I saw this formula today:

$$ \mathbb{P} \left[ X > K \right] = \frac{1}{2} + \frac{1}{\pi} \int_0^\infty Re\left( \frac{e^{-iuK}\varphi(u)}{iu} \right) du $$

Where $\varphi(u)$ is the characteristic function of the random variable $X\in L^1$. I must definitely be missing something since I can only get to:

$$ \mathbb{P} \left[ X > K \right] = \frac{1}{2\pi} \int_{-\infty}^\infty \frac{e^{-iuK}\varphi(u)}{iu} du $$

Do any of you out there know how to show this? Any help would be appreciated!


1 Answer 1


I will just consider the case when $X$ has a pdf.

First let me define the Fourier transform of a function $f(x)$ to be:

$$ \hat{f}(w) = \int_{\mathbb{R}} f(x)e^{-iwx}dx $$

Note that we have the identity:

$$ \langle f, g \rangle = \frac{1}{2\pi}\langle \hat{f}, \hat{g} \rangle $$ This is one version of Parseval's identity which assumes $f,g \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ (at least in the version I'm familiar with).

Ok, now assume that $X$ has some density function $p(x)$. We're looking at: \begin{align*} \int_K^{\infty} p(x) dx &= \int_{\mathbb{R}} \chi_{[K,\infty)}(x) p(x) dx \\ &= \int_{\mathbb{R}} \chi_{[0,\infty)}(x-K) p(x) dx \end{align*} Where $\chi$ is the usual indicator function so that $\chi_{[0,\infty)}$ is the Heaviside step function. Using the above identity, we have: \begin{align*} \int_{\mathbb{R}} \chi_{[K,\infty)}(x) p(x) dx &= \frac{1}{2\pi} \int_{\mathbb{R}} \left( \frac{1}{iw} + \frac{1}{\pi}\delta(w) \right) e^{-iKw} \hat{p}(w) dw \\ &= \frac{1}{2} + \frac{1}{2\pi} \int_{\mathbb{R}} \frac{e^{-iKw} \hat{p}(w)}{iw} dw \end{align*} Note that the Fourier transform of the step is a distribution (see here for instance), and $\chi$ is certainly not in $L^1$ or $L^2$. So this isn't as airtight as it could be at this point.

Finally, consider the second term:

\begin{align*} \int_{\mathbb{R}} \frac{e^{-iKw} \hat{p}(w)}{iw} dw &= \int_0^{\infty} \frac{e^{iKw}\hat{p}(-w)}{-iw} + \frac{e^{-iKw}\hat{p}(w)}{iw} dw \\ &= 2\int_0^{\infty} Re\left( \frac{e^{-iKw}\hat{p}(w)}{iw} \right) dw \end{align*}

Since $\hat{p}(-w) = \hat{p}(w)^*$, the complex conjugate of $\hat{p}(w)$.

  • $\begingroup$ What about the case that $X$ does not have a density function? $\endgroup$
    – saz
    Commented Mar 19, 2015 at 15:56
  • $\begingroup$ hmm, yes, well I suppose a different argument would have to be made. I added a disclaimer to note that I'm just considering this case :) $\endgroup$
    – Chester
    Commented Mar 19, 2015 at 17:54

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