In rstudio/bookdown: Authoring Books and Technical Documents with R Markdown

knitr::opts_chunk$set(echo = TRUE)

Examples

The characteristic function of a random variable $X$ is defined by

$$\varphi _{X}(t)=\operatorname {E} \left[e^{itX}\right], \; t\in\mathcal{R}$$

We derive the characteristic function of $X\sim U(0,1)$ with the probability density function $f(x)=\mathbf{1}_{x \in [0,1]}$.

\begin{equation*}
\begin{split}
\varphi _{X}(t) &= \operatorname {E} \left[e^{itX}\right]\\
 & =\int e^{itx}f(x)dx\\
 & =\int_{0}^{1}e^{itx}dx\\
 & =\int_{0}^{1}\left(\cos(tx)+i\sin(tx)\right)dx\\
 & =\left.\left(\frac{\sin(tx)}{t}-i\frac{\cos(tx)}{t}\right)\right|_{0}^{1}\\
 & =\frac{\sin(t)}{t}-i\left(\frac{\cos(t)-1}{t}\right)\\
 & =\frac{i\sin(t)}{it}+\frac{\cos(t)-1}{it}\\
 & =\frac{e^{it}-1}{it}
\end{split}
\end{equation*}

Note that we used the fact $e^{ix}=\cos(x)+i\sin(x)$ twice.

```{lemma, chf-pdf} For any two random variables $X_1$, $X_2$, they both have the same probability distribution if and only if

$$\varphi {X_1}(t)=\varphi {X_2}(t)$$

```{theorem, chf-sum}
If $X_1$, ..., $X_n$ are independent random variables, and $a_1$, ..., $a_n$ are some constants, then the characteristic function of the linear combination $S_n=\sum_{i=1}^na_iX_i$ is

$$\varphi _{S_{n}}(t)=\prod_{i=1}^n\varphi _{X_i}(a_{i}t)=\varphi _{X_{1}}(a_{1}t)\cdots \varphi _{X_{n}}(a_{n}t)$$

The distribution of the sum of independent Poisson random variables $X_i \sim \mathrm{Pois}(\lambda_i),\: i=1,2,\cdots,n$ is $\mathrm{Pois}(\sum_{i=1}^n\lambda_i)$.

The characteristic function of $X\sim\mathrm{Pois}(\lambda)$ is $\varphi _{X}(t)=e^{\lambda (e^{it}-1)}$. Let $P_n=\sum_{i=1}^nX_i$. We know from Theorem \@ref(thm:chf-sum) that

\begin{equation*}
\begin{split}
\varphi _{P_{n}}(t) & =\prod_{i=1}^n\varphi _{X_i}(t) \\
& =\prod_{i=1}^n e^{\lambda_i (e^{it}-1)} \\
& = e^{\sum_{i=1}^n \lambda_i (e^{it}-1)}
\end{split}
\end{equation*}

This is the characteristic function of a Poisson random variable with the parameter $\lambda=\sum_{i=1}^n \lambda_i$. From Lemma \@ref(lem:chf-pdf), we know the distribution of $P_n$ is $\mathrm{Pois}(\sum_{i=1}^n\lambda_i)$.

In some cases, it is very convenient and easy to figure out the distribution of the sum of independent random variables using characteristic functions.

The characteristic function of the sum of two independent random variables $X_1$ and $X_2$ is the product of characteristic functions of $X_1$ and $X_2$, i.e.,

$$\varphi _{X_1+X_2}(t)=\varphi _{X_1}(t) \varphi _{X_2}(t)$$

```{exercise, name="Characteristic Function of the Sample Mean"} Let $\bar{X}=\sum_{i=1}^n \frac{1}{n} X_i$ be the sample mean of $n$ independent and identically distributed random variables, each with characteristic function $\varphi _{X}$. Compute the characteristic function of $\bar{X}$.

```{solution}
Applying Theorem \@ref(thm:chf-sum), we have

$$\varphi _{\bar{X}}(t)=\prod_{i=1}^n \varphi _{X_i}\left(\frac{t}{n}\right)=\left[\varphi _{X}\left(\frac{t}{n}\right)\right]^n.$$

{hypothesis, name="Riemann hypothesis"} The Riemann Zeta-function is defined as $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s}$$ for complex values of $s$ and which converges when the real part of $s$ is greater than 1. The Riemann hypothesis is that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part $1/2$.