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CaptainBlack · #2 November 22nd, 2008, 01:13 AM

Quote:

Originally Posted by maths_123

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

THANKS

From the definition of expectation we have:

$E(A)=\sum_{i=0}^n i p(A=i)$

Now:

$p(A\ge k)= \sum_{i=k}^n p(A=i)$

So:

$\sum_{k=1}^n p(A \ge k)$

contains $p(A=1)$ once, $p(A=2)$ twice, etc

hence:

$\sum_{k=1}^n p(A \ge k)=\sum_{k=1}^n kp(A=k) = \sum_{k=0}^n kp(A=k)=E(A)$

CB