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Why is E(X^3) = E(X) = 0, if X is symmetrically distributed about 0?

  1. Sep 20, 2012 #1
    This is not a homework problem but is rather something I'm curious about. I apologize if the answer is very simple, but I am having trouble coming up with an absolute and strict proof.

    * X is a discrete random variable that is symmetrically distributed about 0. Hence, E(X) = 0
    * Why is E(X^3) = 0 but E(X^2) is not necessarily zero?
    * My intuition is that this is because X^3 and X are both odd functions. How would I write this out in equation form? Do I have to apply the law of total expectation (aka. law of iterated expectations)?

    Any thoughts and suggestions would be appreciated. Thank you!
  2. jcsd
  3. Sep 20, 2012 #2


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    Your intuition is correct. Consider how you calculate the expected value of, say, [itex]X^3[/itex]:

    [tex]E[X^3] = \sum_{n}n^3 p(n)[/tex]
    where [itex]p(n)[/itex] is the probability mass function for X. Now, [itex]n^3[/itex] is an odd function of [itex]n[/itex]. What kind of function is [itex]p(n)[/itex]?

    By the way, your claim "X is a discrete random variable that is symmetrically distributed about 0. Hence, E(X) = 0" is not always true unless you have an additional constraint on the distribution. For example, if [itex]p(n) = k/n^2[/itex] for [itex]n \neq 0[/itex] and [itex]p(0) = 0[/itex] and [itex]k[/itex] is chosen to make [itex]p(n)[/itex] sum to 1, then [itex]p(n)[/itex] is a perfectly valid and symmetric probability mass function, but [itex]E[X][/itex], and all higher moments, do not exist. Loosely speaking, [itex]p(n)[/itex]'s tails are too heavy.
  4. Sep 20, 2012 #3


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    P.S. However, it is true that if the random variable is symmetrically distributed about zero and its k'th moment exists (for odd k), then that moment must be zero.
  5. Sep 20, 2012 #4
    Thank you for your help! I'm still a little confused about some things, which I'll go through.

    [itex]p(n)[/itex] is an even function since probabilities are positive. Would it be possible for me to derive [itex]E(X^3) = E(X) = 0[/itex] without using the summation formula you noted?

    Does the simple example that I noted also require this additional constraint? Does it have an implicit constraint?

    How can I prove this condition? I think this would help me concretize my intuition and prove that [itex]E(X^3) = E(X) = 0[/itex]. I think that essentially, I'm trying to prove that:

    [tex]\sum_{x}x^3 = \sum_{x}x[/tex]

    Given that [itex]p(x)[/itex] is constant and x is symmetrically distributed around 0. If this is the right thing to be proving, how would I approach the proof?

    Thank you again for your help!
  6. Sep 20, 2012 #5


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    Well, the summation formula I noted is just the definition of expected value for a discrete random variable. What definition are you using?

    You didn't mention what probability distribution (aka probability mass function if the random variable is discrete) is being assumed for X, other than that it is symmetric. The example I gave is a discrete random variable with a symmetric distribution, which does not have any moments. (The sum doesn't converge.)

    The following is not true without further constraint on the distribution: "A discrete random variable with symmetric distribution has E[X] = 0." There are actually two possibilities: either E[X] does not exist, or E[X] = 0. Same is true for E[X^3] and all the other odd moments.

    If you have a "nice" symmetric distribution, i.e., one that decays quickly enough as [itex]|n| \rightarrow \infty[/itex], then the moments will exist and the odd moments will be zero. A sufficient condition for all of the moments to exist is that the distribution decays more rapidly than (1/q(n)) where q is any polynomial. Certainly any probability mass function that is zero outside of some finite interval will be OK, as will any distribution with exponential decay, such as a Gaussian.

    No, those sums are not expected values. You have omitted the probability mass function (pmf). The definition of [itex]E[X][/itex] is
    [tex]E[X] = \sum_{n} n p(n)[/tex]
    and the definition of [itex]E[X^3][/itex] is
    [tex]E[X^3] = \sum_{n} n^3 p(n)[/tex]
    where [itex]p(n)[/itex] is the probability that [itex]X = n[/itex], i.e., the pmf.
    Last edited: Sep 20, 2012
  7. Sep 20, 2012 #6
    Thank you again for your help. I'm sorry if my questions are really basic. I just started learning about basic probability and statistics, so I'm a bit weak on general concepts and proofs.

    Thank you for clarifying this. I'm not trying to prove this for a specific probability distribution. All I know is that X is a discrete random variable that is symmetric around 0, and I'm trying to come up with a general proof that [itex]E(X) = E(X^3)[/itex]. Would this be possible? My friend brought up this problem and said that it is possible to prove this using just the information provided, although he didn't get a chance to explain further.

    I'm not quite sure what you mean by "decay". Could you please explain? How would this apply to the distribution that I mentioned in this example?

    I think I understand what you're saying here. What I'm still confused about is how I can prove that for my specific case,

    [tex]\sum_{n} n p(n) = \sum_{n} n^3 p(n)[/tex]
  8. Sep 20, 2012 #7


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    I guess I'm confused by these two, seemingly contradictory statements:
    You didn't mention a particular distribution in this example. All you said is that it is symmetric. And as I explained above, that is not enough to guarantee the result you want. The reason it is not enough is that the moments can fail to exist for some symmetric distributions. I gave an example in my first response.

    So in other words, if the theorem you are trying to prove is "if X has a symmetric distribution, then E[X] = 0 and E[X^3] = 0", then the theorem is false.

    However, if you add an additional constraint like so: "if X has a symmetric distribution, and E[X] and E[X^3] exist, then E[X] = 0 and E[X^3] = 0," then the theorem is true.

    To see how you might prove the theorem, let's return to the definition of E[X]:

    [tex]E[X] = \sum_n n p(n)[/tex]
    Now, [itex]n[/itex] is an odd function of [itex]n[/itex], and [itex]p(n)[/itex] is an even function of [itex]n[/itex] (definition of symmetric). What can you say about the product of an odd and an even function?
  9. Sep 20, 2012 #8
    Thank you for the clarification. Let the constraint be that E[X] and E[X^3] exist for X, which is a symmetric (center 0) discrete RV. I don't know any more information about the specifics of the distribution.

    The product of an odd and an even function is an odd function. Hence, E[X] is an odd function. The same is true for E[X^3], which would also be an odd function [since [itex]n^3[/itex] is an odd function but [itex]p(n)[/itex] is an even function]. Where do I go from here? Intuitively I can see the answer, but I don't know how to prove that given that both E[X] and E[X^3] exist, E[X^3] = E[X] = 0.
  10. Sep 20, 2012 #9


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    No, [itex]E[X][/itex] is not a function of [itex]n[/itex] at all. It is the SUM of the function [itex]f[/itex] defined by [itex]f(n) = n p(n)[/itex]. Since [itex]f[/itex] is the product of an odd and an even function, [itex]f[/itex] is an odd function. So the problem reduces to proving that the sum of an odd function is zero, i.e.
    [tex]E[X] = \sum_{n = -\infty}^{\infty} f(n) = 0[/tex]
    Here is where we need to assume that the moment exists. Saying that the moment exists is equivalent to saying that
    [tex]\sum_{n = -\infty}^{\infty} f(n)[/tex]
    converges. As soon as we know that, we are free to break it up as follows:
    [tex]\sum_{n = -\infty}^{\infty} f(n) = \sum_{n = -\infty}^{-1} f(n) + f(0) + \sum_{n = 1}^{\infty}f(n)[/tex]
    and we are safe in the knowledge that both sums on the right hand side are FINITE numbers.

    Now you need to take advantage of the fact that [itex]f[/itex] is an odd function. I'll let you handle that part, except I'll give you the hint that [itex]f(0) = 0[/itex] for any odd function. (Why?)
    Last edited: Sep 20, 2012
  11. Sep 20, 2012 #10
    [itex]f(0) = 0[/itex] for any odd function because the only number that does not change when it is multiplied by -1 is 0:

    [itex]f(-x) = -f(x)[/itex]
    [itex]f(-0) = -f(0)[/itex]
    [itex]f(0) = - f(0)[/itex]

    Since [itex]f[/itex] is an odd function,
    [tex]\sum_{n = -\infty}^{-1} f(n) = \sum_{n = 1}^{\infty}f(n)[/tex]

    [tex]\sum_{n = -\infty}^{\infty} f(n) = \sum_{n = -\infty}^{-1} f(n) + f(0) + \sum_{n = 1}^{\infty}f(n)[/tex]

    [tex]= \sum_{n = -\infty}^{-1} f(n) + 0 - \sum_{n = -\infty}^{-1} f(n)[/tex]

    [tex] = 0[/tex]

    Is this the proof for E(X)? And then, the same would follow for E(X^3):
    [tex]\sum_{n = -\infty}^{\infty} x^3 = \sum_{n = -\infty}^{-1} x^3 + f(0) - \sum_{n = -\infty}^{-1} x^3[/tex]

    What I'm curious about is why you set the limits as running from [-∞ to -1] and then from [1 to ∞]. X is a discrete RV but instead of 1 and -1, couldn't we have just used any numbers as long as we had f(0) as the middle term on the RHS?

    Why is this important?
  12. Sep 20, 2012 #11


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    You're missing a minus sign here, but you got it right in the rest of the proof.
    [tex]\sum_{n = -\infty}^{-1} f(n) = \sum_{n = 1}^{\infty} f(-n) = - \sum_{n = 1}^{\infty} f(n)[/tex]
    Yes, same exact proof. Although you should be summing over [itex]f(n) = n^3p(n)[/itex], not [itex]x^3[/itex].

    Well, the sum has to include all of the integers. You can split it up anywhere you like, but the most convenient way, if you want to apply the fact that the function is odd, is to split it at zero. And I didn't want to include zero in one of the sums, because then it will be in only one of the sums and I won't be able to make it cancel with the other one.
    (i.e. why is it important that the sums are finite numbers) - It's important because otherwise you would have an expression involving [itex]\infty - \infty[/itex], which is undefined, and is NOT zero.
  13. Sep 21, 2012 #12

    Ray Vickson

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    You say "I don't know any more information about the specifics of the distribution." Sure you do: you know it is discrete and symmetric, so for ANY odd function f(x) you have
    Ef(X) = 0, because
    [tex] E f(X)= \sum_{x} p(x) f(x) = p(0)f(0) + \sum_{x>0} p(x) f(x) + \sum_{x < 0} p(x) f(x).[/tex]
    By symmetry, for every x > 0 there is an x = -|x| < 0 with the same probability, so you can write
    [tex] E f(X) = p(0) f(0) + \sum_{x>0} [p(x) -p(-x)] f(x) = 0[/tex] because f(0) = 0, f(-x) = -f(x) and p(x) = p(-x) for all x > 0.

  14. Sep 21, 2012 #13
    jbunniii and Ray, thank you again for all of your help! I understand the proof now.
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