probability theory - Let $Xge 0$ be a random variable. Show that for any $Mgt 0$ that $Msum_{k=1}^infty P(Xgt kM)le E(X)le Msum_{k=0}^infty P(Xgt kM)$.

Let $X\ge 0$ be a random variable. Show that for any $M\gt 0$ that
$$M\sum_{k=1}^\infty P(X\gt kM)\le E(X)\le M\sum_{k=0}^\infty P(X\gt kM).$$

Attempt

We know, $E(X)=\int_0^\infty(1-F(x))dx$ and $P(X\gt kM)=1-F(kM)$, where $F$ is distribution function. So by using this I get, $$M\sum_{k=1}^\infty 1-F(kM)\le \int_0^\infty(1-F(x))dx \le M\sum_{k=0}^\infty 1-F(kM).$$

I was trying to approach like this but can't proceed further. If I am on right track then how to proceed further and if not then how to solve this problem?

Thanks

Answer

Another solution is this:

Since $\Bbb{P}[X>x]=1-F(x)$, we can write
$$\begin{eqnarray} \Bbb{E}[X] & = & \int_0^\infty \Bbb{P}[X>x]dx\\ & = & \int_0^\infty \Bbb{P}[X/M>x/M]dx\\ & = & M\int_0^\infty \Bbb{P}[X/M>u]du \end{eqnarray}$$
with $u=x/M$.
Now, notice that for each $k$, and for all $u\in [k,k+1]$,
$$\Bbb{P}[X/M>(k+1)]\leq \Bbb{P}[X/M>u]\leq \Bbb{P}[X/M>k]$$
hence
$$\Bbb{P}[X/M>(k+1)]\leq \int_{k}^{k+1}\Bbb{P}[X/M>u]du\leq \Bbb{P}[X/M>k]$$
hence
$$\sum_{k}\Bbb{P}[X/M>(k+1)]\leq \sum_{k}\int_{k}^{k+1}\Bbb{P}[X/M>u]du\leq \sum_{k}\Bbb{P}[X/M>k]$$
which is the same as

$$\sum_{k=1}^\infty P(X\gt kM)\le \frac{1}{M}E(X)\le \sum_{k=0}^\infty P(X\gt kM)$$