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

  • Thread starter slakedlime
  • Start date
  • Tags
    Distributed
In summary, the conversation discusses the properties of a discrete random variable X that is symmetrically distributed about 0. The expected value of X is 0, but the expected value of X^3 is also 0, while the expected value of X^2 is not necessarily 0. This is because X^3 and X are both odd functions. The conversation also touches on the law of total expectation and the existence of moments for a symmetric distribution. The proof for E(X) = E(X^3) is derived using the definition of expected value and the probability mass function.
  • #1
slakedlime
76
2
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!
 
Physics news on Phys.org
  • #2
slakedlime said:
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.
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.
 
  • #3
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.
 
  • #4
Thank you for your help! I'm still a little confused about some things, which I'll go through.

jbunniii said:
What kind of function is [itex]p(n)[/itex]?

[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?

jbunniii said:
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.

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

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.

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!
 
  • #5
slakedlime said:
[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?
Well, the summation formula I noted is just the definition of expected value for a discrete random variable. What definition are you using?

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

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]
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:
  • #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.

jbunniii said:
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.)

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.

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.

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?

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.

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]
 
  • #7
I guess I'm confused by these two, seemingly contradictory statements:
slakedlime said:
I'm not trying to prove this for a specific probability distribution.
How would this apply to the distribution that I mentioned in this example?
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?
 
  • #8
jbunniii said:
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?

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.
 
  • #9
slakedlime said:
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.
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:
  • #10
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?)

[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]

jbunniii said:
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]

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

Hence,
[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?

we are safe in the knowledge that both sums on the right hand side are FINITE numbers

Why is this important?
 
  • #11
slakedlime said:
[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]
Correct.
Since [itex]f[/itex] is an odd function,
[tex]\sum_{n = -\infty}^{-1} f(n) = \sum_{n = 1}^{\infty}f(n)[/tex]
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]
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]
Yes, same exact proof. Although you should be summing over [itex]f(n) = n^3p(n)[/itex], not [itex]x^3[/itex].

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?
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.
Why is this important?
(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.
 
  • #12
slakedlime said:
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.

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.

RGV
 
  • #13
jbunniii and Ray, thank you again for all of your help! I understand the proof now.
 

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

1. Why is the expected value of X^3 equal to 0 if X is symmetrically distributed about 0?

The expected value of a random variable X is defined as the sum of all possible outcomes of X, each multiplied by its respective probability. For a symmetric distribution, the probabilities of positive and negative outcomes are equal, resulting in a cancellation of positive and negative values when calculating the expected value of X^3. This cancellation results in an expected value of 0.

2. How does symmetry affect the expected value of X^3?

Symmetry in a distribution means that the probabilities of positive and negative outcomes are equal. This results in a cancellation of positive and negative values when calculating the expected value of X^3, leading to an expected value of 0.

3. Can the expected value of X be anything other than 0 if X is symmetrically distributed about 0?

Yes, the expected value of X can be any value other than 0. The expected value of X is only equal to 0 when calculating the expected value of X^3. In other cases, the expected value of X will depend on the specific distribution and its parameters.

4. How does the symmetry of X affect the shape of the distribution?

Symmetry in a distribution means that the probabilities of positive and negative outcomes are equal. This results in a distribution that is centered around 0 and has a symmetrical shape, with the same number of data points on either side of 0. This symmetry can be observed in visual representations of the distribution, such as a histogram or a probability density function.

5. Is it possible for X to be symmetrically distributed about 0 and still have a non-zero expected value?

Yes, it is possible for X to be symmetrically distributed about 0 and have a non-zero expected value. The symmetry of the distribution does not necessarily mean that the expected value will be 0. It is important to consider the specific distribution and its parameters when calculating the expected value of X.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
212
  • Calculus and Beyond Homework Help
Replies
7
Views
3K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
200
  • Calculus and Beyond Homework Help
Replies
13
Views
2K
Back
Top