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Math Help - random variable proof

  1. #1
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    Exclamation random variable proof

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

    THANKS
    Attached Thumbnails Attached Thumbnails random variable proof-untitled.jpg  
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  2. #2
    Grand Panjandrum
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    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:

     <br />
p(A\ge k)= \sum_{i=k}^n p(A=i)<br />

    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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  3. #3
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    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:

     <br />
t P(A \ge t)= \sum_{i=t}^n t P(A=i) \le \sum_{i=t}^n i P(A=i)<br />

    since in the sum each i is \ge t.

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

     <br />
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)<br />

    Hence:

     <br />
P(A \ge t) \le \frac{E(A)}{t}, \ \ \ t=0,1,.., n<br />

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    For the second part we can observe:

     <br />
t P(A \ge t)= \sum_{i=t}^n t P(A=i) \le \sum_{i=t}^n i P(A=i)<br />

    since in the sum each i is \ge t.

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

     <br />
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)<br />

    Hence:

     <br />
P(A \ge t) \le \frac{E(A)}{t}, \ \ \ t=0,1,.., n<br />

    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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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by maths_123 View Post
    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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  6. #6
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    Thumbs up

    Thanks for the help captain black.
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