chaos world

Useful random variable distribution

0-1 Distribution

If random variable $X$ has distribution below:

$0<p<1$, then $X$ obey 0-1 distribution. The expectation of $X$ is $mathrm{E}(X)=p$. The square deviation of $X$ is $mathrm{D}(X)=p(1-p)$.

Binomial Distribution

If random variable $X$ obeys distribution law:

$$mathrm{P}(X=k)=mathrm{C}_n^kp^kq^{n-k},k=0,1,2,cdots,n$$

$0<p<1$, $mathrm{C}_n^k$ refers to combination number $frac{n!}{k!(n-k)!}$, then $X$ obeys binomial distribution, wrote as $X sim mathrm{B}(n,p)$, especially when $n=1$, $X$ obey 0-1 distribution: $X sim mathrm{B}(1,p)$.

The expectation of $X$ is $mathrm{E}(X)=np$. The square deviation of $X$ is $mathrm{D}(X)=np(1-p)$. You can regard it as the sum of $n$ independent random variables which obey 0-1 distribution.

Geometric Distribution

If random variable $X$ obeys distribution law:

$$mathrm{P}{X=k}=pq^{k-1},k=1,2,cdots,$$

where $0<p<1,q=1-p$, then $X$ follows the Geometric Distribution. The expectation of $X$ is

$$mathrm{E}(X)=sum_{k=1}^{infty}kpq^{k-1}=psum_{k=1}^{infty}kq^{k-1}=frac p{(1-q)^2}=frac1p$$

and the deviation of $X$ is

$$mathrm{D}(X)=sum_{k=1}^{infty}(k-frac1p)^2pq^{k-1}=sum_{k=1}^{infty}(k^2-frac{2k}p+frac1{p^2})pq^{k-1}=frac{1-p}{p^2}$$

Hypergeometric Distribution

If random variable $X$ obeys distribution law:

$$mathrm{P}{X=k}=frac{mathrm{C}_M^kmathrm{C}_{N-M}^{n-k}}{mathrm{C}_N^n}, k=l_1,cdots,l_2$$

where $l_1=mathrm{max}(0, n-N+M), l_2=mathrm{min}(M,n)$, then $X$ follows Hypergeometric Distribution.

Poisson Distribution

If random variable $X$ obeys distribution law:

$$mathrm{P}{X=k}=frac{lambda^k}{k!}mathrm{e}^{-lambda},k=0,1,2,cdots,$$

$lambda>0$ is a constant, then the $X$ obeys Poisson Distribution with parameter $lambda$, wrote as $X sim mathrm{P}(lambda)$. The expectation of $X$ is

$$mathrm{E}(X)=sum_{k=0}^{infty}kfrac{lambda^k}{k!}mathrm{e}^{-lambda}=lambda mathrm{e}^{-lambda}sum_{k=0}^{infty}frac{lambda^k}{k!}=lambda.$$

The deviation of $X$ is

$$mathrm{D}(X)=sum_{k=0}^{infty}(k-lambda)^2frac{lambda^k}{k!}mathrm{e}^{-lambda}=sum_{k=0}^{infty}(k^2-2klambda+lambda^2)frac{lambda^k}{k!}mathrm{e}^{-lambda}=lambda.$$

. Especially, if $X sim P(lambda),Y sim P(lambda)$, and $X,Y$ are independent of each other, then $X+Y sim P(2lambda)$.

Uniform Disrtibution

If continuous random variable $X$ has possibility density function like below:

$$% &lt;![CDATA[ f(x)= begin{cases} frac1{b-a}&amp; ale xle b, \ \0 &amp; others, end{cases} %]]&amp;gt;$$

, then we claim that $X$ obeys Uniform Distribution in range $[a, b]$, wrote as $X sim mathrm{U}[a,b]$.

If continuous random variable $X$ has possibility density function like below:

$$% &lt;![CDATA[ f(x)= begin{cases} frac1{b-a} &amp; a&lt; x&lt; b, \ \0 &amp; others, end{cases} %]]&amp;gt;$$

, then we claim that $X$ obeys Uniform Distribution in range $(a, b)$, wrote as $X sim mathrm{U}(a,b)$.

No matter $X sim mathrm{U}(a,b)$ or $X sim mathrm{U}[a,b]$, their distribution functions both are

$$% &lt;![CDATA[ F(x)= begin{cases} 0 &amp; x&lt;a,\ \ frac{x-a}{b-a} &amp;ale x &lt;b,\ \ 1 &amp;ble x end{cases} %]]&amp;gt;$$

. So the expectation of $X$ is

$$mathrm{E}(X)=int_a^bfrac x{b-a}mathrm{d}x=frac{a+b}2$$

, and the deviation of $X$ is

$$mathrm{D}(X)=int_a^bfrac{(x-frac{a+b}2)^2}{b-a}mathrm{d}x=frac{(b-a)^2}{12}$$

Exponential Distribution

If continuous random variable $X$ has possibility density like below:

$$% &lt;![CDATA[ f(x)= begin{cases} lambda mathrm{e}^{-lambda x} &amp; x&gt;0, \ 0 &amp; xle0, end{cases} %]]&amp;gt;$$

$lambda > 0$, then $X$ follows Exponential Distribution with parameter $lambda$, wrote as $X sim E(lambda)$. So the distribution function is

$$% &lt;![CDATA[ F(x)= begin{cases} 1-mathrm{e}^{-lambda x} &amp; x&gt;0, \0 &amp; x le 0, end{cases} %]]&amp;gt;$$

. Easily we can get $mathrm{P} { X>t }=mathrm{e}^{-lambda t}$, so the conditional possibility

$$mathrm{P}{X&gt;t+sbig|X&gt;t}=frac{mathrm{P}{X&gt;t+s}}{mathrm{P}{X&gt;s}}=frac{mathrm{e}^{-lambda(t+s)}}{mathrm{e}^{-lambda s}}=mathrm{e}^{-lambda t}=mathrm{P}{X&gt;t}$$

which is regarded as the “lack in memory” character of exponential distribution. The expectation of $X$ is

$$mathrm{E}(X)=int_0^{+infty}xlambda mathrm{e}^{-lambda x}mathrm{d}x=-xmathrm{e}^{-lambda x}bigg|_0^{+infty}+int_0^{+infty}mathrm{e}^{-lambda x}mathrm{d}x=frac1{lambda}$$

. And the deviation of $X$ is

$$mathrm{D}(X)=int_0^{+infty}(x-frac1{lambda})^2lambda mathrm{e}^{-lambda x}mathrm{d}x=frac1{lambda ^2}$$

Normal Distribution

If continuous random variable $X$ has possibility density function like below:

$$f(x)=frac1{sqrt{2pi}sigma}mathrm{e}^{-frac{(x-mu)^2}{2sigma^2}}$$
$$(i.e. int_{-infty}^{+infty}mathrm{e}^{-x^2}mathrm{d}x=sqrt{int_{-infty}^{+infty}mathrm{e}^{-x^2}mathrm{d}xint_{-infty}^{+infty}mathrm{e}^{-y^2}mathrm{d}y}=sqrt{int_0^{2pi}mathrm{d}thetaint_0^{+infty}mathrm{e}^{-r^2}rmathrm{d}r}=sqrt{pi})$$

$,-infty<x<+infty$, then we say that the $X$ obeys Normal Distribution, wrote as $X sim mathrm{N}(mu,sigma^2)$, where its distribution function is

$$F(x)=frac1{sqrt{2pi}sigma}int_{-infty}^x mathrm{e}^{-frac{(t-mu)^2}{2sigma^2}}mathrm{d}t$$

. Especially when $mu=0,sigma=1$, as $X sim mathrm{N}(0,1)$, $X$ obeys standard normal distribution, we often use $varphi(x)=frac1{sqrt{2pi}}mathrm{e}^{-frac{x^2}2}$ to represent its possibility density, and $Phi(x)=frac1{sqrt{2pi}}int_{-infty}^x mathrm{e}^{-frac{t^2}2}mathrm{d}t$ to denote its distribution. It can be demonstrated that if $Y sim mathrm{N}(mu,sigma^2)$, then $Y$’s distribution function is $Phi(frac{x-mu}{sigma})$. Also we can find that the possibility density function is even, the left and right parts of $x=mu$ are symmetric. So the conclusion below is explicit:

$$Phi(-x)=1-Phi(x),Phi(0)=frac12,mathrm{P}{|X|le a}=2Phi(a)-1,X sim mathrm{N}(0,1)$$

. Finally the expectation and deviation of $X sim mathrm{N}(mu,sigma^2)$ are:

$$mathrm{E}(x)=mu,mathrm{D}(X)=sigma^2.$$

2-Dimensional Normal Distribution

If 2-dimensional continuous random variables $(X,Y)$ has possibility density function like:

$$f(x,y)=frac1{2pisigma_1sigma_2sqrt{1-rho^2}}exp{ {-frac1{2(1-rho^2)}[frac{(x-mu_1)^2}{sigma_1^2}-frac{2rho(x-mu_1)(y-mu_2)}{sigma_1sigma_2}+frac{(y-mu_2)^2}{sigma_2^2}]} }$$

where $mu_1,mu_2,sigma_1>0,sigma_2>0,-1<rho<1$ are all constants, then $(X,Y)$ follows 2-Dimensional Normal Distribution with parameters $mu_1,mu_2,sigma_1,sigma_2,rho$, wrote as $(X,Y) sim mathrm{N}(mu_1,mu_2;sigma_1^2,sigma_2^2;rho)$. The sufficient and necessary condition of $rho=0$ is that $X$ and $Y$ are independent of each other. There is a significant property of 2-Dimensional Normal Distribution as below:

$$aX+bY sim mathrm{N}(amu_1+bmu_2,a^2sigma_1^2+2absigma_1sigma_2rho+b^2sigma_2^2)$$

, where $(X,Y) sim mathrm{N}(mu_1,mu_2;sigma_1^2,sigma_2^2;rho)$.

$chi^2$ Distribution

Assume $X_1, X_2, cdots,X_n$ are independent and obeys standard normal distribution, define

$$chi^2=X_1^2+X_2^2+cdots+X_n^2$$

, then $chi^2$ obeys $chi^2$ distribution whose DOF is $n$, wrote as $chi^2 sim chi^2(n)$. The expectation of $chi^2(n)$ is

$$mathrm{E}[chi^2(n)]=nmathrm{E}[chi^2(1)]=nE(X^2)=n$$

and the deviation of $chi^2(n)$ is

$$mathrm{D}[chi^2(n)]=nmathrm{D}[chi^2(1)]=n[mathrm{E}(X^4)-mathrm{E}(X^2)]= \ n[frac1{sqrt{2pi}}int_{-infty}^{+infty}x^4mathrm{e}^{-frac{x^2}2}mathrm{d}x-mathrm{E}(X)^2-mathrm{D}(X)]=n[3-1]=2n$$

. Assume $chi^2 sim chi^2(n)$, for given $0<alpha<1$, which satisfies the condition:

$$mathrm{P}{chi^2&gt;chi_alpha^2(n)}=int_{chi_alpha^2(n)}^{+infty}f(x)mathrm{d}x=alpha$$

, then the $chi_alpha^2(n)$ is the quantile of $alpha$ on $chi^2(n)$ distribution. If $chi_1^2 sim chi^2(n_1)$ and $chi_2^2 sim chi^2(n_2)$, in addition that $chi_1^2$ and $chi_2^2$ are independent, then $chi_1^2+chi_2^2 sim chi^2(n_1+n_2)$.

$t$ Distribution

If random variable $X$ and $Y$ are independent of each other, assume $X sim mathrm{N}(0,1), Y sim chi^2(n)$, define random variable

$$T=frac{X}{sqrt{Y/n}}$$

, then $T$ obeys $t$ distribution with DOF $n$, denoted as $T sim t(n)$. Because the possibility density function of $t$ distribution is even, when $n to infty$, $t(n)$ is similar to $N(0,1)$. Assume $T sim t(n)$, for given $0<alpha<1$, which satisfies the condition:

$$mathrm{P}{T&gt;t_alpha(n)}=int_{t_alpha(n)}^{+infty}f(x)mathrm{d}x=alpha$$

, then the $t_alpha(n)$ is the quantile of $alpha$ on $t(n)$ distribution. Obviously $t_alpha(n)=-t_{1-alpha(n)}$

$F$ Distribution

If random variable $X$ and $Y$ are independent of each other, assume $X sim chi^2(n_1), Y sim chi^2(n_2)$, define random variable

$$F=frac{X/n_1}{Y/n_2}$$

, then $F$ obeys $F$ distribution with DOF $(n_1,n_2)$, denoted as $F sim F(n_1,n_2)$, where $n_1, n_2$ are the first and second DOF. Assume $F sim F(n_1,n_2)$, for given $0<alpha<1$, which satisfies the condition:

$$mathrm{P}{F&gt;F_alpha(n_1,n_2)}=int_{F_alpha(n_1,n_2)}^{+infty}f(x)mathrm{d}x=alpha$$

, then the $F_alpha(n_1,n_2)$ is the quantile of $alpha$ on $F(n_1,n_2)$ distribution. Obviously $frac1F sim F_alpha(n_2,n_1)$ and

$$F_{1-alpha}(n_1,n_2)=frac1{F_alpha(n_2,n_1)}$$

Sampling Distribution of Normal Variable

If $X sim mathrm{N}(mu,sigma^2)$, $X_1, X_2, cdots, X_n$ are samples from $X$, the average of samples is $bar X=frac1nsum_{i=1}^nX_i$, the variance of samples is $S^2=frac1{n-1}sum_{i=1}^n(X_i-bar X)^2$, then we have the following conclusion:

$$bar X sim mathrm{N}(mu, frac{sigma^2}n), U=frac{bar X-mu}{sigma/sqrt{n}} sim mathrm{N}(0,1);$$
$$chi^2=frac{(n-1)S^2}{sigma^2} sim chi^2(n-1);$$
$$T=frac{bar X-mu}{S/sqrt{n}} sim t(n-1);$$
$$chi^2=frac1{sigma^2}sum_{i=1}^n(X_i-mu)^2 sim chi^2(n)$$

, moreover the $bar X$ and $S^2$ are independent of each other.