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Expected Value of reciprocal (Sorry for reposting)

  1. Nov 4, 2011 #1
    Hi all
    Sorry for reposting, the previous post wasn't clear enough, it's my mistake, I'll make the question more clear, I found lot of people asking the same question on the Internet.

    Given that X is random variable that takes values:

    0..............H-1

    The PMF of X is unknown, but I can tell what is the expected value which is [itex]\bar{X}[/itex]

    There is event Y when calculated it gives the value:

    [itex]P(Y)=E[\frac{1}{X+1}][/itex]

    The QUESTION: Is there a way to find expected value [itex]\bar{Y}[/itex] in the terms of [itex]\bar{X}[/itex]? regarding that: the PMF of X is unknown we know just the expected value.

    It's wrong to say that (just if you can confirm it will be great):
    [itex] E[\frac{1}{X+1}]=\frac{1}{E[X]+1}[/itex]
    Thanks and sorry for repost
     
    Last edited by a moderator: Apr 5, 2012
  2. jcsd
  3. Nov 4, 2011 #2

    mathman

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    You need the distribution function for X (the mean is not enough) to get the mean of 1/(X+1).
     
  4. Nov 5, 2011 #3
    thanks apparently I do
     
  5. Nov 6, 2011 #4
    If X is strictly positive, you can apply Jensen's inequality etc. to get 1 >= E[1/(X+1)] >= 1/(E[X]+1).
     
  6. Nov 7, 2011 #5

    Stephen Tashi

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    What does that notation mean? Is Y some event ( like "A red bird lights on the window") and P(Y) is its probability?
     
  7. Apr 5, 2012 #6
    hi giglamesh, have you had the answer so far? Im having exactly problem like you
     
  8. Apr 5, 2012 #7
    hi all
    yes P(Y) is another event which probability is the expected value of other function of random variable.
    Applying Jenesen Inequality does not help because it gives the lower bound.
    So I decided to work on the problem to get X distribution to calculate the E[1/(1+X)]

    but few days later I modified the problem to another distribution P(Y)=E[1/X] in another post.
    Greetings
     
  9. Apr 5, 2012 #8
    did you get the answer for E(1/X) as well?
     
    Last edited by a moderator: Apr 5, 2012
  10. Apr 5, 2012 #9
    yes just find the distribution of X, the PMF (discret case)
    then calulate the probability like this:
    P(Y)=E[1/X]=sum_{i=1}^{i=n}{(1/i)*P(X=i)}
    using Jenesen inequality here doesn't help because the funtion is defined to be 0 at 0 so we can't consider it convex.
    hope that would help
     
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