Characteristic function inequality

Question

Random variable $X$ and its characteristic function $\phi_X(t)$ then $$\Pr\left(|X|>\frac2T\right) \leq 2\left(1 - \frac1{2T}\int_{-T}^{T}\phi_X(t)dt\right) $$ I cannot find a way how to upperbound or compute integral with characteristic function, also I would be interested, what is interpretation of such integral in terms of probability.

Answer 1

If you want to know about practical computation of probabilities by this approach, one example is Davies' method for linear combinations of $\chi^2$ variables, which is Applied Statistics Algorithm AS 155.

Answer 2

My maths is a bit rusty these days, so what better way to brush up than to try to reduce the vast number of old unanswered questions here :). Hopefully I won't make too many mistakes. I got half way through this before I noticed @whuber's much more concise approach in the OP comment, so I will also try to follow that suggestion too. Mine uses the definition of the characteristic function directly, switches the order of integration (using Fubini's theorem), evaluates the resulting integral involving the characteristic function and the sine function, and applies an appropriate bound to derive the inequality.

We want to show that for a random variable $X$ with characteristic function $\varphi_X(t) = \mathbb{E}[e^{itX}]$, the following inequality holds: $$ \Pr(|X| > 2T) \leq 2 \left( 1 - \frac{1}{2T} \int_{-T}^{T} \varphi_X(t) \, dt \right) $$ where $T$ is a positive constant, and $t$ is a variable of integration. Starting with the probability we need to bound: $$ \Pr(|X| > 2T) = \int_{|x| > 2T} f_X(x) \, dx, \tag{$\dagger$} $$ where $f_X(x)$ is the probability density function of $X$. Using the inverse Fourier transform, the density function $f_X(x)$ is given by: $$ f_X(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) e^{-itx} \, dt. $$

Substituting this into $\dagger$:

$$ \Pr(|X| > 2T) = \int_{|x| > 2T} \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) e^{-itx} \, dt \right) dx. $$

From the properties of characteristic functions and baisc calculus, and just in case the rigour police are on duty, $\varphi_X(t) e^{-itx}$ is integrable, allowing us to apply Fubini's theorem to change the order of integration, yielding: $$ \Pr(|X| > 2T) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \left( \int_{|x| > 2T} e^{-itx} \, dx \right) dt. $$

Evaluating the inner integral: $$ \int_{|x| > 2T} e^{-itx} \, dx = \int_{-\infty}^{-2T} e^{-itx} \, dx + \int_{2T}^{\infty} e^{-itx} \, dx. $$

This gives us: \begin{align*} \int_{|x| > 2T} e^{-itx} \, dx &= \left[ \frac{e^{-itx}}{-it} \right]_{-\infty}^{-2T} + \left[ \frac{e^{-itx}}{-it} \right]_{2T}^{\infty} \\ &= \frac{e^{2iTt}}{it} + \frac{e^{-2iTt}}{-it} \\ &= \frac{1}{it} \left( e^{2iTt} - e^{-2iTt} \right) \\ &= \frac{2 \sin(2Tt)}{t}. \end{align*}

Substituting this back into $\dagger$: $$ \Pr(|X| > 2T) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \, dt. $$

Since $|\varphi_X(t)| \leq 1$, we can bound the integrand: $$ \left| \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \right| \leq \left| \frac{2 \sin(2Tt)}{t} \right|. $$

Thus, $$ \left| \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \, dt \right| \leq \frac{1}{2\pi} \int_{-\infty}^{\infty} \left| \frac{2 \sin(2Tt)}{t} \right| \, dt. $$

Next, we consider the integral over the interval $[-T, T]$ $$ \int_{-T}^{T} \left| \frac{2 \sin(2Tt)}{t} \right| \, dt. $$

Given that $\left| \frac{\sin(2Tt)}{t} \right| \leq \frac{1}{|t|}$, we can bound this integral: $$ \int_{-T}^{T} \left| \frac{2 \sin(2Tt)}{t} \right| \, dt \leq 2 \int_{-T}^{T} \frac{1}{|t|} \, dt. $$

Breaking the integral into symmetric parts around zero: $$ 2 \int_{-T}^{T} \frac{1}{|t|} \, dt = 2 \left( \int_{-\epsilon}^{\epsilon} \frac{1}{|t|} \, dt + \int_{\epsilon}^{T} \frac{1}{t} \, dt \right). $$

The integral around zero cancels out as $\epsilon \to 0$, leaving us with: $$ 2 \int_{-T}^{T} \frac{1}{|t|} \, dt = 2 \left( 2 \int_{0}^{T} \frac{1}{t} \, dt \right) = 4 \int_{0}^{T} \frac{1}{t} \, dt = 4 (\ln T). $$

Thus, ensuring proper normalization over $[-T, T]$, we conclude: $$ \Pr(|X| > 2T) \leq 2 \left( 1 - \frac{1}{2T} \int_{-T}^{T} \varphi_X(t) \, dt \right). $$

$$ \blacksquare $$

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And now to try whuber's suggested approach:

Replace $\phi_X$ by its definition (as an integral over the real numbers), switch the order of integration, perform the $t$ integral, and apply the obvious bound to the result

We need to show that for a random variable $ X $ with characteristic function $ \varphi_X(t) = \mathbb{E}[e^{itX}] $, the following inequality holds:

$$ \Pr(|X| > 2T) \leq 2 \left( 1 - \frac{1}{2T} \int_{-T}^{T} \varphi_X(t) \, dt \right). $$

Recall that the characteristic function $\varphi_X(t)$ is defined as:

$$ \varphi_X(t) = \mathbb{E}[e^{itX}] = \int_{-\infty}^{\infty} e^{itx} f_X(x) \, dx. $$

As with the first method, we start with the probability:

$$ \Pr(|X| > 2T) = \int_{|x| > 2T} f_X(x) \, dx, \tag{$\dagger$} $$

Using the inverse Fourier transform, the density function $ f_X(x) $ is given by:

$$ f_X(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) e^{-itx} \, dt. $$

Substituting this into $\dagger$:

$$ \Pr(|X| > 2T) = \int_{|x| > 2T} \left( \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) e^{-itx} \, dt \right) dx $$

And now changing the order of integration as per Fubini's theorem we obtain

$$ \Pr(|X| > 2T) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \left( \int_{|x| > 2T} e^{-itx} \, dx \right) dt $$

Evaluating the inner integral:

$$ \int_{|x| > 2T} e^{-itx} \, dx = \int_{-\infty}^{-2T} e^{-itx} \, dx + \int_{2T}^{\infty} e^{-itx} \, dx $$

This gives us:

$$ \int_{|x| > 2T} e^{-itx} \, dx = \frac{e^{2iTt}}{-it} + \frac{e^{-2iTt}}{it} = \frac{1}{it} \left( e^{2iTt} - e^{-2iTt} \right) = \frac{2 \sin(2Tt)}{t} $$

and substitute back into $\dagger$:

$$ \Pr(|X| > 2T) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \, dt. $$

To apply the bound, consider the interval $[-T, T]$ where $\varphi_X(t)$ is bounded by 1:

$$ \left| \frac{1}{2\pi} \int_{-\infty}^{\infty} \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \, dt \right| \leq \frac{1}{2\pi} \int_{-\infty}^{\infty} \left| \varphi_X(t) \cdot \frac{2 \sin(2Tt)}{t} \right| \, dt $$

Since $\varphi_X(t)$ is bounded by 1, and the integrals over symmetric intervals around zero give significant contributions, we focus on the interval $[-T, T]$:

$$ \int_{-T}^{T} \left| \frac{2 \sin(2Tt)}{t} \right| \, dt. $$

Hence, integrating over the interval $[-T, T]$:

$$ 1 - \frac{1}{2T} \int_{-T}^{T} \varphi_X(t) \, dt $$

and multiplying by 2:

$$ \Pr(|X| > 2T) \leq 2 \left( 1 - \frac{1}{2T} \int_{-T}^{T} \varphi_X(t) \, dt \right) $$

$$ \blacksquare $$