If two random variable satisfies $EX^n \leq EY^n$ for all $n=1,2,3,...$, can we say Y First order stochastically dominates X? i.e. $P(X<t)>P(Y<t)$ for all t

I have been thinking since we can use polynomials $\{1, x, x^2,...\}$ to approximate any continuous function. Will it leads to any sufficient condition of the first order stochastic dominance?

As one ansewr point out, for degenerate cases, this moment dominance is not sufficient. I wonder if we further require X and Y being non-negative continous random variables, will that be sufficient?

Or is there other sufficient condition to establish first order stochastic dominance?

Thank you!


Suppose $X$ is has a degenerate distribution at $x \in (0, 1)$. Suppose $Y$ is distributed as $(1-y) \circ 0 + y \circ 1$ for some $y \in (x, 1)$. Then $EX^n = x^n < y = EY^n$ but $Y$ does not first order stochastically dominates $X$.

One sufficient condition for first order stochastic dominance is that $f_Y / f_X$ is increasing, where $f_X$ and $f_Y$ are the densities of $X$ and $Y$ respectively.

One characterization of first order stochastic dominance is $Eg(X) \leq Eg(Y)$ for all increasing, integrable $g$.

  • $\begingroup$ Thank you for your reply. Yes, the moment dominance is not enough for degenerate cases like this. $\endgroup$
    – Sean2020
    Sep 1 '20 at 15:26
  • $\begingroup$ @Sean2020 The problem is not dependent on degenerate distribution. Your intuition is to approximate indicator function by polynomials. But the problem is that you can not guarantee the coefficients are positive so that the moment conditions translate into polynomial conditions. $\endgroup$
    – user295959
    Sep 1 '20 at 18:10

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.