random variable proof

maths_123 · #1 $Old$ November 21st, 2008, 06:58 AM

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

THANKS

CaptainBlack · #2 $Old$ November 22nd, 2008, 12:13 AM

Quote:

Originally Posted by maths_123 $View Post$

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

CaptainBlack · #3 $Old$ November 22nd, 2008, 12:27 AM

Quote:

Originally Posted by maths_123 $View Post$

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

maths_123 · #4 $Old$ November 23rd, 2008, 09:21 AM

Quote:

Originally Posted by CaptainBlack $View Post$

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

Hi sorry to be a pain, but how do you do part ii as you have helped me on part i and iii but not ii