1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability Theory - Expected Value.

  1. Oct 8, 2005 #1
    Hey.. im having some problems with this homework question, i thought perphaps somebody can help me on it.

    -- Let X be a random variable with probability mass function
    x P(x)
    -1 p
    1 1 - p

    Find the value of the constant C not equal to 1 such that E[c^X] = 1

    So this is my work so far. We have E[X] = 1 -2p and E[c^X] = 1 ...

    E[X] = 1 -2p
    E[c^X] = 1

    then c^X = c^(1-2p) = 1 so...

    c^(1-2p) = 1

    I do the logarithms....

    (1-2p)logc = log1
    (1-2p)logc = 0
    log c = 0
    c = 1

    but that is no true, since c cannot equal 1. What am i doing wrong?

    The only other solution i can see is 1^(1-2p)^-1 = 1/(1-2p)
  2. jcsd
  3. Oct 8, 2005 #2
    sorry.. the mass funtion looks awkward above... its actually...

    x p(x)
    -1 p
    1 p-1
  4. Oct 8, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    You seem to have a completely wrong idea about what's going on here!
    No, c^X is not c^(1-2p). That would be cE(x) which is irrelevant.

    What your probability function says is that x can have one of only two values: -1 and 1, with probabililty p and 1-p respectively. That means that c^x can have one of only two values: c-1= 1/c and c, with probability p and 1-p respectively. The "expected value" is the sum of the each possible value times its probability: p/c+ c(1-p)= 1. Solve that for c in terms of p.
  5. Oct 9, 2005 #4
    Hey... I see the error I made now....
    for solving... i managed to simplyfy the equation to 0 = (1-p)c^2 - c + p

    This is the same as... 0 = (1-p)x^2 - x + p and the idea is to solve for x or p. Believe me, I tried EVERYYTHING to solve this, even the quadratic. I just keep finding my self at a ded end. I know that c=1 works just fine, but the question says that c cannot equal to one.
    Anybody have any ideas?
  6. Oct 9, 2005 #5
    Ive been stuck on this for the whole day... how on earth do i solve for c?

    (1-p)C^2 - C + p = 0 //or// p/C + C(1-p) = 1

    I just keep getting 1 as an answer and the questions says C not equal to 1
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Probability Theory - Expected Value.
  1. Probability theory. (Replies: 4)

  2. Probability Theory (Replies: 2)

  3. Probability Theory (Replies: 3)

  4. Probability theory (Replies: 3)