1. ## random variable proof

THANKS

2. Originally Posted by maths_123

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

3. Originally Posted by maths_123

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

4. Originally Posted by CaptainBlack
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

5. Originally Posted by maths_123
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
$If E(A) <1$

$1> E(A) \ge p(A \ge 1)$

But $P(A=0)+P(A\ge 1)=1$, so $P(A \ge 1)<1$ implies $P(A=0)>0$

CB

6. Thanks for the help captain black.