Handout Ch 5

Bernoulli Distribution A random variable X has a Bernoulli distribution if Pr(X = 1) = p and Pr(X = 0) = 1– p = q The p.m.f. of X can be written as

Binomial Distribution If the random variable X1, …, Xn form n Bernoulli trials with parameter p, and if , then X has a binomial distribution. The p.m.f. of X can be written as If X1, …, Xk are independent random variables and if Xi has a binomial distribution with parameters ni and p, then the sum has a binomial distribution with parameters and p.

Example 1 (5.2.8) A certain electronic system contains 10 components. Suppose that the probability that each individual component will fail is 0.2 and that the components fail independently of each other. Given that at least one of the components has failed, what is the probability that at least two of the components have failed?

Solution The number X of components that fail will have a binomial distribution with parameters n=10 and p=0.2. Therefore,

Poisson Distribution X has a Poisson distribution with mean l if the p.m.f. of X has:

Poisson Distribution The moment generating function

Poisson Distribution If the random variables X1, …, Xk are independent and if Xi has a Poisson distribution with mean , then the sum has a Poisson distribution with mean Proof: Let denote the m.g.f. of Xi and denote the m.g.f. of the sum Example 5.4.1: The mean number of customers who visit the store in one hour is 4.5. What is the probability that at least 12 customers will arrive in a two-hour period? X = X1 + X2 has a Poisson distribution with mean 9.

Poisson Approximation to Binomial Distribution When the value of n is large and the value of p is close to 0, the binomial distribution with parameters n and p can be approximated by a Poisson distribution with mean np. Proof: For a binomial distribution with l= np, we have As , then Also,

羅必達法則 當x→a時，函數f(x)及g(x)都趨於零； 在點a的附近鄰域內，f’(x)及g’(x)都存在，且g’(x) ≠0 存在(或為無窮大)， 則 各種形式：0/0，∞/∞，0× ∞， ∞- ∞，00， ∞0，1∞

Example 2 (5.4.8) Suppose that X1 and X2 are independent random variables and that Xi has a Poisson distribution with mean (i=1,2). For each fixed value of k (k=1,2,…), determine the conditional distribution of X1 given that X1+X2=k

Solution

Example 3 (5.4.14) An airline sells 200 tickets for a certain flight on an airplane that has only 198 seats because, on the average, 1 percent of purchasers of airline tickets do not appear for the departure of their flight. Determine the probability that everyone who appears for the departure of this flight will have a seat

Solution

Geometric Distribution Suppose that the probability of a success is p, and the probability of a failure is q=1 – p. Then these experiments form an infinite sequence of Bernoulli trials with parameter p. Let X = number of failures to first success. f ( x | p ) = pqx for x = 0, 1, 2, … Let Y = number of trials to first success. f ( y | p ) = pqy–1

The m.g.f of Geometric Distribution If X1 has a geometric distribution with parameter p, then the m.g.f. It is known that

Example 4 假設某工廠產出不良品的機率為0.1，請問 (1)在發現第一個不良品前有10個良品產出的機率為何？ (2)直到產出10 個或10個以上產品才發現第一個不良品之機率 為何？

Solution 令X 表示發現第一個不良品時檢查出良品的產品數，則 (1)發現第一個不良品前有10個良品產出的機率為 (2)直到產出10 個或10個以上產品才發現第一個不良品之 機率， Or

Normal Distribution There are three reasons why normal distribution is important Mathematical properties of the normal distribution have simple forms Many random variables often have distributions that are approximately normal Central limit theorem tells that many sample functions have distributions which are approximately normal The p.d.f. of a normal distribution

極座標 直角座標與極座標的轉換 ∫ ∫dxdy=rdrdθ X=r*cosθ，y=r*sinθ Ex: x2+y2≦1 (x,y) r

The m.g.f. of Normal Distribution

Properties of Normal Distribution If the random variables X1, …, Xk are independent and if Xi has a normal distribution with mean mi and variance si2, then the sum X1+ . . .+ Xk has a normal distribution with mean m1 + . . .+ mk and variance s12 + . . .+ sk2. Proof: The variable a1x1 + . . .+ akxk+ b has a normal distribution with mean a1m1 + . . .+ akmk + b and variance a12s12 + . . .+ ak2sk2 Suppose that X1, …, Xn form a random sample from a normal distribution with mean m and variance s2 , and let denote the sample mean. Then has a normal distribution with mean m and s2/n.

Example 5 (5.6.11) Suppose that a random sample of size n is to be taken from a normal distribution with mean μ and standard deviation 2. Determine the smallest value of n such that

Solution

Example 6 Suppose that the joint p.d.f. of two random variables X and Y is as follows. Show that these two random variable X and Y are independent.

Solution And Therefore, X and Y are independent Recall that suppose X and Y are random variables that have a continuous joint p.d.f. Then X and Y will be independent if and only if, for and And Therefore, X and Y are independent

Exponential Distribution A random variable X has an exponential distribution with parameters b has: Memoryless property of exponential distribution

Life Test Suppose X1, …, Xn denote the lifetime of bulb i and form a random sample from an exponential distribution with parameter β. Then the distribution of Y1=min{X1, …, Xn} will be an exponential distribution with parameter n β. Proof: Determine the interval of time Y2 between the failure of the first bulb and the failure of a second bulb. Y2 will be equal to the smallest of (n-1) i.i.d. r.v., so Y2 has an exponential distribution with parameter (n-1) β. Y3 will have an exponential distribution with parameter (n-2) β. The final bulb has an exponential distribution with parameter β.

Physical Meaning of Exponential Distribution An exponential distribution is the time required to have for the 1st event to occur, i.e., where β is rate of event. In a Poisson process, both the waiting time until an event occurs and the period of time between any two successive events will have exponential distributions. In a Poisson process, the waiting time until the nth occurrence with rate b has a gamma distribution with parameters n and b.

Example 7 (5.9.11) Suppose that n items are being tested simultaneously, the items are independent, and the length of life of each item has an exponential distribution with parameter β. Determine the expected length of time until three items have failed. Hint: The required value is E(Y1+Y2+Y3)

Solution