Can you please help me with this proof. Look at the picture.
THANKS
From the definition of expectation we have:
$\displaystyle E(A)=\sum_{i=0}^n i p(A=i)$
Now:
$\displaystyle
p(A\ge k)= \sum_{i=k}^n p(A=i)
$
So:
$\displaystyle \sum_{k=1}^n p(A \ge k)$
contains $\displaystyle p(A=1)$ once, $\displaystyle p(A=2)$ twice, etc
hence:
$\displaystyle \sum_{k=1}^n p(A \ge k)=\sum_{k=1}^n kp(A=k) = \sum_{k=0}^n kp(A=k)=E(A)$
CB
For the second part we can observe:
$\displaystyle
t P(A \ge t)= \sum_{i=t}^n t P(A=i) \le \sum_{i=t}^n i P(A=i)
$
since in the sum each $\displaystyle i$ is $\displaystyle \ge t$.
But as $\displaystyle P(A=i)$ and $\displaystyle i$ for $\displaystyle i=0, .. n$ are all non-negative:
$\displaystyle
t P(A \ge t) \le \sum_{i=t}^n i P(A=i)\le \sum_{i=0}^n i P(A=i)=E(A)
$
Hence:
$\displaystyle
P(A \ge t) \le \frac{E(A)}{t}, \ \ \ t=0,1,.., n
$
CB