---
title: 機率TP8
---
# Theory of Probability：<br>Continuous Random Variables - Basics

NTNU 機率論

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###### tags: `NTNU` `CSIE` `必修` `Theory of Probability`

## Normality is Preserved by Linear Transformations
If $X$ is a normal random variable with mean $\mu$ and variance $\sigma^2$, and if $a(a\neq 0)$ and $b$ are scalars, then the random variable
$$
Y = aX+b
$$
is also normal with mean and variance
$$
E[Y] = a\mu+b\\
\text{var}(Y)=a^2\sigma^2
$$

## Standard Normal Random Variable
- A normal random variable $Y$ with zero mean $\mu = 0$ and unit variance $\sigma^2=1$ is said to be a **standard normal**
$$
f_Y(y) = {1\over \sqrt{2\pi}}e^{-{y^2\over 2}},\ -\infty \leq y \leq \infty
$$
- Normalization Property
$$
\int^\infty_{-\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{y^2}{2}}dy=1
$$
- The standard normal is symmetric around $y=0$

### The PDF of a Random Variable Can be Arbitratily Large
- Consider a random variable $X$ with PDF
$$
f_X(x)=\begin{cases}
\frac{1}{2\sqrt{x}},&\text{if }0<x\leq1,\\
0,&\text{otherwise},
\end{cases}
$$
- The PDF value becomes infinite large as $x$ approaches zero
- Normalizatino Property
$$
\int^1_0f_X(x)dx=\int^1_0\frac{1}{2\sqrt{x}}dx=\sqrt{x}|^1_0=1
$$

## Expectation of a Continuous Random Variable
- Let $X$ be a continuous random variable with PDF $f_x$
- The expectaion of $X$ is defined by
$$
E[X]=\int^\infty_{-\infty}x\cdot f_X(x)dx
$$
- The expectation of a function $g(X)$ has the form
$$
E[g(X)]=\int^\infty_{-\infty}g(x)\cdot f_X(x)dx
$$
- The variance of $X$ is defuned by
$$
\text{var}(X)=E[X^2]-(E[X])^2\geq0
$$
- we also have
$$
\text{var}(X)=E[X^2]-(E[X])^2\geq0
$$
- if $Y = aX+b$, where $a$ and $b$ are given scalars, then
$$
E[Y]=aE[X]+b\\
\text{var}(Y)=a^2\text{var}(X)
$$

### Expaonential Random Variable
- An **expaonential** random variable $X$ has a PDF if the form
$$
f_X(x)=\begin{cases}\lambda e^{-\lambda x},&\text{if }x\geq 0,\\
0,&\text{otherwise}
\end{cases}
$$
- Normalization Property
$$
\int_{-\infty}^\infty f_X(x)dx=\int_0^\infty\lambda e^{-\lambda x}dx=-e^{-\lambda x}|^\infty_0=1
$$
- The probability that $X$ exceeds a certain value decreasese exponentially
$$
P(X\geq a)=\int_a^\infty\lambda e^{-\lambda x}dx=e^{-\lambda a}
$$

## Properties of a CDF
- The CDF $F_X(x)$ is *monotonically(單調) non-decreasing*
- if $x_i\leq x_j$, then $F_X(x_i)\leq F_X(x_j)$
- The CDF $F_X(x)$ tends to 0 as $x\to-\infty$, ans to 1 as $x\to\infty$
- If $X$ is deicrete, then $F_X(x)$ is a *piecewise constant* function(分段常值函數) of $x$.
- If $X$ is continuous, then $F_X(x)$ is a *continous* function of $x$.

- If $X$ is descrete and takes integer values, the PMF and the CDF can be obtained from each other by summing or differencing.
$$
\begin{align}
F_X(k)&=P(X\leq k)&=\sum_{i=-\infty}^kp_X(i),\\
p_X(k)&=P(X\leq k)-P(X\leq k-1)&=F_X(k)-F_X(k-1)
\end{align}
$$
- If X is continous, the PDF and the CDF can be obtained from each other by integration or differentiation
$$
F_X(x)=P(X\leq x)=\int_{-\infty}^\infty f_X(t)dt,\\
f_X(x)=\frac{dF_X(x)}{dx}
$$
- The second equality is valid for those $x$ for which the CDF has a derivative (or at which the PDF is continuous)

## Example(slide(TP8) p25)
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P(X_1>k), P(X_2>k), P(X_3>k) independent
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## CDF of the **Standard Normal**
- The CDF of the standard normal $Y$, denoted as $\phi(y)$, is recorded in a table and is a very useful tool for calculating various probabilities, including normal variables.
$$
\phi(y)=P(Y\leq y)=P(Y<y)=\int_{-\infty}^y\frac{1}{\sqrt{2\pi}}e^{\frac{-t^2}{2}}dt
$$
- The table only provides the value of $\phi(y)$ for $y\geq 0$
- Because the symmetry of the PDF, the CDF at negative values of $Y$ can be computed from corresponding positive ones
$$
\begin{align}
\phi(-0.5)&=P(Y\leq -0.5)=1-P(Y\leq 0.5)\\
&=1-\phi(0.5)=1-0.6915\\
&=0.3085
\end{align}
$$
> $\phi(-y)=1-\phi(y), \forall y$
## Table of the CDF of **Standard Normal**
## CDF Calculation of the **Normal**
- The CDF of a normal random variable $X$ with mean $\mu$ and variance $\sigma^2$ is obtained using the standard normal table as
$$
P(X\leq x)=P\left(\frac{X-\mu}{\sigma}\leq\frac{x-\mu}{\sigma}\right)=P\left(Y\leq\frac{x-\mu}{\sigma}\right)=\phi\left(\frac{x-\mu}{\sigma}\right)
$$
- Let $Y=\frac{X-\mu}{\sigma}$, since $X$ is normal and $Y$ is a linear function of $X$, $Y$ hence is also normal (with mean 0 and variance 1).
$$
E[Y]=\frac{E[X]-\mu}{\sigma}=0, \text{var}(Y)=\frac{\text{var}(X)}{\sigma^2}=1
$$

## Relation between the **Geometric** and **Exponential**
- The CDF of the geometric
- The CDF of the exponential
- Compare the above two CDFs and let

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