+ Reply to Thread
Results 1 to 10 of 10

Thread: Chebyshev's Inequality problem

  1. #1
    Points: 5,567, Level: 48
    Level completed: 9%, Points required for next Level: 183

    Location
    Cincinnati, OHIO
    Posts
    48
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Chebyshev's Inequality problem




    Let X ~ Poisson (lambda). With Chebyshev's Inequality, show that

    P(X >=2 lambda) <= 1/lambda

    The way I thought to interpret this theorem was to determine what percentage of the data is clustered near the mean when faced with a non-normal distribution like a Poisson distribution that determines the probability of a given number of events within a fixed interval. This is where I get tripped up when attempting to lay out the proof with a symbol (lambda) as opposed to a number. Any assistance would be greatly appreciated. Thanks.

  2. #2
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem

    Note that the mean and standard deviation of the Poisson random variable are both lambda. So, look back at Chebyshev and see where these values fit in.

    P(|x-mu|<k*(sigma)) >= 1-(1/k^2)

    If we subtract this probability from 1 we should get P([-k*(sigma) >= x-mu] or [x-mu >= k*(sigma)]) ...

    Excuse my excessive use of brackets.
    Last edited by Buckeye; 09-27-2016 at 06:29 AM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  3. #3
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem

    Pardon me, the variance is lambda
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  4. #4
    Points: 5,567, Level: 48
    Level completed: 9%, Points required for next Level: 183

    Location
    Cincinnati, OHIO
    Posts
    48
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Re: Chebyshev's Inequality problem

    I thought the formula stated for r.v. X with E(X) = mu and alpha > 0 ,

    P[|x - mu| >= alpha] <= Var(X)/alpha^2

    This is where I get confused about plugging in the Poisson distribution characteristics.

    Please clarify. Thanks.

  5. #5
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem

    I think you are very close. There are a few ways to see the same inequality https://en.wikipedia.org/wiki/Chebyshev%27s_inequality

    Look under the section "Probabilistic statement"
    Perhaps yours is correct. I haven't fiddled around with it too much

    Our goal is to get the absolute value bars out of the problem. So, we need to restate/rearrange this inequality to deal with that. If you want, you can treat the lambda as a number. Then switch back later.
    Last edited by Buckeye; 09-27-2016 at 02:36 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  6. #6
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem

    Let alpha = lambda in your version. almost qed
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  7. #7
    Points: 5,567, Level: 48
    Level completed: 9%, Points required for next Level: 183

    Location
    Cincinnati, OHIO
    Posts
    48
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Re: Chebyshev's Inequality problem

    Ok, P(|X - mu| >=k*sigma) <= 1/k^2

    So if the variance for Poission is lambda, let's plug in.

    P(|X-mu| >=2(lambda)k) <= 1/k^2

    P(|X-mu| <= (lambda)k) >= 1- 1/k^2

    P(-k(lambda) >= X - mu

    Let me know if I did the proof right.

    Thanks.

  8. #8
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem

    Quote Originally Posted by greg6363 View Post
    P[|x - mu| >= alpha] <= Var(X)/alpha^2
    I was thinking: P(|X-\lambda|\geq\lambda)\leq\frac{\lambda}{\lambda^{2}}

    P.S. I hope \lambda is your favorite Greek letter.
    Last edited by Buckeye; 09-27-2016 at 07:03 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

  9. #9
    Points: 5,567, Level: 48
    Level completed: 9%, Points required for next Level: 183

    Location
    Cincinnati, OHIO
    Posts
    48
    Thanks
    0
    Thanked 0 Times in 0 Posts

    Re: Chebyshev's Inequality problem

    Here goes.

    P(|X - lambda| >= lambda) <= lambda/lambda^2

    P(|X - lambda| >= 2 lambda) <= 1/lambda

    P(|X - lambda| <= lambda >= 1 - 1/lambda

    P(-lambda >= X - lambda)

    Let me know. Thanks.

  10. #10
    Points: 1,821, Level: 25
    Level completed: 21%, Points required for next Level: 79
    Buckeye's Avatar
    Location
    Ohio
    Posts
    102
    Thanks
    31
    Thanked 4 Times in 4 Posts

    Re: Chebyshev's Inequality problem


    |X-\lambda|\geq\lambda is equivalent to X-\lambda\geq\lambda or X-\lambda\leq-\lambda
    Last edited by Buckeye; 09-27-2016 at 09:51 PM.
    "I have discovered a truly remarkable proof of this theorem which this margin is too narrow to contain." Pierre de Fermat

+ 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