Timeline for Question about characteristic function with independence assumption

7 events

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Apr 22, 2014 at 13:04	comment	added	Thomas Rippl		@Zeno44: again you are right. Edited the answer accordingly.
Apr 22, 2014 at 13:03	history	edited	Thomas Rippl	CC BY-SA 3.0	corrected the statement according to zeno44's idea. structure is clearer now.
Apr 20, 2014 at 19:10	comment	added	user45183		but "$a^n-b^n$ equals zero (for real $a$ and $b$) only when $a$ and $b$ are equal" clearly is not true. Consider $a=1, b=1$ and $n=2$. This is exactly my counter example in my comment above :)
Apr 19, 2014 at 14:31	history	edited	Thomas Rippl	CC BY-SA 3.0	corrected according to zeno44's suggestion.
Apr 19, 2014 at 14:28	comment	added	Thomas Rippl		@Zeno44 you are right. I corrected Claim 1 so that it can be used. a,b>0 was not necessary, only that $a$ and $b$ are not the same. $a^n-b^n$ equals zero (for real $a$ and $b$) only when $a$ and $b$ are equal.
Apr 17, 2014 at 13:07	comment	added	user45183		But your proof of the second claim has a mistake. You say at the end that you can use Claim 1. But what if $\frac{v_2}{v_1}$ is negative? Claim 1 only considers positive $a,b >0$. Here is an example: Consider $X_1 \sim N(0,\sigma_1^2)$ and $X_2 \sim N(0,\sigma_2^2)$ as well as $v = (1,1)$ and $w=(1,-1)$. Then $\langle v,X_1 \rangle = X_1 + X_2 \sim N(0,\sigma_1^2 + \sigma_2^2)$ and $\langle w,X_1 \rangle = X_1 + X_2 \sim N(0,\sigma_1^2 + \sigma_2^2)$. You cannot reconstruct $\sigma_1$ and $\sigma_2$ uniquely here.
Apr 14, 2014 at 20:52	history	answered	Thomas Rippl	CC BY-SA 3.0