Chapter 3. Julian Chan. June 29, 2012

Transcription

1 Chapter 3 Julian Chan June 29, 202 Continuous variables For a continuous random variable X there is an associated density function f(x). It satisifies many of the same properties of discrete random variables except that instead of having a summation we have an intergral. To be clear it satisifies: f(x) 0 for all x. f(x)dx =. The CDF is given by F (a) = P (X a) = a f(x)dx. P (b X a) = a b f(x)dx = F (a) F (b). By the fundamental theorem of calculus F (x) = f(x). The main difference between continuous and discrete random variables is that instead of a sum there is an intergral. The is illistrated with the poperty of expectation: E[g(x)] = Consequently the mean µ is computed as µ = E[X] = g(x)f(x)dx. and the moment generating function is given by: M x (t) = E[e tx ] = xf(x)dx, e tx f(x)dx. The higer moments are computed in the usual way( either with derivatives or expectation), and the variance is also, but in particular: var(x) = E[x 2 ] (E[x]) 2 = x 2 f(x)dx ( xf(x)dx) 2.

2 Example. Suppose that X is a continuous random variable whose probability density function is given by: { c (4x 2x f(x) = 2 ) 0 x 2 0 else What is the value of c that makes this a density? From the second property of a density we have: 2 0 c (4x 2x2 )dx =. After some calculus we find that f(x)dx = Find the P (X > ). We find that: 2 c = P (X > ) = 8 (4x 2x2 )dx =.5. Example. You have contacted customer service to contact you. You are told you will be given a call between 8 in the morning and 8 hours. When you ask what hour block would be more likely to expect the call you are told that all are equally likely. The density function is that of a uniform distribution which is given by: { f(x) = x 8 0 else where x denotes the hours of the day. Find the probability that the package is delivered between 9 to 2, and 5 to 8 hours. while We compute the mean: P (9 < X < 2) = P (9 < X < 2) = µ = x dx = 3. 0 dx =.3 0 dx =.3. 0 The variance is 8.33, andthemomentgeneratingfunctionise 0t 0. Example. Consider the density function given { 4x 3 f(x) = 5 x 2 0 else Find the expectation and variance. 2

3 2 Uniform A random variable is said to follow a uniform random variable on [a, b] if it has density given by: { f(x) = b a a x b 0 else In this instance the mean, the variance and moment generating function are particularly easy to compute. To check your understanding you should compute them. 3 NORMAL The normal density is given by: f(x) = (x µ)2 exp ( 2πσ 2σ 2 ) for < x <. It has a mean of µ and a variance of σ 2. The moment generating function is given by: M x (t) = exp (µt + σ 2 t 2 /2). This is perhaps one of THE most important proabability density functions known. It has many important properties. Theorem. If X is a normal random variable with mean µ and standard deviation σ then the transformed random variable x µ σ is a STANDARD normal random variable (ie it has mean µ and standard deviation ). Example. An expert witness in a peternity suit testifies that the length in days of a pregnancy (time from impregnation to the delivery of the child) is approximately normally distributed with parameters µ = 270 and σ 2 = 00. The defendant in the suit is able to prove that he was out of the country a period that began 290 days before the birth of the child and ended 240 days before the birth. If the defandant was in fact the father of the child what is the probability that the mother could have had the very long or very short pregnancy indicated by the testimony? The probability we we seek is: P (X < 240 or X > 290) since these probabilities are disjoint we have P (X < 240 or X > 290) = P (X < 240) + P (X > 290) = P (Z < ) + P (Z > ) = 00 3

4 Example. Most galaxies take the form of a flattened disk with the shape of a flattened disk with the majority of the light coming from a thin fundamental plane. The degree of flatteneing differs from galaxy to galaxy. Let X denote the perpendicular distance from the center of the milkyway to a gaseous mass. If X is normally distributed with mean zero and standard deviation 00 parses (9.2 trillion miles) find the probability a gaseous mass is more than 250 parces from the center? The probability we seek is given by: P (X > 250) = P (Z > ) =.9938 = At what distacne does only 20% of the masses lie further than this distance? This is a bit of reverse thinking. The number we seek is a, and it satisifies the property that: P (X > a) =.2 We can find a by using the standard normal table that is we translate it as follows: P (X > a) = P (Z > a 0 00 ) =.2. Using the standard normal tables we find that: a 0 00 =.85. We remark that in this instance the moment generating function is given by M x (t) = exp (µt + σ 2 t 2 /2) = exp (00 2 t 2 /2). We have yet to explain why we have diverged from discrete random random variable. Continuous random variables in part are needed to compute discrete probabilities for large sample sizes. This is illistrated with the following example: Example. Consider a class with 500 students. For this particular class the pass rate for each student is p =.4 and we assume that the students are independent and identically distributed (ie they have roughly the same study patterns and intelligence). Let x be the random variable that a certain number of students pass. We see that this is a binomial distribution since each out come is either a pass or a fail (a Bernoulli trial). What is the probability that exactly 300 pass? P (X = 300) = ( ) However this number is impossible for most calculators to calculate!!! This number is so close to zero that for all practicle purposes it is zero. What is the probability that between 00 and 300 students pass? This is given by: P (00 < X < 300) = 300 X=00 4 ( 500 X ).4 X X.

5 If there is no hope in computing one term of this series what hope is there in computing 200? Fortunately the Binomial can be approximated by a normal random variable with mean np and variacne np( p) thanks to the theorem below. We compute the probability as follows: P (00 < X < 300) P ( < Z < ) = What is the probability that more than 300 pass? This is approximated by P (X > 300) P (Z < ) = 0. Theorem. Let X be a binomial random variable with parameters n and p. For large n (there are a few guide lines for what large n is) the random variable X can be approximated by a normal random variable with mean np and variacne np( p). There are two things to note about this theorem. The first is what is large n, and the second is that a continunity correction error can be used. The following fact is one that we will use later on and is related to hypothesis testing, but can be used in several ways. Let X be a normal random varible with mean µ and variance σ then P ( σ < X µ < σ) =.68, and P ( 2σ < X µ < 2σ) =.95, P ( 3σ < X µ < 3σ) =.997. This is often refered to the 68%, 95%, 99, 7% rule. Example. The time until death of a specific disease is normally distributed. It is found that the mean is given by µ = 8days and the standard deviation is given by.5 days. Find a range for which 95% of all individuals will live once infected. so P ( 2σ < X µ < 2σ) =.95 P ( 2.5 < X 8 < 2.5) =.95, P (5 < X < ) =.95, thus we find that 95% of all individuals will live between 5 and 8 days. Later we will see that this is an example of what is called a confidence interval. 5

6 4 Chebyshev s inequality We would like a rune that is similar to the 68%, 95%, 99, 7% rule for random variables that may not be normally distributed. The best that we can do is what is called Chebyshev s inequality: 5 Gamma P ( X µ ) > kσ) > k 2. The gamma distribution is a two parameter family of continuous probability distribution. For different values of the parameter we associate a different distribution. There are two equivalent (via a U subsitution) form of the gamma distribution. The density of each is given by: Γ(α)β α xα e x β The function Γ(α) is called the gamma function (not a distribution) and is given by: Γ(α) = 0 e t t α dt. Upon integrating by parts one can see that it has the property that: Γ(α) = (α ) Γ(α ). Thus for positive integer values of α this function is exactly the factorial function!!! We also note the important calculation that Γ( 2 ) = π. The gamma distribution has a mean, variance, and moment generating function given by: µ = αβ, σ 2 = αβ 2, and M x (t) = ( βt) α for t < β. When α = we obtain the exponential distribution which we now discuss in more detail. 6 Exponential The exponential distribution is obtained from the gamma distribution with α =. It thus has the following properties: β xe x β The exponential distribution has a mean, variance, and moment generating function given by: µ = β, σ 2 = β 2, and M x (t) = ( βt). 6

7 Example. The time until failure x of a light bulb in days is exponentially distributed with β = /30. The store offers a 0 day warrenty on the poduct. Find the proportion of light bulbs that can be returned? This is given by P (X < 0) = 0 The mean time until failure is given by 30 days. 0 /30 e x/30 dx = e /3 =. There is a fundamental connection to the poisson process as you can guess since both functions involve e. The relationship is: Theorem. Consider a poisson process with paramer λ. Let W be the random varialble time of the occurance of the first event then W is an expoential distribution with parameter β = /λ. Proof. F (W ) = P (W < X) = P (W > X) let y be the number of occurances of the event in the interval [0, X]. Now Y is a poisson distribution with parameter X so we have P (W > x) = P (Y = 0) = e x (x) 0 0! = e x. Example. The average number of lightning strikes on transformers durning a severe thunder storm season in a given area is 2 per week. If the number of lightning strikes follows a poisson process find the probability that durning the next storm season one must wait at most one week in order to see the first transformer strike. We have that λ = / 2 = 2 we now find that: P (X > ) = P (X < ) = 0 2 e x 2dx = ( e 2) = e 2. 7 Chi-squared distribution We briefly mention the Chi-squared distribution denoted χ γ since it will be used in future sections. The χ γ ditribution is defined in terms of a transformation which is the topic which will will talk about at the next of this section, and again in the next chapter. It is one of the most important concepts in probability theory. The transformation is as follows: let X be a gamma random variable with β = 2, and α = γ 2 > 0. The parameter γ determines the distribution, and as γ X converges to a normal random variable. Table 4 of appendix A allows us to work with this probability distribution. 7

8 8 Weibull The Weibull distribution is given by the density function: αβx β e αxβ with parameters α > 0, and β > 0. The Weibull distribution is most commonly used in the application of reliability. However it has several other applications, and is obtained from the gamma distribution. The mean and variacne are given by: µ = α β Γ( + β ) and σ 2 = α 2 β Γ( + 2 β ) µ2. Example. The length of time in hours that a rechargeable battery will hold it s charge is a random variable distributed according to a weibull distribution with α =.0 and β = 2. We first find that the density function is given by: 50xe x2 00 What is the probability that the battery keeps it s charge for at least 0 hours? This is given by:// P (x > 0) = 0 50xe x2 00 dx =. What is the mean length of time a battery will hold it s charge? µ =.0 2 Γ( + 2 ) = 5 π 9 Transformations of random variables In this section I will discuss the CDF method of transformations, and in the book the equivalent PDF method is given. The idea is this if x is a random variable then so is y = x 2, y = x, y = x +. If y = f(x) then one would like to know what is the density of y? For a definite answer as to what the density is one would need a specific function and density for x. These next few examples should be illimunating. Example. Let x be a standard normal random variable and determine the denisty of y = x 2. We start with the CDF of y: F y (a) = P (Y < a) = P (x 2 < a) = P ( a a < x < a) 8

9 The point being here that we do not necessarily know how to calculate P (Y < a) however we can calculate P ( a a < x < a) since we know the density of x, and we do so as follows P ( a a < x < a) = a a 2π exp ( x2 2 ) = F x( a) F ( a). Recall the connection between a CFD and a PDF is given by F (z) = f(z). Thus we have This gives us the pdf of y: F y (a) = F x ( a) F ( a). f y = d da (F x( a) F ( a)) = f x ( a) 2 a f x( a) 2 a = 2aπ exp ( a 2 ). This formula is not good for all values of a only ones for which 0 < a <. Do you see why? Example. Let X be an exponential distribution with parameter one, and y = ln(x). Find the density for y. The same steps (and words apply) so we start with the CDF of y: F y (a) = P (Y < a) = P (ln(x) < a) = P (x < e a ) The point being here that we do not necessarily know how to calculate P (Y < a) however we can calculate P (x < e a ) since we know the density of x, and we do so as follows P (x < e a ) = F x (e a ). Recall the connection between a CFD and a PDF is given by F (z) = f(z). Thus we have The density is given by for < a <. F y (a) = F x (e a ). f y = f x (e a )e a = e ea e a 9

10 Example. Let X be a uniform distribution on [0, ], and y = ln(x). Find the density for y. The same steps (and words apply) so we start with the CDF of y: F y (a) = P (Y < a) = P ( ln(x) < a) = P (x > e a ) = P (x < e a ) The point being here that we do not necessarily know how to calculate P (Y < a) however we can calculate P (x < e a ) since we know the density of x, and we do so as follows P (x < e a ) = F x (e a ). Recall the connection between a CFD and a PDF is given by Thus we have The density is given by F (z) = f(z). F y (a) = F x (e a ). f y = 0 f x (e a ) e a = e a for 0 < a <. Note this this is just the exponential distribution! As we conclude this chapter we will again be asked to calculated probabilities involving transformations as they occur naturally. In addition we will also be asked to determine the distribution of one of the most famous transformations of random variables: x +...+x n n. 0