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I have recently come across a result, which states that if $F$ is the distribution function of a random variable $X$ with characteristic function $C$, then:

$\sum\left(\nabla{F(x)}\right)^{2}=\lim_{T\to\infty}\frac{1}{2T}\int^{T}_{-T}\left|C(t)\right|^{2}\,\mathrm{d}t$

where $\nabla{F(x)}$ the $\mathbb{P}(X=x)$. Well, if I straightforward accept this, then I've proven that the integrability of $\left|C(t)\right|^{2}\Longrightarrow{F}$ is continuous! Can someone give an elementary proof of the result I've stated above (in bold-italics)?

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    I've added $LaTeX$ formatting to your question, but I am not sure if that's what you meant to write... please check to make sure.2012-10-18

1 Answers 1

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Both sides of the identity are $\mathbb P(X=Y)$, where $X$ and $Y$ are independent random variables with characteristic function $C$ and CDF $F$.

Note first that, for every $t$, $|C(t)|^2=\mathbb E(\mathrm e^{\mathrm it(X-Y)})$ hence, for any nonzero $T$, the term in the limit in the RHS is $ \mathbb E\left(\frac1{2T}\int_{-T}^T\mathrm e^{\mathrm it(X-Y)}\,\mathrm dt\right)=\mathbb E\left(\mathbf 1_{X=Y}+u(T(X-Y))\right),\qquad u(t)=\frac{\sin(t)}{t}\,\mathbf 1_{t\ne0}. $ When $T\to+\infty$, $u(T(X-Y))\to0$ almost surely, and $|u(t)|\leqslant1$ for every $t$ hence, by dominated convergence, the RHS of the identity is $\mathbb E\left(\mathbf 1_{X=Y}\right)=\mathbb P(X=Y)$.

On the other hand, let $v(x)=\mathbb P(Y=x)=\mathbb P(X=x)$ for every $x$. By independence of $X$ and $Y$, $\mathbb P(X=Y)=\mathbb E(\mathbb P(X=Y\mid X))=\mathbb E(v(X))=\mathbb E(v(Y))$. Writing $ v=\sum_{x\in A}v(x)\mathbf 1_{x},\qquad A=\{x\mid v(x)\ne0\}, $ one gets $ \mathbb P(X=Y)=\sum_{x\in A}v(x)\mathbb E(\mathbf 1_{X=x})=\sum_{x\in A}v(x)\mathbb P(X=x)=\sum_{x\in A}v(x)^2, $ which is the LHS of the identity.