다양한 이산 랜덤 변수

3.5 Important Discrete Random Variables

• Bernoulli R.V
• Binomial R.V
• Geometric R.V
• Uniform Random Variable
• Poisson Random Variable

 

3.5.1 Bernoulli Random Variable

Indicator function	$$I_A\left(\xi \right)=\begin{cases}0\ \ \ \ \xi \  A\\1\ \ \ \ \zeta \in A\end{cases}$$
Sample Space	$$S_X=\left\{{0,\ 1}\right\}$$
PMF	$$p_I\left(0\right)=1-p,\ \ P_I\left(1\right)=p\ \ where\ P\left[A\right]=p$$
Mean	$$m_I=E\left[I_A\right]=0\cdot \left(1-p\right)+1\cdot p=p$$
Variance	$$VAR\left[I_A\right]=E\left[I_A^2\right]-E\left[I_A\right]^2=0^2\cdot \left(1-p\right)+1^2\cdot p-p^2=p\left(1-p\right)$$

3.5.2 Binomial Random Variable

Definition	Number of successes in n Bernoulli trials
Sample Space	$$S_X=\left\{{0,\ 1,\ ...,\ n}\right\}$$
PMF	$$p_k=\begin{pmatrix}n\\k\end{pmatrix}p^k\left(1-p\right)^{n-k}$$
Mean	$$E\left(X\right)=np$$ $$E\left(X\right)=\sum _{k=0}^nkp_X\left(k\right)=\sum _{k=0}^nk\begin{pmatrix}n\\k\end{pmatrix}p^k\left(1-p\right)^{n-k}=np\sum _{k=1}^n\frac{\left(n-1\right)!}{\left(k-1\right)!\left(n-k\right)!}p^{k-1}\left(1-p\right)^{n-k}$$ $$np\sum _{j=0}^{n-1}\frac{\left(n-1\right)!}{j!\left(n-1-j\right)!}p^j\left(1-p\right)^{n-1-j}=np$$
Variance	$$E\left[X^2\right]=\sum _{k=0}^nk^2\ \frac{n!}{k!\left(n-k\right)!}p^k\left(1-p\right)^{n-k}=\sum _{k=1}^nk\ \frac{n!}{\left(k-1\right)!\left(n-k\right)!}p^k\left(1-p\right)^{n-k}=np\sum _{k=0}^{n-1}\left(k'\ +1\right)\begin{pmatrix}n-1\\k'\end{pmatrix}p^{k'}\left(1-p\right)^{n-1-k}$$ $$np\left\{\sum _{k'=0}^{n-1}\left(k'+1\right)\begin{pmatrix}n-1\\k'\end{pmatrix}p^{k'}\left(1-p\right)^{n-1-k'}+\sum _{k'=0}^{n-1}\begin{pmatrix}n-1\\k'\end{pmatrix}p^k\left(1-p\right)^{n-1-k'}\right\}$$ $$np\left\{\left(n-1\right)p+1\right\}=np\left(np+q\right)$$ $$\sigma _X^2=E\left[X^2\right]-E\left[X\right]^2=np\left(np+q\right)-\left(np\right)^2=npq=np\left(1-p\right)$$

3.5.3 Geometric Random Variable

Definition	Number M of independent Bernoulli trials until first success
PMF	$$P\left[M=k\right]=p_M\left(k\right)=\left(1-p\right)^{k-1}p,\ \ k=1,\ 2\ ,3\ ,\ ...$$
Mean	$$E\left[X\right]=\sum _{k=1}^{\infty }kpq^{k-1}=p\ \frac{1}{\left(1-q\right)^2}=\frac{1}{p}\ \ \ \left({\because }\sum _{k=0}^{\infty }kx^{k-1}=\frac{1}{\left(1-x\right)^2}\right)$$ or $$E\left(X\right)-\left(1-p\right)E\left(X\right)=\sum _{k-1}^{\infty }kpq^{k-1}-\sum _{k=1}^{\infty }kpq^k=\left(1-q\right)p\sum _{k=0}^{\infty }kq^k=p^2\ \frac{1}{\left(1-q\right)^2}=1$$ $$E\left(X\right)=\frac{1}{p}$$
Variance	$$E\left[X^2\right]=\sum _{k=1}^{\infty }k^2pq^{k-1}=\sum _{k=0}^{\infty }\left(k+1\right)^2pq^k$$ $$\left(1-p\right)E\left[X^2\right]=\left(1-p\right)\sum _{k=1}^{\infty }k^2pq^{k-1}=\sum _{k=0}^{\infty }k^2pq^k$$ $$E\left[X^2\right]-\left(1-p\right)E\left[X^2\right]=\sum _{k=0}^{\infty }\left(2k+1\right)pq^k=2pq\sum _{k=0}^{\infty }kq^{k-1}+p\sum _{k=0}^{\infty }q^k=\frac{2pq}{\left(1-q\right)^2}+\frac{p}{1-q}=\frac{\left(1+q\right)}{p}$$ $$E\left[X^2\right]-E\left[X\right]^2=\frac{\left(1-p\right)}{p^2}$$
Probability of M<=k
Memoryless property	$$P\left[M\ge k+j\mid M>j\right]=P\left[M\ge k\right]\ for\ all\ j,k>1$$ $$P\left[M\ge k+j\right]=q^{k+j-1},\ \ P\left[M>j\right]=p\left[M\ge j+1\right]=q^j$$ $$\frac{P\left[M\ge k+j\right]}{p\left[M>j\right]}=\frac{q^{k+j-1}}{q^j}=q^{k-1}=P\left[M\ge k\right]$$

3.5.3-1 Conditional PMF

설명 필요

3.5.4 Uniform Random Variable

3.5.5 Poisson Random Variable