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I am studying characteristic functions in probability theory and I am struggling to understand the following equality.

$$\int_{-\infty}^{\infty}e^{itX}dF_X(x)=\int_{-\infty}^{\infty}e^{itX}f_X(x)dx$$

Why is this transformation true?

Wikipedia states that $F_X$ is the cumulative distribution function of $X$, and the integral is of the Riemann–Stieltjes kind. But we haven't learned that yet. How do I have to understand this equity without using Riemann–Stieltjes?

What I also don't understand is what the true meaning of integrating by $dF(x)$ is.

Please explain that in terms of someone who has only visited introductions to Measure Theory, Lebesgue Integrals, and Probability Theory.

Thank you very much for your efforts!

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From the expression you have stated you may see that $dF_X(x) = f_X(x) dx$... –  Fabian Feb 7 '12 at 15:18
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Where I come from, we introduced the notation $\int f dF$, where $F$ is a cumlative distribution function to be just a shortcut for $\int f dP$ if $P$ is the probability measure corresponding to $F$, without any mention of the Rieman-Stieltjes integral (Probably to not carry two letters around for basically the same thing). Maybe your course does that, too? –  Jens Feb 8 '12 at 10:57
    
@Jens Does this really mean $\int X dF_X=\int X dP_X$ if X is a realvalued measurable function where $F_X$ is the corresponding cumulative distribution function to the probability measure $P_X$ of the random variable $X$? –  Aufwind Feb 8 '12 at 13:32
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As I stated, that was my definition for $dF$, so yeah. (Note though, that $\int XdP_X$ does not really make sense unless $X$ lives on its image, you'd usually use $\int X dP$ or $\int x dP_X(x)$. You've made the same mistake in the original question, compare your equation to the Wikipedia.) –  Jens Feb 8 '12 at 15:31
    
@Jens, sorry for the hassle, but I have to know for sure: I can treat the integrals as $\int XdP=\int xdP_X(x)=\int xdF(x)=\int xf_X(x)dx$? So you mean I have to correct $\int_{-\infty}^{\infty}e^{itX}dF_X(x)$ to $\int_{-\infty}^{\infty}e^{itx}dF_X(x)$? –  Aufwind Feb 8 '12 at 17:22
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1 Answer

up vote 2 down vote accepted

In cases in which the probability distribution has a density function, one has $$ \mathbb E(e^{itX}) = \int_{-\infty}^\infty e^{itx} f_X(x)\; dx = \int_{-\infty}^\infty e^{itx} \; dF(x). $$ For discrete probability distributions (where all the probability is accounted for by point-masses), the first integral makes no sense but the second is still valid.

For "continuous singular" probability distributions like the Cantor distribution, the first integral makes no sense but the second one does. Such distributions have no point masses. Equivalently, their cumulative distributions functions are continuous. I was surprised the first time I read that the Cantor function is continuous, but think about it carefully: it's a monotonic function and has no jumps.

The Riemann--Stieltjes integral is defined as the limit as the partition grows finer, of $$ \sum f(x^*) \; \Delta F(x) = \sum f(x^*) (F(x+\Delta x) - F(x)) $$ where $x^*$ is between $x$ and $x+\Delta x$.

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