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---
title: "Theorem and Proof env using Pandoc's fenced div"
documentclass: book
output:
bookdown::pdf_document2:
keep_tex: true
bookdown::html_document2: default
---
```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```
# Examples
::: {.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.
:::
::: {.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
$$\varphi _{S_{n}}(t)=\prod_{i=1}^n\varphi _{X_i}(a_{i}t)=\varphi _{X_{1}}(a_{1}t)\cdots \varphi _{X_{n}}(a_{n}t)$$
:::
::: {.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.
:::
::: {.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.,
$$\varphi _{X_1+X_2}(t)=\varphi _{X_1}(t) \varphi _{X_2}(t)$$
:::
::: {.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 $1/2$.
:::
# Referencing
You can see Theorem \@ref(thm:chf-sum) and Lemma \@ref(lem:chf-pdf)
# Special features
Containing Markdown syntax
::: solution
`code`
:::
::: lemma
**bold** and *emph*
:::
```{asis, echo = knitr::is_latex_output()}
:::theorem
\textbf{bold in latex}
:::
```
```{asis, echo = knitr::is_html_output()}
:::theorem
<strong>bold in html</strong>
:::
```
::: {#exr-1 .exercise}
1) Open RStudio.
2) Write `1 + 9` to console.
:::
::: {.exercise}
Do this:
1) Open RStudio.
2) Write `1 + 9` to console.
:::
::: {.exercise}
```r
library(knitr)
kable(mtcars)
```
Copy and paste the chunk above
:::
::: {.solution #sol-2}
Copy and paste the chunk below
```r
library(knitr)
kable(mtcars)
```
:::
|
