Continuous Distributions 1

1. Uniform Distribution

1.1 Definition

A continuous random variable $X$ is said to follow the uniform distribution on an interval $[a,b]$ with $a<b$ denoted by $\mathcal{U}(a,b)$ if:

$$f_X(x)=\begin{cases} \frac{1}{b-a} &x\in[a,b]\\ 0 & \text{otherwise} \end{cases}$$

1.2 Significance

In probability theory and statistics, the continuous uniform distribution or rectangular distribution is a family of symmetric probability distributions. The distribution describes an experiment where there is an arbitrary outcome that lies between certain bounds. The bounds are defined by the parameters, a and b, which are the minimum and maximum values. The interval can either be closed (e.g. $[a,b]$) or open (e.g. $]a,b[$).

1.2 Standard Uniform distribution

It is the uniform distribution on the interval $[0,1]$:

$$\mathcal{U}=\mathcal{U}(0,1)$$

1.2 Opposite of Uniform Random Variable

Let $X \sim \mathcal{U}(a,b)$

$$\begin{align*} \forall x\in\mathbb{R},F_{-X}(x)&=\mathcal{P}(-X<x)\\ &=\mathcal{P}(X>-x)\\ &=1-F_X(-x)\\ \forall x\in\mathbb{R},f_{-X}(x)&=f'_X(-x)\\ &=\frac{1}{b-a}\mathbb{1}_{[a,b]}(-x)\\ &=\frac{1}{(-a)-(-b)}\mathbb{1}_{[-b,-a]}(x)\\ \implies -X&\sim\mathcal{U}(-b,-a) \end{align*}$$

1.3 Linear Transformation of a Uniform Random Variable

• Let $\alpha \in\mathbb{R}_+^*,\beta\in\mathbb{R}$
• Let $a,b\in\mathbb{R}$ with $a<b$
• Let $X\sim \mathcal{U}(a,b)$ and $Y=\alpha X+b$

$$\begin{align*} \forall x\in\mathbb{R},F_Y(x)&=\mathcal{P}(Y< x)\\ &=\mathcal{P}(\alpha X< x-\beta)\\ &= \mathcal{P}(X<\frac{x-\beta}{\alpha})\\ &=F_X\left(\frac{x-\beta}{\alpha}\right)\\ \implies \forall x\in\mathbb{R},F_Y(x)&=\frac{1}{a}F_X'\left(\frac{x-\beta}{\alpha}\right)\\ &=\frac{1}{\alpha}f_X(\frac{x-\beta}{\alpha})\\ &= \frac{1}{\alpha(b-a)} \mathbb{1}_{[a,b]}(\frac{x-\beta}{\alpha}) \\ &= \frac{1}{\alpha b-\alpha a}\mathbb{1}_{[\alpha a +\beta,\alpha b +\beta]}(x) \\ &= \frac{1}{(\alpha b +\beta) - (\alpha a +\beta)}\mathbb{1}_{[\alpha a +\beta,\alpha b +\beta]}(x) \\ \implies &Y\sim \mathcal{U}(\alpha a+\beta,\alpha b+\beta) \end{align*}$$

In particular:

$$\boxed{X\sim\mathcal{U}(a,b)\iff \frac{X-a}{b-a}\sim\mathcal{U}(0,1)}$$

For $\alpha < 0$, We have $Y=\alpha X+\beta=-\left(-\alpha X-\beta\right).$

We have:

$$-\alpha X -\beta \sim \mathcal{U}(-\alpha a-\beta,-\alpha b-\beta) \implies \alpha X+\beta \sim\mathcal{U}(\alpha b+\beta,\alpha a+\beta)$$

1.4 Moments & Central Moments

1.4.1 Moments

$$\begin{align*} \forall n\in\mathbb{N}^*,\quad \mathbb{E}[X^n]&=\int_{a}^b \frac{x^n}{b-a} \text{dx}\\ &=\frac{b^{n+1}-a^{n+1}}{(n+1)(b-a)} \end{align*}$$

In particular, the expected value $\mathbb{E}[X]$ is

$$\boxed{\mathbb{E}[X]=\frac{a+b}{2}}$$

1.4.1 Central Moments

For $n\in\mathbb{N}^*$, the $n^\text{th}$-central moment of $X$ is the $n^\text{th}$-moment of $X-\mathbb{E}[X]$

But $X-\mathbb{E}[X]\sim \mathcal{U}(\frac{a-b}{2},\frac{b-a}{2})$

$$\begin{align*} \forall n \in\mathbb{N}^*,\quad \mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^n\right]&=\frac{(\frac{b-a}{2})^{n+1}-(\frac{a-b}{2})^{n+1}}{(n+1)(\frac{b-a}{2}-\frac{a-b}{2})} \\ &=\frac{1-(-1)^{n+1}}{2^n(n+1)}\cdot\frac{(b-a)^n}{b-a} \end{align*}$$

In particular, the variance $\mathbb{V}[X]$ is:

$$\boxed{\mathbb{V}[X]=\frac{(b-a)^2}{12}}$$

2. Exponential Distribution

2.1 Definition

A continuous random variable $X$ is said to follow the exponential distribution with paramter $\lambda\in\mathbb{R}_+^*$ if:

$$f_X(x)=\begin{cases} \lambda e^{-\lambda x} &x \in\mathbb{R}_+ \\ 0 & \text{otherwise} \end{cases}$$

We denote it by:

$$X \sim \mathcal{E}(\lambda)$$

2.2 Significance

The exponential distribution is the probability distribution of the time between events in a Poisson point process. It is the continuous analogue of the geometric distribution, and it has the key property of being memoryless.

It is used to model radioactive decay.

2.3 Moments

2.3.1 Raw Moments

$$\begin{align*} \forall n\in\mathbb{N},\quad \mathbb{E}[X^n]&=\int_{\mathbb{R}_+}\lambda t^ne^{-\lambda t} \text{dt}\\ &= \int_{\mathbb{R}_+}\left(\frac{u}{\lambda}\right)^ne^{-u} \text{du} \quad \text{with} \space u=\lambda t,\quad \text{du}=\lambda\text{dt}\\ &=\lambda^{-n}\int_{\mathbb{R}_+}u^ne^{-u} \text{du}\\ &=\frac{\Gamma(n+1)}{\lambda^n}\\ &=\frac{n!}{\lambda^n} \end{align*}$$

In particular, the expected value $\mathbb{E}[X]$ is:

$$\boxed{\mathbb{E}[X]=\frac{1}{\lambda}}$$

2.3.2 Central moments

$$\begin{align*} \forall n\in\mathbb{N},\quad \mathbb{E}\left[\left(X-\mathbb{E}[X]\right)^n\right]&=\sum_{k=0}^n(-1)^{n-k}{n \choose k}\mathbb{E}[X^k]\mathbb{E}[X]^{n-k}\\ &=\sum_{k=0}^n(-1)^{n-k}{n \choose k}\frac{k!}{\lambda^n}\\ &=\frac{1}{\lambda^n}\sum_{k=0}^n(-1)^{n-k}\frac{n!}{(n-k)!} \end{align*}$$

In particular, the variance $\mathbb{V}[X]$ is:

$$\boxed{\mathbb{V}[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}}$$

2.4 Memoryless

Memory-less is a fundamental property in the exponential distribution, It states:

$$\forall T,r\in\mathbb{R}_+,\quad \mathcal{P}(X \ge T+r \mid X\ge T)=\mathcal{P}(X\ge r)$$

The proof is as follow:

$$\begin{align*} \forall T,r\in\mathbb{R}_+,\quad \mathcal{P}(X \ge T+r \mid X\ge T)&=\frac{\mathcal{P}(X \ge T+r)}{\mathcal{P}(X\ge T)}\\ &=\frac{\int_{T+r}^{+\infty}\lambda e^{-\lambda u}\text{du}}{\int_{T}^{+\infty}\lambda e^{-\lambda u}\text{du}} \\ &=\frac{e^{-(T+r)\lambda}}{e^{-T\lambda}}\\ &=e^{-\lambda r}\\ &=\mathcal{P}(X \ge r) \end{align*}$$

2.5 Scaling

• Let $k\in\mathbb{R}_+^*$
• Let $X\sim \mathcal{E}(\lambda)$ and $Y=kX$

We will calculate the probability distribution function of $Y:$

$$\begin{align*} \forall x\in\mathbb{R}_+,\quad f_Y(x)&=\frac{1}{k}f_X\left(\frac{x}{k}\right)\\ &=\frac{\lambda}{k}e^{-\frac{\lambda}{k}x} \end{align*}$$

By that:

$$\boxed{\forall k\in\mathbb{R}_+^*,\quad X\sim \mathcal{E}(\lambda) \iff kX\sim \mathcal{E}\left(\frac{\lambda}{k}\right)}$$