Prof. Thistleton MAT 505 Introduction to Probability Lecture 13

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1 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 Sections from Text and MIT Video Lecture: Sections 5.4, probabilisticsystems-analysis-and-applied-probability-fall-2/video-lectures/lecture-8-continuousrandomvariables/ Topics from Syllabus: Gamma Random Variables Review and Looking Ahead We now know how to describe the normal distribution in terms of its probability density function, how the parameters in the PDF relate to the mean and standard deviation of the normal distribution, how to use a table to obtain areas under the standard normal distribution, and, hence, to obtain areas (probabilities) for any normal distribution. The key to these types of problems is to express the probability of your desired event in notation, then either a. convert your event into an equivalent event and use the standard normal distribution. b. use a facility such as pnorm() in R. Who am I and why do you care? More Examples The following examples are taken from Introductory Probability and Statistical Applications by Paul L. Meyer and from Probability: An Introduction with Statistical Applications by John J. Kinney and perhaps other books I can t think of right now. Suppose X N(μ = 3, σ 2 = 4). Find a number c such that P(X > c) = 2P(X c) Since the area under the PDF accumulates (upon integration) to, we can say that {P(X > c) = 2P(X c)} {P(X c) = 3 } Now make a quick call to R: qnorm(/3, 3, 2) # returns SUNY POLY Page

2 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 This is your quantile function in action. You give it a left tail area (together with the mean and standard deviation so we know which normal distribution we are working with) and it returns the x value associated with that tail area. If you drive a Studebaker and like to go to the hop and listen to Bill Haley and the Comets, you can always look up in your z table to find the z value that has an area to the left of.3333 or so. This should give you z =.43 : Convert from standard units z = x μ σ back to your problem s units x = μ + z σ = 3 + (.43) 2 = 2.4. If you insist on using this antiquated approach, you can find your table on the very last page of your text. Suppose that the breaking strength of a cotton fabric (in pounds), say X, has a distribution N(μ = 65, σ 2 = 9). A sample of this fabric is considered to be defective if X < 62. What is the probability that a sample chosen at random from a large lot will be defective? This is fairly straightforward. Modern approach: pnorm(62, 65, 3) #returns Old fashioned approach: z = =, Area =.587 A bag of cement labeled as 4 pounds actually comes from a distribution which is normal with mean of 39. pounds and standard deviation 9.4 pounds. Find the probability that 2 out of 5 randomly selected bags weigh less than 4 lbs. We need to think about this as a normal distribution giving us the basic probability, then use the binomial random variable to finish the problem. P(X < 4) = (use: pnorm(4, 39., 9.4)) P( exactly 2 out of 5) = ( 5 2 ) ( )2 ( ) 3 = SUNY POLY Page 2

3 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 (use choose(5,2)* ( )^2 *( )^3) Capacitors from a manufacturer are normally distributed with mean 5μfand standard deviation.4 μf. An application requires 4 capacitors between 4.3 μf and 5.9 μf. If the manufacturer ships 5 randomly selected capacitors, what is the probability that a sufficient number of capacitors will be within specifications? For an individual capacitor, P( 4.3 < X < 5.9) = pnorm(5.9,5,.4) pnorm(4.3,5,.4) = p = We want P(4 good or 5 good) = choose(5,4) p 4 ( p) + choose(5,5) p 5 ( p) P(4 good or 5 good) = = The Gamma Γ(α, β) distribution: Generalizing the Exponential in the Most Obvious Way There are a wide variety of phenomena whose histograms look sort of mound shaped, but with some rightward skew. We will see soon how to obtain a PDF from theoretical grounds to model these types of phenomena (we need the concept of sums of independent random variables and convolutions). As a quick for instance, you might want to model service times with an exponential distribution. We saw that, when the number of customers grows quite large, a normal distribution is a good mode. What about when the number is not so large? If we let T be the service time of the first customer and T 2 be that of the second, then total service time may be expressed as T = T + T 2 You should take a moment to obtain a histogram of this new random variable. It turns out that we can generalize the exponential distribution to model our situation. We will start with an exponential distribution with λ =, i.e. SUNY POLY Page 3

4 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 g(x) = λ e λx = e x for x We then "tie down" the left side by multiplying g(x) by x. f(x) = x e x Is this still a valid probability distribution? Can we fix this? That is, can we normalize to obtain area under the curve equal to? x e x dx = e x ( x + )] = So we take f(x) = x e x I (,) Now multiply g(x) by x 2 instead of x. Is this still a valid probability distribution? Can we fix this? That is, can we normalize to obtain area under the curve equal to? x 2 e x dx = e x ( x(x + 2) + 2)] = 2 So we take f(x) = 2 x2 e x I (,) Can you stand one more? Multiply g(x) by x 3. Is this still a valid probability distribution? Can we fix this? That is, can we normalize to obtain area under the curve equal to? x 3 e x dx = e x ( x(x(x + 3) + 6) 6)] = 6 SUNY POLY Page 4

5 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 So we take f(x) = 6 x3 e x I (,) Now develop a formula to generate probability distributions of the form f(x) = A n x n e x = n! xn e x I (,) What we have done is to construct a type of distribution called a gamma distribution. In its most general form this distribution is characterized by parameters α and β. To see why these are called gamma distributions consider the following function, called a gamma function: Γ(α) x α e x dx This function has several interesting properties. First, evaluate Γ(), Γ(2), Γ(3), etc. You will find Γ(n) = (n )! Also, you can relate the integral defining Γ ( ) to the integral 2 x2 e 2 dx (hint: use a u = x substitution). This is useful when we talk about χ 2 distributions. As a further hint, you might note that, since the standard normal random variable is a legitimate density x 2 2π e 2 dx =, so substitute u = x, dx = 2 du to get 2 e u2 du = π We can now build a proper probability density function using these ideas. Note again that Γ(α) x α e x dx SUNY POLY Page 5

6 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 and so we can obtain a valid pdf by defining f(x) as f(x) Γ(α) xα e x I [,) This is the gamma distribution with (shape) parameter α. In order to introduce the other (scale) parameter, β, first evaluate the integral I = x α e x β dx Build on the answer from above and define the gamma distribution with parameters α and β f Γ (x) Γ(α)β α xα e x β I [,) Mean and Variance of a Gamma Distribution We can easily show (without really integrating) that the mean of a gamma distribution is μ = α β. Start out by looking at the definition of E[X] E[X] = x Γ(α)β α xα e x β dx If you think about it, this looks almost like a Γ(α + ) distribution. Write out the density f Γ(α+,β) = Γ(α + )β α+ xα e Do a little book keeping. We ll be overly explicit because this is a common technique. We know This means that f Γ(α+,β) dx =, or SUNY POLY Page 6 x β Γ(α + )β α+ xα e x β dx = Γ(α + ) = α Γ(α) and x α = x x α and β α+ = β β α = Γ(α + )β α+ xα e x β dx = α Γ(α)β β α xxα e x β dx

7 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 One more time: α β = x Γ(α) β α xα e x β dx = E[X] Now, you show that the variance of a gamma distribution is σ 2 = α β 2 Examples. One of the reasons why the gamma distribution is so important has to do with the following property: Suppose that X and X 2 are independent (technical term- let s just assume that the outcome of one has no influence on the outcome of another), identically distributed (iid) random variables which have gamma distributions with parameters α and β. Then X + X 2 is also a gamma random variable with parameters 2α and β. (The alphas add up, the betas remain the same.) Demonstrate (do not prove yet) that the sum of two independent gamma distributions is again a gamma distribution. Use R and histograms. Take α = 3 and β = 2. We now have seen a common technique for verifying a theoretical result- produce random data and then superimpose a histogram that you believe is appropriate. Here we will generate gamma deviates, add them, and superimpose an appropriate gamma distribution. We have been told that X, X 2 iid Γ(α = 3, β = 2) X + X 2 Γ(6, 2) Here s our code (with one million deviates) data=rgamma(e6, 3, 2) data2=rgamma(e6, 3, 2) hist(data+data2, 5, freq=false, col='yellow') x=seq(min(data+data2), max(data+data2),.) points( x, dgamma(x, 6, 2 ), type='l', lwd=3) SUNY POLY Page 7

8 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 And, here s our plot. We have a terrific match! 2. Suppose that the lifetimes of two components are exponentially distributed with μ = days. The first is put into operation. As soon as the first fails, the second will be put into operation. What is the probability distribution for the total lifetime of the first component together with its backup? What is the likelihood the system lasts at least 5 days? For each component, its lifetime X and X 2 are exponentially distributed. The exponential distribution is just a special case of the gamma, and so we have X, X iid 2 exp (λ = =.), same as Γ(α =, β = μ = ) μ For the system lifetime X we have So, X Γ(2, ) alphas add up, betas stay the same P(X > 5) = F X (5) = pgamma(5, 2,.) = SUNY POLY Page 8

9 Prof. Thistleton MAT 55 Introduction to Probability Lecture 3 Rayleigh Distribution There are so very many distributions that people have used to model phenomena they find interesting. One of these, the Rayleigh distribution, is used as a model for vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The PDF depends upon a parameter θ as follows: f(x; θ) = x x2 e 2θ θ2 2 I [,) Do you recognize this random variable? Almost like a Gamma, but there is an x 2 term in the exponent. This will pinch your tail down more quickly. But it s not a Gaussian, because there is an x term out front and the support is on the non-negative reals. SUNY POLY Page 9