\lim_{n\to \infty}}\left(\dfrac{k}{k+\delta}\right)^n=0$

    so

    $P[Y \geq k + \delta] \leq \displaystyle{\lim_{n\to \infty}}\left(\dfrac{k}{k+\delta}\right)^n=0$

    and as probability is always positive it must be we have equality.
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  5. #5
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    Re: Markov's inequality

    As romsek suggested above - probabilities are always positive.
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