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Take different $D_i \in \mathbb{R} \rightarrow \mathbb{R}$ functions $f_1, f_2$ (i.e. $\exists x : f_1(x) \neq f_2(x)$). We have

$E[f_1(x)] = E[f_2(x)]$

Are there conditions that $f_1, f_2$ must satisfy for this to happen?

I translated this problem to integral form as $ \int_{E_1} x dg_1 = \int_{E_2} x dg_2$

$g_1, g_2$ being the probability measures of $f_1(x)$ and $f_2(x)$, which can be easily calculated and $E_i$ the corresponding domains. Now, while the domains may be different, they are "similar", so we don't want to just fix domains conveniently -- instead, we want to study $f_1$ and $f_2$. Maybe there's a measure-theoretical backdoor into this, because every lead takes me to functional equations territory, which I can't handle at all.

Disclaimer: Not a homework problem. So yes, I'll be profiting indirectly from the solution, even though it's a tiny piece to a large, mostly non-mathematical puzzle. Also, I hope I'm making myself clear and following the local etiquette. I'm not a native english person, and this is my first post on MO.

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    $\begingroup$ No conditions on $f_1$ and $f_2$ will be simpler than just the condition $E f_1 = E f_2$. It's like asking, what condition guarantees that $f_1(x) = f_2(x)$ at a particular point $x$. $\endgroup$
    – John Jiang
    Commented Sep 1, 2010 at 20:53
  • $\begingroup$ You can't handwave intractability like that. For one, there's such a thing as the theory of functional equations. Cox's theorem derives the laws of probability from an associativity condition and two functional equations, $f(f(x))=x$ and $y f\left(\frac{f(z)}{y}\right) = z f \left( \frac{f(y)}{z} \right)$. More to the point, Brouwer's fixed-point theorem implies that there's $x*$ such that $g \circ f (x*) = f(x*)$, and thus there's $f*(x)=f(x)$ at $x=f^{-1}(x)$. This can't be applied here, though, because moment conditions involve more than one point. $\endgroup$
    – user8948
    Commented Sep 2, 2010 at 0:34

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Suppose you take the uniform distribution on [-1,1], and the standard normal distribution on $\mathbf{R}$. Then both these distributions have expectation 0. Similarly, if you consider the Bernoulli random variable taking values -1 and 1 with equal probability, the it has expectation 0. All these examples are quite different in my opinion. (Okay, the uniform distribution is somewhat similar to the Bernoulli distribution, but still they differ significantly- I am leaving it a little vague here, as I am sure you know what I mean). So, it seems that there are no connections as you are hoping. Am I missing something?

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