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CaptainBlack · #3 November 22nd, 2008, 01:27 AM

Quote:

Originally Posted by maths_123

Can you please help me with this proof. Look at the picture.

THANKS

For the second part we can observe:

$t P(A \ge t)= \sum_{i=t}^n t P(A=i) \le \sum_{i=t}^n i P(A=i)$

since in the sum each $i$ is $\ge t$ .

But as $P(A=i)$ and $i$ for $i=0, .. n$ are all non-negative:

$t P(A \ge t) \le \sum_{i=t}^n i P(A=i)\le \sum_{i=0}^n i P(A=i)=E(A)$

Hence:

$P(A \ge t) \le \frac{E(A)}{t}, \ \ \ t=0,1,.., n$

CB