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Let $X$ be any nonnegative random variable. Then show that $$E[X]-1\leq \sum_{n=1}^{\infty}P[X\geq n]\leq E[X].$$

I was able to show the second inequality by $$\sum_{n=1}^{\infty}P[X\geq n]=\sum_{n=1}^{\infty}\int_{ X\geq n}dP=\int_\Omega \sum_{1\leq n\leq X}1dP\leq \int_\Omega X dP =E[X].$$

However, the first inequality seems just impossible to show to me. Maybe there is a trick to show I guess. My first impression is that, since $P[X\geq 0]=1$, it may be possible so tackle with $$E[X]\leq \sum_{n=0}^{\infty}P[X\geq n]$$ but could not figure out. I will thank to any suggestion or solution. Thank you!

Lev Bahn
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You've already shown that $$\sum_{n=1}^{\infty}P[X\geq n] = \int_\Omega \sum_{1\leq n\leq X}1dP$$ You used the fact that $\sum_{1 \leq n \leq X} 1 \leq X$ to get the upper bound. For the lower bound, we just notice that $X - 1 \leq \sum_{1 \leq n \leq X} 1$ to get that $$\sum_{n=1}^{\infty}P[X\geq n] \geq \int_\Omega X - 1 dP = \mathbb{E}[X]-1.$$

Rhys Steele
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  • Thank you! (+1 ) But I cannot see how $X-1\leq \sum_{1\leq n \leq X}1$. Could you give me a bit more detail? – Lev Bahn May 30 '20 at 17:28
  • The right hand side is the smallest integer less than or equal to $X$. There is always an integer in $[X-1, X]$ and so the smallest integer less than or equal to $X$ is always at least $X-1$. – Rhys Steele May 30 '20 at 17:30
  • Ah... I did not see that.. Thank you! – Lev Bahn May 30 '20 at 17:31
  • In fact, what the equalities you write at the start shows is that $\sum_{n=1}^\infty P[X \geq n] = E[\operatorname{floor}(X)]$. The inequalities are really then just the fact that $X - 1 \leq \operatorname{floor}(X) \leq X$. – Rhys Steele May 30 '20 at 17:33