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Infinite mean?

  1. Jan 10, 2007 #1
    Is it possible to have a distribution of a rv with infinite mean?
    Techinically, mean is the expected value so.... if the integral/summation does not converge?
    Does anyone have a specific example of such a distribution?
    Thanks!
     
  2. jcsd
  3. Jan 10, 2007 #2

    quasar987

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    For the sake of giving you quick answer, yes, it's possible. We've seen an exemple of a continuous rv that did not have a converging mean in my prob class last semester but I lended my notes so I can't check what the density function was.
     
  4. Jan 10, 2007 #3

    Hurkyl

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    You could try constructing one yourself!

    Try working backwards. Let's do the discrete case for simplicity.

    Write down a sum you know doesn't converge, and assume that what you wrote is the sum that calculates expected value. Once you do that, it's easy to compute the weight of each individual event.

    Of course, to be a probability distribution, the sum of the weights has to be 1. Try computing it. You probably won't get 1... but if you get a finite value, do you see an easy way to modify your distribution so that it's a probability distribution?


    (more in white: highlight to see)
    -----------------------
    Well, let's try an example. I know that the sum:

    0 + 1 + 4 + 9 + 16 + ... = sum_{i = 0 .. infinity} i^2

    doesn't converge. Well, if this sum is computing the expected value, then

    sum_{i = 0 .. infinity} i^2 = sum_{i = 0 .. infinity} i * P(i)

    so, P(i) = i for all events i. Now, to see what the total weight is:

    sum_{i = 0 .. infinity} P(i) = +infinity

    Oh well, this one doesn't work. We have to try something else.

    ------------------------
     
    Last edited: Jan 10, 2007
  5. Jan 11, 2007 #4

    Fredrik

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    Here's an explicit example. A friend tells you he will give you 2^k dollars for some non-negative integer k, and that the method he will use to determine the amount is such that the probability P(2^k) that you will get 2^k dollars is 2^k/3^(k+1) for k=0,1,2,... The probabilities add up to 1:

    [tex]\sum_{k=0}^{\infty}P(2^k)=\sum_{k=0}^{\infty}\frac{2^k}{3^{k+1}}=\frac{1}{3}\cdot\frac{1}{1-\frac{2}{3}}=1[/tex]

    But the expectation value

    [tex]\sum_{k=0}^{\infty}P(2^k)2^k[/tex]

    is infinite, so your friend needs to be infinitely rich to pull this off.
     
    Last edited: Jan 11, 2007
  6. Jan 13, 2007 #5
    thanks for the help!
     
  7. Jan 21, 2007 #6
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