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Old November 21st, 2008, 07:58 AM
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Exclamation random variable proof

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

THANKS
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Old November 22nd, 2008, 01:13 AM
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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
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Old November 22nd, 2008, 01:27 AM
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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
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Old November 23rd, 2008, 10:21 AM
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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
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Old November 23rd, 2008, 12:58 PM
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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
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Old November 24th, 2008, 12:09 PM
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Thanks for the help captain black.
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