Approximate Moment Conditions

Question

It is known in classical probability that if two random variables $X$ and $Y$ obeys $$\mathbb{E} X^k = \mathbb{E}Y^k, \ \forall \ k \geq 1$$ with additional condition that $\mathbb{E}X^k$ does not grow too fast, then $X$ is equal to $Y$ in distribution. I was wondering that has anybody studied the same problem with the equality of moments replaced by approximation? Specifically, suppose $$|\mathbb{E} X^k - \mathbb{E}Y^k| \le \alpha_k, \ \forall \ k \geq 1$$ for some small $\alpha_k$, can we give a descent bound of $$|\mathbb{P}(X \le t) - \mathbb{P}(Y \le t)|$$ for some $t$ or uniformly in $t$ in terms of the gaps $\alpha_k$. We may assume that both $X$ and $Y$ are supported in $[0, 1]$.

Many thanks!

John

Answer 1

Here is one approach you can use. The differences $a_k$ provide you with bounds on the difference between the characteristic functions of the two distributions. This may be easy or not depending on your example. Then you can apply known bounds on the difference between two distributions based on the difference between their characteristic functions, for example Lemma 2 in Chapter XVI Section 3 of Feller, An Introduction to Probability Theory and its Applications, Vol II.

Answer 2

Here's a very simple example which illustrates one of the difficulties of this approach.

Consider $t=1/2$ and random variables $X$ which is $1/2$ with probability $1$ and $Y_\epsilon$ which is $1/2 + \epsilon$ with probability $1$, where $\epsilon > 0$. We have $P(X \le 1/2) = 1$ while $P(Y_\epsilon \le 1/2) = 0$. On the other hand, $E X^k = (1/2)^k$ and $E Y_\epsilon^k = (1/2 + \epsilon)^k$. In order to be able to rule out $P(Y \le 1/2) = 0$, you'll need that for every $\epsilon > 0$ there is some $k$ with $\alpha_k < (1/2 + \epsilon)^k - 1/2^k$, i.e. $$ \inf_k \left((\alpha_k + 1/2^k)^{1/k} - 1/2\right) \le 0$$ In particular, you need control of infinitely many moments to get any decent bound.