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Lets have $f(t)$ be this characteristic function. I am told that $f'(t)=-t \cdot f(t)$ and that this can be proven, I found using partial integration and the dominated convergence theorem. I am aware of these theorems. I unfortunately do not know how to do this. The attempts tried are not of value worth sharing.

$$X:N(0,1)$$ $$f(t)=\int_{-\infty}^{\infty}e^{itx}\frac{1}{2\pi}e^{-\frac{x^2}{2}}dx$$

Bozo Vulicevic
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  • I don't understand your question. The second function you have is just $ce^{itx}$ with a real number $c$, and it does not satisfy $f'(t) = t f(t)$. Rather, the only functions that do are of the form $f(t) = c e^{t^2/2}$. –  Jul 18 '16 at 20:46
  • the question is proving that $f(t)=ce^{-\frac{t^2}{2}}$ – Bozo Vulicevic Jul 18 '16 at 21:01
  • Check out what Byron Schmuland commented here, in the answer: http://math.stackexchange.com/questions/29299/characteristic-function-of-the-normal-distribution – Bozo Vulicevic Jul 18 '16 at 21:02
  • See this answer: http://math.stackexchange.com/a/270580/36150 – saz Jul 20 '16 at 06:44
  • "The attempts tried are not of value worth sharing." Hmmm... – Did Jul 21 '16 at 19:12

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