+ Reply to Thread
Results 1 to 6 of 6

Thread: Bounded in Probability proof

  1. #1
    Points: 740, Level: 14
    Level completed: 40%, Points required for next Level: 60

    Posts
    31
    Thanks
    11
    Thanked 1 Time in 1 Post

    Bounded in Probability proof




    Hi all,

    I could use some help understanding the following proof so please feel free to pitch in!

    Consider any random variable X with distribution function F_X(x). Then given \epsilon >0 we can bound X in the following way. Because the lower limit of F_X(x) is 0 and its upper limit is 1, we can find y_1\ and\ y_2 such that:

    F_X(x)<\epsilon/2 for x\leq y_1 and F_X(x)>1-\epsilon/2 for x\geq y_2

    Let y=max\left\{|y_1|,|y_2|\right\} then

    \displaystyle P[|X|\leq y]=F_X(y)-F_X(-y-0)\geq 1-\epsilon/2-\epsilon/2=1-\epsilon

    What I need help understanding is where the inequality in the last line comes from precisely. I know it has something to do with the bounds we set up above but I cannot figure it. I am a beginner so please explain in detail. Thanks.

  2. #2
    TS Contributor
    Points: 12,501, Level: 73
    Level completed: 13%, Points required for next Level: 349

    Posts
    951
    Thanks
    0
    Thanked 103 Times in 100 Posts

    Re: Bounded in Probability proof

    Note that
    1a) y \geq y_1, so -y \leq -y_1
    Thus,
    2a) F_X(-y) < \epsilon/2, or equivalently -F_X(-y) > -\epsilon/2

    On the other hand,
    1b) y \geq y_2
    Thus,
    2b) F_X(y) > 1-\epsilon/2

    Add (2a) and (2b)
    All things are known because we want to believe in them.

  3. The Following User Says Thank You to Mean Joe For This Useful Post:

    JohnK (10-02-2013)

  4. #3
    Points: 740, Level: 14
    Level completed: 40%, Points required for next Level: 60

    Posts
    31
    Thanks
    11
    Thanked 1 Time in 1 Post

    Re: Bounded in Probability proof

    Quote Originally Posted by Mean Joe View Post
    Note that
    1a) y \geq y_1, so -y \leq -y_1
    Thus,
    2a) F_X(-y) < \epsilon/2, or equivalently -F_X(-y) > -\epsilon/2

    On the other hand,
    1b) y \geq y_2
    Thus,
    2b) F_X(y) > 1-\epsilon/2

    Add (2a) and (2b)
    Wow, thanks a lot. One thing though, in 1a) don't you mean -y\leq y_1? That is also implied by the inequality and I think this is the result we need.
    Last edited by JohnK; 10-02-2013 at 07:18 AM.

  5. #4
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Bounded in Probability proof

    I think it should be like

    -y \leq -|y_1| \leq y_1

  6. The Following User Says Thank You to BGM For This Useful Post:

    JohnK (10-02-2013)

  7. #5
    Points: 740, Level: 14
    Level completed: 40%, Points required for next Level: 60

    Posts
    31
    Thanks
    11
    Thanked 1 Time in 1 Post

    Re: Bounded in Probability proof

    Quote Originally Posted by BGM View Post
    I think it should be like

    -y \leq -|y_1| \leq y_1
    Why the absolute value in y_1?

  8. #6
    TS Contributor
    Points: 22,410, Level: 93
    Level completed: 6%, Points required for next Level: 940

    Posts
    3,020
    Thanks
    12
    Thanked 565 Times in 537 Posts

    Re: Bounded in Probability proof


    It is because

    y = \max\{|y_1|, |y_2|\} \geq |y_1| \Rightarrow -y \leq -|y_1|

  9. The Following User Says Thank You to BGM For This Useful Post:

    JohnK (10-02-2013)

+ Reply to Thread

           




Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts






Advertise on Talk Stats