recap from last week.
independent Bernoulli trails
\(X\sim Ber(p)\), \(E(X)=P\), \(Var(X)=p(1-p)\), \(P(X=1)=p\)
\(Y\sim Bin(n,p)\), \(P(X=k)=\begin{pmatrix}n\\k\end{pmatrix}p^k(1-p)^{1-k},k=0,1,2,3,...n\), \(E(X)=np\), \(Var(X)=np(1-p)\)
\(X\sim Geometric(p)\), \(P(X=k)=(1-p)^{k-1}p\), \(k=1,2,3,..\infty\), \(E(X)={1\over p}\), \(Var(X)={1-p\over p^2}\)
\(Y\sim NB(r,p)\) Negative binomial distribution, \(P(X=k)={\begin{pmatrix}k-1\\r-1\end{pmatrix}}p^{r-1}(1-p)^{k-r},k=r,r+1,....\infty\), \(E(X)={r\over p}\), \(Var(X)={r(1-p)\over p^2}\)
p changed, event dependent on each other
\(X\sim HyperG (M_1,M_2,n)\), \(P(X=k)={\begin{pmatrix}M_1\\k\end{pmatrix}\begin{pmatrix}M_2\\n-k\end{pmatrix}\over\begin{pmatrix}M_1+M_2\\n\end{pmatrix} },k=min(0,n-M_2),1,2,3,...max(n,M_1)\), \(E(X)={nM_1\over N}\), \(Var(X)={nM_1\over N}(1-{M_1\over N})({N-n\over N-1})\), n is the selected items, N-n is the number of non-selected items.
The hypergeometric distribution can sometimes be approximated using the binomial distribution.
For large population size N, the difference between sampling with and without replacement is very small.
Suppose $X \sim Hyp(M_1,M_2,n)$ and n $N \le 0.05$ where $N = M_1 + M_2$. Then $P(X = x) \simeq P(Y = x)$ where $Y ∼Bin(n,p = M_1 N )$.Example: Suppose M1 = 100, M2 = 900, and n = 25. Find P(X = 3).
#Using the hypergeometric distribution:
dhyper(3,100,900,25)## [1] 0.229574#Using the binomial approximation:
dbinom(3,25,0.1)## [1] 0.2264973The Poisson distribution is used to model the probability that a number of events occur in an interval of time or space.
The Poisson random variable X denotes the number of events that occurred.
Notation \(X\sim Poisson (\lambda)\)
The PMF:
\[ p(x)=P(X=x)={e^{-\lambda}\lambda^x\over x!},~~~~~~~x=0,1,2,3...\lambda>0 \]
The mean and variance for \(X\sim Poisson(\lambda)\) are
\[ E(X)=\lambda~~~~~~~\sigma_X^2=\lambda \]
R will compute the PMF, P(X = x): dpois(x,λ)
R will also compute the CDF, P(X ≤x): ppois(x,λ)
R will provide a sample of n Poisson random variables: rpois(n,λ)
Example: Suppose that a person taking Vitamin C supplements contracts an average of three colds per year, and that this average increases to five colds per year for persons not taking Vitamin C supplements. Suppose further that the number of colds a person contracts in a year is a Poisson random variable.
\(X_1\)= number of colds for patient taking VC, \(X_1\sim Poisson(3)\)
\(X_2\)= number of colds for patient not taking VC, \(X_2\sim Poisson(5)\)
\(P(X_1\leq)=P(X_1=0)+P(X_1=1)+P(X=2)=\)
## first is x, second is lambda
ppois(2,3)## [1] 0.4231901\(P(X_2\le 2)=0.125\)
ppois(2,5)## [1] 0.124652\(A\) = taking VC, \(A^c\)= not taking VC
X = number of codes of a randomly selected person
\[ \begin{aligned} P(X\le 2)&=P(x\le2,A)+P(X\le2,A^c)\\ &=P(X\le 2|A)P(A)+P(X\le 2|A^c)P(A^c)\\ &=0.423\times 0.7+0.125\times 0.3\\ &=0.334 \end{aligned} \]
The exponential distribution is often used to model lifetimes of equipment or waiting times until events occur.
Notation \(X\sim Exp(\lambda)\)
PDF:
\[ f(x)=\begin{cases}\lambda e^{-\lambda x},~~~x\geq0\\0,~~~~~~~~~~~x<0\end{cases} \]
The mean and variance for \(X\sim Poisson(\lambda)\) are
\[ E(X)=\lambda~~~~~~~\sigma^2_X=\lambda \]
CDF:
For \(0\leq X\)
\[ F(X)=P(X\le x)\int^X_{-\infty}f(x)dx=\int^x_0 \lambda e^{-\lambda t}dt\\ let~u=\lambda t =\int^{\lambda x}_0 e^{-u}du\\ =-e^{\lambda x}-1-e^0\\ =1-e^{-\lambda x} \]
the expected value and variance of \(X\sim Exp(\lambda)\) are
\[ E(X)={1\over \lambda}~~~~~~~\sigma_X^2={1\over \lambda^2} \]
Example: Let X be the amount of time (in minutes) a postal clerk spends with his or her customer. The time spent has an exponential distribution with λ = 0.25.
\[ E(X)={1\over 0.25}=4 \]
\[ P(2\le X<4)=\int^4_2 \lambda e^{-\lambda x}dx=\int_2^4 0.25e^{-0.25x}=0.239\\ P(2\le X<4)=F(4)-F(2)=(1-e^{-0.24\times 4})=0.239 \]
An exponential random variable X has the memory-less property:
\[ P(X > s + t |X > s) = P(X > t) \]