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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 - PowerPoint PPT Presentation
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The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Review of Exam 2
Sections 4.6 – 5.6
Jiaping Wang
Department of Mathematical Science
04/01/2013, Monday
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Outline
Negative Binomial, Poisson, Hypergeometric Distributions and Moment Generating Function
Continuous Random Variables and Probability Distribution
Uniform, Exponential, Gamma, Normal Distributions
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 1. Negative Binomial, Poisson, Hypergeometric Distributions and MGF
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Negative Binomial Distribution
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,
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 UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Poisson Distribution
The Poisson probability function: P(X=x)=p(x)=, x= 0, 1, 2, …., for λ> 0The 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) = λ for Poisson random variable.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Hypergeometric Distribution
The probability function is: P(X=x) = p(x) = Which is called hypergeometric probability 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?
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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.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Part 2. Continuous Random Variables and Probability Distribution
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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.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
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,
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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
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Variance
Definition 5.4: For a random variable X with probability density function f(x), the variance of X is given by VWhere μ=E(X).
For constants a and b, we have
E(aX+b)=aE(X)+bV(aX+b)=a2V(X)
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Part 3. Uniform, Exponential, Gamma, Normal Distributions
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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
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Uniform Distribution – CDF
The distribution function for a uniformly distributed X is given by
For (c, c+d) contained within (a, b), we haveP(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.
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Uniform Distribution -- Mean and Variance
-which depends only on the length of the interval [a, b].
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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 = 2θ = 1/2
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Cumulative Distribution Function
The exponential CDF is given as
θ = 2θ = 1/2
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Mean and Variance
Then we have V(X)=E(X2)-E2(X)=2θ2- θ2= θ2.
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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.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
=
Similary , we can find , so
Suppose with being independent Gamma variables with parameters α and β, then
.
The UNIVERSITY of NORTH CAROLINA at CHAPEL HILL
Probability Density Function
In general, the normal density function is given byhere the parameters μ and σ are constants (σ >0) that
determines the shape of the curve.
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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.
𝑓 (𝑧 )= 1
√2𝜋exp (− 𝑧 22 ) ,−∞< 𝑧<∞
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Mean and Variance
Then we have V(Z)=E(Z2)-E2(Z)=1.As Z=(X-μ)/σX=Zσ+μE(X)=μ, V(X)=σ2.
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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