1
$\begingroup$

Say you have two (bounded) random variables, $X$ and $Y$, on the same discrete probability space, such that $E(X)=E(Y)$ but $Var(X) < Var(Y)$. Do I know that, for any $k \geq 0$, $$ \text{Prob}(|X-E(X)|\geq k) \leq \text{Prob}(|Y - E(Y)|\geq k). $$ Note that this is no duplicate of this question.

I know Chebyshev's inequality relates the above probabilities to the variance, but it doesn't say anything about the above, at least to my understanding.

Finally, if the above is true, how does it change for $|E(X) - E(Y)|$ being small but strictly positive?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ It is true. However, for different expected values, it is not. $\endgroup$ – Carl Feb 2 '19 at 14:18
  • $\begingroup$ It's easy to come up with a counter example. Just choose $X$ and $Y$ to have different kurtosis. It's possible to pick $X$ as a leptokurtic distribution so that the tails are heavy enough that a large value of $k$ produces an imbalance. $\endgroup$ – AdamO Feb 2 '19 at 14:27
1
$\begingroup$

Take $X$ to have support -3, 0, 3 with probabilities 0.1, 0.8, and 0.1. Then the expectation of $X$ is 0 and the variance of $X$ is 1.8.

Take $Y$ to have support -2, and 2 with probabilities 0.5, and 0.5. The variance of $Y$ is 4.

Pr(Y > 2) = 0. Pr(X > 2) > 0. So the statement is not true.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks @AdamO, that makes a lot of sense. I guess I'm actually concerned with the special case of log-likelihood, i.e. when $Prob(X= log(p_i)) = p_i$ (and hence the mean is the negative of the Shannon entropy). In this case, it's not clear to me that a counter-example can be constructed in the fashion you propose. But I guess I should add a new question about this. $\endgroup$ – Paul Feb 2 '19 at 17:09
1
$\begingroup$

As I understand the question in a current form, the question is about whether the inequality holds for every $k$.

Then it is certainly not true.

Consider two random variables $X$ and $Y$ on discrete space $\{1, \dots, 9\}$

Here is a histogram of $X$

enter image description here

And here is a histogram of $Y$

enter image description here

Since $Y$ has fatter tails, it has higher variance.

And here is the important plot of deviation from mean for various $k$

where $X$ is denoted as blue and $Y$ as red. We can see the inequality doesn't hold for $k = 1$

enter image description here

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks, Lukasz. I chose AdamO's answer as the correct one, because it's clear and concise and the variances can be calculated. But this was also very helpful! $\endgroup$ – Paul Feb 2 '19 at 17:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.