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Let $G: \mathbb{R}\rightarrow [0,1]$ be a cumulative distribution function (CDF) symmetric about zero, i.e., $G(x)=1-G(-x)$ at each $x\in \mathbb{R}$.

Take some real numbers $\mu_1,\mu_2$.

Consider the two CDFs

(1) $F_1:x\mapsto G(x-\mu_1)$

(2) $F_2: x\mapsto G(x-\mu_2)$

The functions (1) and (2) are still symmetric but not about zero (because of the shifting by $\mu_1,\mu_2$)

Question: suppose $\mu_1\neq \mu_2$. Can we say something about how the moments of $F_1,F_2$ will differ? In other words, is there a deterministic relation (depending on $\mu_1,\mu_2$) between the moments of $F_1$ and the moments of $F_2$?

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  • $\begingroup$ $G(x)=1-G(-x)$ is not symmetric about $0$. $\endgroup$ – Sauhard Sharma Dec 28 '18 at 10:37
  • $\begingroup$ I think it is. see here e.g., arxiv.org/pdf/0708.0499.pdf $\endgroup$ – STF Dec 28 '18 at 10:45
  • $\begingroup$ The very trivial relationship will be they have the identical central moments. For their moment about origin, it can be in terms of $\mu_1, \mu_2$, and other central moments, e.g. $E[X_i^2] = Var[X_i] + \mu_i^2$, therefore $E[X_1^2] - E[X_2^2] = \mu_1^2 - \mu_2^2$. Not sure if you want those relationship. Note that we have not used the symmetric condition here yet. $\endgroup$ – BGM Dec 28 '18 at 12:35
  • $\begingroup$ @BGM Thanks. I'm not sure I got your claim: (1) we are assuming $\mu_1\neq \mu_2$ (2) let $m_1\equiv \int t dF_1(t)$ and $m_2\equiv \int t dF_2(t)$ be the two first moments: is there any relation between $m_1$ and $m_2$? If yes, which one? (3) consider now the second central moments $v_1\equiv \int (t-m_1)^2 dF_1(t)$ and $v_2\equiv \int (t-m_2)^2 dF_2(t)$: is there any relation between $v_1$ and $v_2$? If yes, which one? $\endgroup$ – STF Dec 28 '18 at 13:45
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    $\begingroup$ What I claiming is the trivial fact that $X_1 - \mu_1$ and $X_2 - \mu_2$ has the same distribution as the original one without shift. Therefore the central moments should be equal: $E[(X_1 - \mu_1)^n)] = E[(X_2 - \mu_2)^2]$ whenever they exist. From this you can expand it and manipulate them to obtain some relationship between the moments about the origin. The symmetric assumption may help you to deal with the odd central moments. $\endgroup$ – BGM Dec 28 '18 at 15:04

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