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Let $Z\sim\mathcal{N}(0,1)$ (i.e. a random variable which distribution is the standard normal distribution). Determine the characteristical function of $Z$.

It is $\mathbb{P}_Z=f\lambda$ with $f(x)=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}$ and $\lambda$ the Lebesgue-mesaure. $$ \varphi_Z(t)=\int e^{itz}\, d\mathbb{P}_Z=\int e^{itz}f\, d\lambda $$ Now in a textbook I saw that they simply write $$ \int e^{itz}f\, d\lambda=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{itz}e^{-\frac{z^2}{2}}\, dz. $$

Why can they switch over to the Riemann-integral here?

And why integrating over $(-\infty,\infty)$? I do not see that $Z$ has to be a random variable on $\mathbb{R}$.

mathfemi
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1 Answers1

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The last formula is nothing more than a tautology. Recall that $d\lambda$ is the Lebesgue measure on the line. Also, the integrand function is continuous, hence Riemann integrable. Therefore the Lebesgue integral reduces to a Riemann integral.

Giuseppe Negro
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  • We had two sentences concerning the connection between Lebesgue- and Riemann-integral. (1) The Borel-measurable function $f\colon\mathbb{R}\to\mathbb{R}^+$ is Riemann-integrable on every compact intervall. Then $f\in L_{\lambda}^1\Leftrightarrow$ f is improper Riemann-integrable (2) Let $f\colon [a,b]\to\mathbb{R}$ be a Borel-measurable function. If $f$ is Riemann-integrable, then it is Lebesgue-integrable and both integrals are the same. Which is it here? (1) or (2)? – mathfemi Feb 23 '14 at 11:45
  • Could you please explain how one veryfies (2) here? I am a little bit confused, sorry. – mathfemi Feb 23 '14 at 11:47
  • Yes, you are right, actually I was wrong and you need both 1 and 2. Let's see. In our case the function $f$ is the pdf of the $N(0, 1)$ distribution times the phase term $e^{itz}$. It is a continuous function so: $$\text{(i) it is Riemann-integrable on any compact interval}.$$ Moreover, it decays exponentially, so: $$\text{(ii) it is improperly Riemann integrable on the line}.$$ Now we plug (ii) into (1) to discover that $f$ is $L^1_\lambda$. It remains to prove that the value of its Lebesgue integral coincides with the value of its improper Riemann integral. [...] – Giuseppe Negro Feb 23 '14 at 11:53
  • [...] Here's where (2) kicks in. On any compact interval, we have the equality $$\int_{[a, b]} f, d\lambda = \int_a^b f(z), dz, $$ (LHS is Lebesgue, RHS is Riemann) by means of (2). We now let $a\to -\infty $ and $b\to +\infty$. By definition of improper integral the RHS converges to $$\int_{-\infty}^\infty f(z), dz.$$ We now claim that the LHS converges too and its limit is precisely the Lebesgue integral $$\int_{\mathbb{R}} f, d\lambda.$$ To see this just rewrite the integrand in the LHS as $$f\chi_{[a, b]}$$ and apply the dominated convergence theorem, applicable by previous comment. – Giuseppe Negro Feb 23 '14 at 11:58
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    This kind of argument is standard and usually nobody bothers to do it explicitly. Hope this helps. – Giuseppe Negro Feb 23 '14 at 11:59