A property of almost surely positive random variable

Question

Let $X$ be an almost surely positive random variable.

I must demostrate that:

$\sum_1^\infty P(x \geq n) \leq E[X] \leq \sum_0^\infty P(X> n)$

Could someone help me?

Answer 1

Hint: Use the following formula

$$\mathbb{E}[X] =\int_0^{+\infty}P(X>t)dt$$

Answer 2

Hint : for all $k\geq 1$, \begin{align*} (k-1) 1(k-1\leq X< k)\leq X\times 1(k-1\leq X< k)\leq k\times 1(k-1\leq X< k) \end{align*} Taking the expectation of that gives you \begin{align*} (k-1) \mathbb P(k-1\leq X< k)\leq \mathbb E[X\times 1(k-1\leq X< k)]\leq k\mathbb P(k-1\leq X< k) \end{align*} Taking the sum over all $k\geq 1$ gives you your result (with few steps), by observing that $\sum_{k\geq 1}\mathbb E[X\times 1(k-1\leq X< k)] = \mathbb E[X]$ and rewriting sums.

Answer 3

I used the formula given by Slup; I did not know it, so thank you really.

I think this works:

$\sum_1^\infty P(X\geq n) \leq \int_0^{+\infty}P(X>t)dt \leq \sum_0^\infty P(X> n)$