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Review of Exam 2 Sections 4.6 – 5.6

Review of Exam 2 Sections 4.6 – 5.6. Jiaping Wang Department of Mathematical Science 04/01/2013, Monday. Outline. Negative Binomial, Poisson, Hypergeometric Distributions and Moment Generating Function Continuous Random Variables and Probability Distribution

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Review of Exam 2 Sections 4.6 – 5.6

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  1. Review of Exam 2Sections 4.6 – 5.6 Jiaping Wang Department of Mathematical Science 04/01/2013, Monday

  2. Outline Negative Binomial, Poisson, Hypergeometric Distributions and Moment Generating Function Continuous Random Variables and Probability Distribution Uniform, Exponential, Gamma, Normal Distributions

  3. Part1. Negative Binomial, Poisson, HypergeometricDistributions and MGF

  4. Negative Binomial Distribution What if we were interested in the number of failures prior to the second success, or the third success or (in general) the r-th success? Let X denote the number of failures prior to the r-th success, p denotes the common probability. The negative binomial distribution function: P(X=x)=p(x)=, x= 0, 1, 2, …., q=1-p If r=1, then the negative binomial distribution becomes the geometric distribution. In summary,

  5. PoissonDistribution The Poisson probability function: P(X=x)=p(x)=, x= 0, 1, 2, …., for λ> 0 The distribution function is F(x)=P(X≤x)= Recall that λ denotes the mean number of occurrences in one time period, if there are t non-overlapped time periods, then the mean would be λt. Poisson distribution is often referred to as the distribution of rare events. E(X)= V(X) = λfor Poisson random variable.

  6. Hypergeometric Distribution Now we consider a general case: Suppose a lot consists of N items, of which k are of one type (called successes) and N-k are of another type (called failures). Now n items are sampled randomly and sequentially without replacement. Let X denote the number of successes among the n sampled items. So What is P(X=x) for some integer x? The probability function is: P(X=x) = p(x) = Which is called hypergeometric probability distribution.

  7. Moment Generating Function The k-th moment is defined as E(Xk)=∑xkp(x). For example, E(X) is the 1st moment, E(X2) is the 2nd moment. The moment generating function is defined as M(t)=E(etX) So we have M(k)(0)=E(Xk). For example, So if set t=0, then M(1)(0)=E(X). It often is easier to evaluate M(t) and its derivatives than to find the moments of the random variable directly.

  8. Part 2. Continuous Random Variables and Probability Distribution

  9. Density Function A random variable X is said to be continuous if there is a function f(x), called probability density function, such that Notice that P(X=a)=P(a ≤ X ≤ a)=0.

  10. Distribution Function The distribution function for a random variable X is defined as F(b)=P(X ≤ b). If X is continuous with probability density function f(x), then Notice that F’(x)=f(x). For example, we are given Thus,

  11. Expected Values Definition 5.3: The expected value of a continuous random variable X that has density function f(x) is given by Note: we assume the absolute convergence of all integrals so that the expectations exist. Theorem 5.1: If X is a continuous random variable with probability density f(x), and if g(X) is any real-valued function of X, then

  12. Variance Definition 5.4: For a random variable X with probability density function f(x), the variance of X is given by V Where μ=E(X). For constants a and b, we have E(aX+b)=aE(X)+b V(aX+b)=a2V(X)

  13. Part 3. Uniform, Exponential, Gamma, Normal Distributions

  14. Uniform Distribution – Density Function Consider a simple model for the continuous random variable X, which is equally likely to lie in an interval, say [a, b], this leads to the uniform probability distribution, the density function is given as

  15. Uniform Distribution – CDF The distribution function for a uniformly distributed X is given by For (c, c+d) contained within (a, b), we have P(c≤X≤c+d)=P(X≤c+d)-P(X≤c)=F(c+d)-F(c)=d/(b-a), which this probability only depends on the length d.

  16. Uniform Distribution -- Mean and Variance -which depends only on the length of the interval [a, b].

  17. Probability Density Function In general, the exponential density function is given by Where the parameter θ is a constant (θ>0) that determines the rate at which the curve decreases. θ = 2 θ = 1/2

  18. Cumulative Distribution Function The exponential CDF is given as θ = 2 θ = 1/2

  19. Mean and Variance Then we have V(X)=E(X2)-E2(X)=2θ2- θ2= θ2.

  20. Probability Density Function (PDF) In general, the Gamma density function is given by Where the parameters α and β are constants (α >0, β>0) that determines the shape of the curve.

  21. = Similary , we can find , so Suppose with being independent Gamma variables with parameters α and β,then .

  22. Probability Density Function In general, the normal density function is given by here the parameters μ and σ are constants (σ >0) that determines the shape of the curve.

  23. Standard Normal Distribution Let Z=(X-μ)/σ, then Z has a standard normal distribution It has mean zero and variance 1, that is, E(Z)=0, V(Z)=1.

  24. Mean and Variance Then we have V(Z)=E(Z2)-E2(Z)=1. As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2.

  25. For example, P(-0.53<Z<1.0)=P(0<Z<1.0) +P(0<Z<0.53)=0.3159+0.2019=0.5178 P(0.53<Z<1.2)=P(0<Z<1.2)-P(0<Z<0.53)=0.3849-0.2019 =0.1830 P(Z>1.2)=1-P(Z<1.22)=1-0.3888=0.6112

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