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Show that the Poisson Mass Function has the same value for λ and λ-1

  1. Apr 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that, quite generally, the Poisson Mass Function has the same value at λ (i.e. the average) and at (λ-1).

    2. Relevant equations
    The Poisson Mass Function is
    p(x) = [e^(-λ) * λ^(x)]/(x!)

    3. The attempt at a solution
    I started out by plugging in λ-1 into the equation to get p(x)= [e^(-λ-1) * (λ-1)^(x)]/(x!) but now I've gotten stuck. I'm not sure how to progress and prove this other than plug in different x and λ values. I've seen the graphs and such that indicate that λ and λ-1 share the same value, but I just can't figure out how to mathematically prove it.
     
  2. jcsd
  3. Apr 11, 2016 #2

    Charles Link

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    If I'm not mistaken, ## \lambda ## does not change in what you are trying to show. One thing you can do is let ## p(x-1)=p(x) ## and solve for ## x ##.
     
  4. Apr 11, 2016 #3
    Thank you for the advice. I'm very confused as to how this would actually show that p(x|λ) = p(x|λ-1).
     
  5. Apr 11, 2016 #4

    Charles Link

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    This one is quite simple I think. ## e ^ {-\lambda} * \lambda^ {(x-1)}/(x-1)!=e ^ {-\lambda} * \lambda^x/x! ## . A little algebra and you have the answer. (Note x is an integer here).
     
  6. Apr 11, 2016 #5
    Oh, I see where I'm confused. I'm not looking to solve for x-1, but rather λ-1. x remains x for this equation.
     
  7. Apr 12, 2016 #6
    I reread this, and so I've attached the homework question (and the mentioned graph) to better help describe what I'm getting at. I'm very lost how to prove this.http://imgur.com/NyuuHmT
     
  8. Apr 12, 2016 #7
  9. Apr 12, 2016 #8

    Charles Link

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    ## \lambda ## is a parameter that is pre-selected that gives the shape of the curve. For a given ## \lambda ## there will be two consecutive integers, x-1 and x that have the same value for ## p ## i.e. ## p(x-1)=p(x) ##. That equation, in post #4 is easy to solve. ## x!/(x-1)!=x ##. I could show you the rest, but the Forum rules want the OP to do some of the work. When you get this answer for ## x ## it tells you where this "peak" occurs, i.e. for a given pre-selected ## \lambda ##, where the ## p ## function levels off with two equal values before it drops back down. If ## p(x) ## were a continuous function, you would set ## dp(x)/dx=0 ## to find this peak. Instead, ## x ## has only integer values, so you set ## p ## equal for two consecutive integers.
     
    Last edited: Apr 12, 2016
  10. Apr 12, 2016 #9

    Ray Vickson

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    No: look at the diagram you supplied. The parameter ##\lambda## is fixed, and you are being asked to show that ##p(x|\lambda) = p(x-1|\lambda)## when ##x = \lambda##. That's all.
     
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