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