# Examples

We can use Pandoc fenced div syntax now

::: {.definition} The characteristic function of a random variable $X$ is defined by

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

::: {.example} 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. :::

We can include some Mardown syntax and R code

::: {.lemma #chf-pdf} For any two random variables x, y, you can add then using +, and for example

x = 1
y = 1
x+y


:::

::: {.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

$$\varphi {S{n}}(t)=\prod_{i=1}^n\varphi {X_i}(a{i}t)=\varphi {X{1}}(a_{1}t)\cdots \varphi {X{n}}(a_{n}t)$$ :::

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

::: {.proof} 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 $\mathrm{Pois}(\sum_{i=1}^n\lambda_i)$. :::

::: {.remark} In some cases, it is very convenient and easy to figure out the distribution of the sum of independent random variables using characteristic functions. See \@ref(thm:chf-sum) :::

::: {.corollary} 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.,

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

And a name can be added

::: {.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 $1/2$. :::

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bookdown documentation built on Sept. 5, 2021, 5:19 p.m.