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I would greatly appreciate any help with this problem. If $f_1, f_2 , f_3$ are three characteristic functions (in Statistics, e.g $E(\exp(itX)))$ such that $f_1*f_3=f_2*f_3$ for all $t$ and we are given that $f_3$ is non-zero almost everywhere, how can we show that $f_1 = f_2$ for ALL $t$? It's obvious that $f_1 = f_2$ for $t$ not in the null set where $f_3 $ is non-zero, but how can we conclude that $f_1 = f_2$ for $t$ in the null set where $f_3$ is zero?

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characteristic functions are continuous in $t$, cf. en.wikipedia.org/wiki/… –  uncookedfalcon Sep 8 '12 at 7:25

1 Answer 1

(i) characteristic functions are continuous in $t$: for $t_j \rightarrow t$, we have pointwise convergence $e^{it_jX(\omega)} \rightarrow e^{it X(\omega)}$ for any $\omega$ in your sample space by the continuity of $e^{it}$; since this always has norm one, one can simply invoke Dominated Convergence to finish up.

(ii) with that, do you see why if two continuous functions on $\mathbb{R}$ agree save for possibly on a set $S$ of measure zero, they agree?

Roughly, for a point $s \in S$, for any $\epsilon$, consider $(s - \epsilon, s + \epsilon)$; this can't be all in $S$, so we get a sequence $x_i \in S^c$, with $x_i \rightarrow s$, hence $$f(s) = \lim_i f(x_i) = \lim_i f(g_i) = g(s)$$

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Thanks for clearing that up - I do already have the result that characteristic functions are continuous in R^n. I suspected that the fact they agreed almost everywhere was the key. Sorry, my real analysis is kinda shaky. Am I correct that result (ii) is true in R^n too? –  Ken Dunn Sep 8 '12 at 7:39
    
no worries man-yeah, take the ball of radius $\epsilon$ in that case too –  uncookedfalcon Sep 8 '12 at 7:41

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