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Review of Exam 2

Sections 4.6 – 5.6

Jiaping Wang

Department of Mathematical Science

04/01/2013, Monday

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Outline

Negative Binomial, Poisson, Hypergeometric Distributions and Moment Generating Function

Continuous Random Variables and Probability Distribution

Uniform, Exponential, Gamma, Normal Distributions

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Part 1. Negative Binomial, Poisson, Hypergeometric Distributions and MGF

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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.

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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.

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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.

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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.

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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.

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=

Similary , we can find , so

Suppose with being independent Gamma variables with parameters α and β, then

.

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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