Thread: 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

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.

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

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.

As romsek suggested above - probabilities are always positive.