APPM/MATH 4/5520 Solutions to Problem Set Two. = 2 y = y 2. e 1 2 x2 1 = 1. (g 1
|
|
- Percival Hunt
- 6 years ago
- Views:
Transcription
1 APPM/MATH 4/552 Solutions to Problem Set Two. Let Y X /X 2 and let Y 2 X 2. (We can select Y 2 to be anything but when dealing with a fraction for Y, it is usually convenient to set Y 2 to be the denominator.) We will find the joint distribution of Y and Y 2 and then integrate out Y 2 in order to get the marginal distribution for Y alone. we have y g (x, x 2 ) x /x 2 and y 2 g 2 (x, x 2 ) x 2 which gives us x g (y, y 2 ) y y 2 and x 2 g2 (y, y 2 ) y 2. The Jacobian of this transformation is x x y y 2 y J 2 y y 2 y y 2 Since X and X 2 are independent N(, ) random variables, we can get their joint pdf by multiplying two N(, ) pdfs: f X,X 2 (x, x 2 ) e 2 x2 e 2 x2 2 e 2 (x2 +x2 2 ). f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J e 2 (y2 y2 2 +y2 2 ) y 2 (Note that there are no indicators anywhere because all variables go from to so there is no need to zero out the pdfs anywhere.) Now let s compute the marginal pdf for Y. f Y (y ) f Y,Y 2 (y, y 2 ) dy 2 y 2 e 2 y2 2 (+y2 ) dy 2 π y 2 e 2 y2 2 (+y2 ) dy 2 Method One: Do the integral. To do the integral, just think of + y 2 as a constant. In fact, let s call it c. f Y (y ) π y 2 e 2 cy2 2 dy2 With the substitution u y 2 2 (so du 2y 2 dy 2, and u still goes from to ), f Y (y ) π y 2 e 2 cy2 2 dy2 2y 2 e 2 cy2 2 dy2 e 2 cu du π c π(+y 2 )
2 and y goes from to. This is the pdf of the Cauchy distribution with α and β. Y X /X 2 has a Cauchy(α, β ) distribution. Method Two: Integrate without Integrating y 2 e 2 cy2 2 on (, ) looks like the pdf for a Weibull distribution. (The tip off is seeing e to a power to a power!) Specifically, it looks like the Weibull distribution with α, γ 2 and β 2/c which would be ( ) 2 2 y2 e (y 2/ 2/c) 2 cy 2 e 2 cy2 2. 2/c 2/c So f Y (y ) π y 2 e 2 cy2 2 dy2 π cy 2 e 2 cy2 2 dy2 }{{} πc π(+y 2 ) which again is the pdf of the Cauchy distribution with α and β. 2. We need the joint pdf of Y and Y 2. y g (x, x 2 ) x + x 2 and y 2 g 2 (x, x 2 ) x x 2 x g (y, y 2 ) 2 (y + y 2 ) and x 2 g 2 (y, y 2 ) 2 (y y 2 ) The Jacobian is J x y x y 2 y y 2 /2 /2 /2 /2 /2 Since X and X 2 are independent N(, ) random variables, we can get their joint pdf by multiplying two N(, ) pdfs: f X,X 2 (x, x 2 ) e 2 x2 e 2 x2 2 e 2 (x2 +x2 2 ). f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J e 2[ 4 (y +y 2 ) (y y 2 ) 2 ] /2 4π e 8 (2y2 +2y2 2 ). (Note that there are no indicators anywhere because all variables go from to so there is no need to zero out the pdfs anywhere.)
3 Since this joint pdf factors into a y -part and y 2 -part (indicators, though not here, included), we have that Y and Y 2 independent. (The problem is done but, just for the record, both Y and Y 2 are N(, 2) random variables!) 3. First, note that the joint pdf of X and X 2 is f X,X 2 (x, x 2 ) indep f X (x ) f X2 (x 2 ) Γ(α) βα x α e βx I (, ) (x) βe βx 2 I (, ) (x 2 ) Γ(α) βα+ x α e β(x +x 2 ) I (, ) (x) I (, ) (x 2 ) Let Then y g (x, x 2 ) x 2 x + x 2 and y 2 g 2 (x, x 2 ) x + x 2. x g (y, y 2 ) y 2 y y 2 y 2 ( y ) and x 2 y y 2. The Jacobian of the transformation is x x y y 2 J y y 2 y 2 y y 2 y y y 2 y 2 ( y ) y 2. The new joint pdf is f Y,Y 2 (y, y 2 ) f X,X 2 (g (y, y 2 ), g 2 (y, y 2 )) J Γ(α) βα+ [y 2 ( y )] α e β[y 2( y )+y y 2 ] I (, ) (y 2 ( y )) I (, ) (y y 2 ) y 2 Γ(α) βα+ y α 2 ( y ) α e βy2 y 2 I (, ) (y 2 ( y )) I (, ) (y y 2 ). The indicators tells us that we need y 2 ( y ) > and y y 2 > to hold simultaneously. If you shade these regions in the y and y 2 plane, you will see that it corresponds to the infinite height rectangle with base sitting on the y -axis between and. Thus, the region is also described by the constraints < y < and < y 2 <. Since y 2 >, we can drop the absolute value on the y 2 in the joint pdf above and conclude that it is f Y,Y 2 (y, y 2 ) Γ(α) βα+ y α 2 ( y ) α e βy 2 I (,) (y ) I (, ) (y 2 ). Ultimately, we care about the pdf for Y only. In general, we must integrate out y 2 but here, they appear to be independent. Focusing on the y only, we have ( y ) α I (,) (y ). ()
4 While this is not a proper pdf, the constants from the joint pdf must sort themselves out correctly, leaving us with appropriate constants for the marginal pdfs. From (), we conclude that Y Beta(a, b α). 4. For maxes and mins, always start with cdfs! (Unless you want to plug and chug into a formula you have previously derived using cdfs!) (a) F X() (x) P (X () x) P (min(x, X 2,..., X n ) x) P (min(x, X 2,..., X n ) > x) P (X > x, X 2 > x,..., X n > x) indep P (X > x) P (X 2 > x) P (X n > x) ident [P (X > x)] n unif ( x) n f X() (x) d dx F X () (x) n( x) n ( ) n( x) n. To complete the pdf, we include the support for the minimum which is the same as the support for the original random variables: f X() (x) n( x) n I (,) (x). This is the pdf for the Beta distribution with parameters a and b n. X () Beta(, n) Some notes about matching up the constants for a pdf: (i) You don t have to do it! You found a pdf so it must integrate to. If that s the case and you try to change out that leading n for some other constant, it would no longer integrate to and wouldn t be a proper pdf. The Beta(, n) pdf is a proper pdf and looks like f(x) B(, n) ( x)n I (,) (x). Thus, we must have that n /B(, n)! (ii) Alternatively, you can use the facts that and that, for a positive integer n, So you could directly compute that B(a, b) Γ(a)Γ(b) Γ(a + b) Γ(n) (n )! B(, n) n. ( Excitement, not a factorial.)
5 (b) Now for the max... F X(n) (x) P (X (n) x) P (max(x, X 2,..., X n ) x) indep P (X x) P (X 2 x) P (X n x) ident [P (X x)] n unif x n f X(n) (x) d dx F X (n) (x) nx n. To complete the pdf, we include the support of X (n) : f X(n) (x) nx n I (,) (x). This is the pdf for the Beta distribution with parameters a n and b. X (n) Beta(n, ) 5. To find the mean (expected value) for X (), we need to know its distribution. First, note that the cdf for the P areto(γ) distribution is F (x) x γ ( + u) γ+ dx ( + x) γ. F X() (x) P (X () x) P (min(x, X 2,..., X n ) x) P (min(x, X 2,..., X n ) > x) iid [P (X > x)] n [ ] n (+x) γ (+x) γn This is the cdf of a Pareto distribution with parameter γn. You might recognize this because you used the P areto(γ) cdf in this problem. If you do not recognize it, then take the derivative to find the pdf and look in your table of distributions. we have X () P areto(γn). From the table of distributions, we see that the mean is as long as γn >. E[X () ] γn
6 6. This transformation from uniforms to normals is known as the Box-Muller Transformation and is usually used by software packages to simulate normal random variables. Consider X 2 + X2 2 : X 2 + X2 2 ( 2 ln U cos 2 (U 2 ) + ( 2 ln U 2 ) sin 2 (U 2 ) ( 2 ln U )(cos 2 (U 2 ) + sin 2 (U 2 )) ( 2 ln U ) 2 ln U To solve for U 2, I will use the fact that tan x sin x/ cos x, and consider the ratio X 2 /X : X 2 X 2 ln U sin(u 2 ) 2 ln U cos(u 2 ) we have that sin(u 2) cos(2πu 2 ) tan(u 2) U g (X, X 2 ) e 2 (X2 +X2 2 ) and U 2 g 2 (X, X 2 ) tan and the Jacobian of the transformation is u u x J x u 2 u 2 e 2 (x2 +x2 2 ) x 2 e 2 (x2 +x2 2 ) ( ) ( ) x x x 2 x 2 +x2 2 x 2 x 2 +(x 2 /x ) 2 x e 2 (x2 +x2 2 ) x 2 e 2 (x2 +x2 2 ) x x 2 +x2 2 Therefore, the joint density of X and X 2 is given by f X,X 2 (x, x 2 ) f U,U 2 (g (x, x 2 ), g 2 (x, x 2 )) J +(x 2 /x ) 2 e 2 (x2 +x2 2 ). ( ) X2 I (,) (e 2 (x2 +x2)) ( ( )) 2 I (,) tan x 2 x e 2 (x2 +x2 2 ) Here begins a long note about indicators... brace yourself. Note that the first indicator is always since e 2 (x2 +x2 2 ) lives between and. (Don t be alarmed about the idea that it could equal. That will happen when both x and x 2 are zero. Since they are continuous random variables, this will happen with probability zero. For similar reasons, you could have started with your uniforms on [, ] as opposed to (, ) since those single endpoints don t matter for a continuous random variable.) Since the first indicator is always, we can drop it. Now, we are left with only one indicator that takes the value whenever < tan (x 2 /x ) <. By examination of arctan, this appears to only be true for x 2 /x >, which leads us to quadrants and 3 of the (x, x 2 ) plane. This does not seem to have us heading in the right direction for independent standard normal random variables! The problem here is with the definition of the arctan. y tan(x) is not invertible until we restrict its domain. It is usually restricted to ( π/2, π/2) but doesn t have to be. We may X
7 invert it on different regions. In this problem, the restriction ends up restricting possible values for x and x 2! Indeed, at a previous point in this solution, we had that x 2 x tan(u 2 ). Here, the right-hand side is taking on values from to so the left-hand side should be able to take on these values as well. It appears at this point that x and x 2 can both take on values from to. It was not until we applied the arctan that we got some restriction. Indeed, if one looks at the original definitions of x and x 2, x 2 ln u cos(u 2 ) x 2 2 ln u sin(u 2 ) it is easy to see that any (x, x 2 ) pair is possible. Since this region (the entire plane!) is rectangular, we have that the joint pdf for X and X 2 is f X,X 2 (x, x 2 ) e 2 (x2 +x 2) I (, ) (x ) I (, ) (x 2 ) e 2 x2 I(, ) (x ) e 2 x2 2 I(, ) (x 2 ). Since the joint density factors, we see that X and X 2 are independent. Furthermore, from the form of the desity, we see that they are N(, ) random variables! 7. The pdf for X is f(x) ( p) x p I {,,2,...}. The moment generating function is therefore M(t) E[e tx ] x etx P (X x) x e tx ( p) x p p x [e t ( p)] x This looks like a nice geometric sum of the form n r n r if r <. (Note that the sum blows up if we do not have r <.) For our particular problem, the sum is geometric with r e t ( p). converge for r e t ( p) e t ( p) <. In terms of a range of acceptable t values, this means that we need ( ) t < ln ln( p). p The sum will only we will finish off our moment generating function under this assumtion for t. It is M(t) p [e t ( p)] x p e x t ( p)
8 which, again, is valid for t < ln( p). This matches the mgf given in the table of distributions! Yay! 8. There are slicker ways to do this problem. We will show a method that is in line with the basic concepts we have been talking about so far in class. It is usually convenient to think of distributions of mins and or maxes in terms of cdfs first and then to take derivatives to get pdfs. This problem is no exception. We will consider the joint cdf: P (X () x, X (n) y) y x f X(),X (n) (u, v) du dv. Note that, at this point, we do not have to be careful about the relative position of the arguments x and y for this to be true. Indicators in f X(),X (n) (u, v) will take care of that in the end! However, we can get the joint pdf back from the joint cdf by taking derivatives with respect to x and y. Note that P (X () x, X (n) y) if y x. assume that x < y. Following the ideas from the marginal distributions of X () and X (n), you might want to rewrite the event {X () x, X (n) y} in terms of X, X 2,..., X n. This is difficult to do straight away. However, because we have that {X (n) y} {X () x, X (n) y} {X () > x, X (n) y} with the events on the right-hand side being disjoint. Thus, we have that and therefore that Term : P (X (n) y) P (X () x, X (n) y) + P (X () > x, X (n) y) P (X () x, X (n) y) P (X (n) y) P (X () > x, X (n) y) }{{}}{{} 2 P (X (n) y) P (max(x, X 2,..., X n ) y) P (X y, X 2 x,..., X n x) indep P (X x) P (X 2 x) P (X n x) ident [P (X x)] n [F (x)] n where F is the cdf of any one of the original X, X 2,..., X n. Term 2 : Note that the event {X () > x, X (n) y} is equivalent to the event that all of the indinividual X i are between x and y. (Think about it!)
9 Thus, we have P (X () > x, X (n) y) P (x < X y, x < X 2 y,..., x < X n y). By independence of X, X 2,..., X n, this is equal to P (x < X y) P (x < X 2 y) P (x < X n y). By identicalness this is [x < X y] n [F (x) F (y)] n where F is the cdf of any one of the original X, X 2,..., X n. Putting it all together, we have P (X () x, X (n) y) [F (x)] n [F (x) F (y)] n. Taking the derivative with respect to y first yields (you could do x first instead) n[f (x) F (y)] n f(x) n[f (x) F (y)] n f(x). With respect to x now yields n(n )f(x)[f (x) F (y)] n 2 ( f(y)) n(n )f(x)f(y)[f (x) F (y)] n 2. Recall that this only hold for x < y. The joint pdf is otherwise. the final answer is { n(n )f(x)f(y)[f (x) F (y)] n 2, x < y f X(),X (n) (x, y), otherwise.
APPM/MATH 4/5520 Solutions to Exam I Review Problems. f X 1,X 2. 2e x 1 x 2. = x 2
APPM/MATH 4/5520 Solutions to Exam I Review Problems. (a) f X (x ) f X,X 2 (x,x 2 )dx 2 x 2e x x 2 dx 2 2e 2x x was below x 2, but when marginalizing out x 2, we ran it over all values from 0 to and so
More information(x 3)(x + 5) = (x 3)(x 1) = x + 5. sin 2 x e ax bx 1 = 1 2. lim
SMT Calculus Test Solutions February, x + x 5 Compute x x x + Answer: Solution: Note that x + x 5 x x + x )x + 5) = x )x ) = x + 5 x x + 5 Then x x = + 5 = Compute all real values of b such that, for fx)
More information2. The CDF Technique. 1. Introduction. f X ( ).
Week 5: Distributions of Function of Random Variables. Introduction Suppose X,X 2,..., X n are n random variables. In this chapter, we develop techniques that may be used to find the distribution of functions
More informationEco517 Fall 2004 C. Sims MIDTERM EXAM
Eco517 Fall 2004 C. Sims MIDTERM EXAM Answer all four questions. Each is worth 23 points. Do not devote disproportionate time to any one question unless you have answered all the others. (1) We are considering
More information10.7 Trigonometric Equations and Inequalities
0.7 Trigonometric Equations and Inequalities 857 0.7 Trigonometric Equations and Inequalities In Sections 0. 0. and most recently 0. we solved some basic equations involving the trigonometric functions.
More information10.7 Trigonometric Equations and Inequalities
0.7 Trigonometric Equations and Inequalities 857 0.7 Trigonometric Equations and Inequalities In Sections 0., 0. and most recently 0., we solved some basic equations involving the trigonometric functions.
More information2t t dt.. So the distance is (t2 +6) 3/2
Math 8, Solutions to Review for the Final Exam Question : The distance is 5 t t + dt To work that out, integrate by parts with u t +, so that t dt du The integral is t t + dt u du u 3/ (t +) 3/ So the
More information18.440: Lecture 28 Lectures Review
18.440: Lecture 28 Lectures 18-27 Review Scott Sheffield MIT Outline Outline It s the coins, stupid Much of what we have done in this course can be motivated by the i.i.d. sequence X i where each X i is
More informationJoint Probability Distributions and Random Samples (Devore Chapter Five)
Joint Probability Distributions and Random Samples (Devore Chapter Five) 1016-345-01: Probability and Statistics for Engineers Spring 2013 Contents 1 Joint Probability Distributions 2 1.1 Two Discrete
More information2 Random Variable Generation
2 Random Variable Generation Most Monte Carlo computations require, as a starting point, a sequence of i.i.d. random variables with given marginal distribution. We describe here some of the basic methods
More information(y 1, y 2 ) = 12 y3 1e y 1 y 2 /2, y 1 > 0, y 2 > 0 0, otherwise.
54 We are given the marginal pdfs of Y and Y You should note that Y gamma(4, Y exponential( E(Y = 4, V (Y = 4, E(Y =, and V (Y = 4 (a With U = Y Y, we have E(U = E(Y Y = E(Y E(Y = 4 = (b Because Y and
More informationMATH 18.01, FALL PROBLEM SET #5 SOLUTIONS (PART II)
MATH 8, FALL 7 - PROBLEM SET #5 SOLUTIONS (PART II (Oct ; Antiderivatives; + + 3 7 points Recall that in pset 3A, you showed that (d/dx tanh x x Here, tanh (x denotes the inverse to the hyperbolic tangent
More information3.4 Introduction to power series
3.4 Introduction to power series Definition 3.4.. A polynomial in the variable x is an expression of the form n a i x i = a 0 + a x + a 2 x 2 + + a n x n + a n x n i=0 or a n x n + a n x n + + a 2 x 2
More informationAP Calculus Chapter 9: Infinite Series
AP Calculus Chapter 9: Infinite Series 9. Sequences a, a 2, a 3, a 4, a 5,... Sequence: A function whose domain is the set of positive integers n = 2 3 4 a n = a a 2 a 3 a 4 terms of the sequence Begin
More informationOrder Statistics and Distributions
Order Statistics and Distributions 1 Some Preliminary Comments and Ideas In this section we consider a random sample X 1, X 2,..., X n common continuous distribution function F and probability density
More informationBMIR Lecture Series on Probability and Statistics Fall 2015 Discrete RVs
Lecture #7 BMIR Lecture Series on Probability and Statistics Fall 2015 Department of Biomedical Engineering and Environmental Sciences National Tsing Hua University 7.1 Function of Single Variable Theorem
More informationChapter 11 - Sequences and Series
Calculus and Analytic Geometry II Chapter - Sequences and Series. Sequences Definition. A sequence is a list of numbers written in a definite order, We call a n the general term of the sequence. {a, a
More informationThis exam is closed book and closed notes. (You will have access to a copy of the Table of Common Distributions given in the back of the text.
TEST #3 STA 5326 December 4, 214 Name: Please read the following directions. DO NOT TURN THE PAGE UNTIL INSTRUCTED TO DO SO Directions This exam is closed book and closed notes. (You will have access to
More informationLecture 17: The Exponential and Some Related Distributions
Lecture 7: The Exponential and Some Related Distributions. Definition Definition: A continuous random variable X is said to have the exponential distribution with parameter if the density of X is e x if
More informationlim F n(x) = F(x) will not use either of these. In particular, I m keeping reserved for implies. ) Note:
APPM/MATH 4/5520, Fall 2013 Notes 9: Convergence in Distribution and the Central Limit Theorem Definition: Let {X n } be a sequence of random variables with cdfs F n (x) = P(X n x). Let X be a random variable
More information18.440: Lecture 28 Lectures Review
18.440: Lecture 28 Lectures 17-27 Review Scott Sheffield MIT 1 Outline Continuous random variables Problems motivated by coin tossing Random variable properties 2 Outline Continuous random variables Problems
More information3 Continuous Random Variables
Jinguo Lian Math437 Notes January 15, 016 3 Continuous Random Variables Remember that discrete random variables can take only a countable number of possible values. On the other hand, a continuous random
More informationMAS223 Statistical Inference and Modelling Exercises
MAS223 Statistical Inference and Modelling Exercises The exercises are grouped into sections, corresponding to chapters of the lecture notes Within each section exercises are divided into warm-up questions,
More informationSTAT 3610: Review of Probability Distributions
STAT 3610: Review of Probability Distributions Mark Carpenter Professor of Statistics Department of Mathematics and Statistics August 25, 2015 Support of a Random Variable Definition The support of a random
More information2.2 Separable Equations
2.2 Separable Equations Definition A first-order differential equation that can be written in the form Is said to be separable. Note: the variables of a separable equation can be written as Examples Solve
More informationRandom Variables. Random variables. A numerically valued map X of an outcome ω from a sample space Ω to the real line R
In probabilistic models, a random variable is a variable whose possible values are numerical outcomes of a random phenomenon. As a function or a map, it maps from an element (or an outcome) of a sample
More information1.4 Techniques of Integration
.4 Techniques of Integration Recall the following strategy for evaluating definite integrals, which arose from the Fundamental Theorem of Calculus (see Section.3). To calculate b a f(x) dx. Find a function
More informationSolutions to Exam 2, Math 10560
Solutions to Exam, Math 6. Which of the following expressions gives the partial fraction decomposition of the function x + x + f(x = (x (x (x +? Solution: Notice that (x is not an irreducile factor. If
More information1. (25 points) Consider the region bounded by the curves x 2 = y 3 and y = 1. (a) Sketch both curves and shade in the region. x 2 = y 3.
Test Solutions. (5 points) Consider the region bounded by the curves x = y 3 and y =. (a) Sketch both curves and shade in the region. x = y 3 y = (b) Find the area of the region above. Solution: Observing
More informationMat104 Fall 2002, Improper Integrals From Old Exams
Mat4 Fall 22, Improper Integrals From Old Eams For the following integrals, state whether they are convergent or divergent, and give your reasons. () (2) (3) (4) (5) converges. Break it up as 3 + 2 3 +
More information, find P(X = 2 or 3) et) 5. )px (1 p) n x x = 0, 1, 2,..., n. 0 elsewhere = 40
Assignment 4 Fall 07. Exercise 3.. on Page 46: If the mgf of a rom variable X is ( 3 + 3 et) 5, find P(X or 3). Since the M(t) of X is ( 3 + 3 et) 5, X has a binomial distribution with n 5, p 3. The probability
More informationSampling Distributions
In statistics, a random sample is a collection of independent and identically distributed (iid) random variables, and a sampling distribution is the distribution of a function of random sample. For example,
More information1.1 Review of Probability Theory
1.1 Review of Probability Theory Angela Peace Biomathemtics II MATH 5355 Spring 2017 Lecture notes follow: Allen, Linda JS. An introduction to stochastic processes with applications to biology. CRC Press,
More informationMathematic 108, Fall 2015: Solutions to assignment #7
Mathematic 08, Fall 05: Solutions to assignment #7 Problem # Suppose f is a function with f continuous on the open interval I and so that f has a local maximum at both x = a and x = b for a, b I with a
More information(x 3)(x + 5) = (x 3)(x 1) = x + 5
RMT 3 Calculus Test olutions February, 3. Answer: olution: Note that + 5 + 3. Answer: 3 3) + 5) = 3) ) = + 5. + 5 3 = 3 + 5 3 =. olution: We have that f) = b and f ) = ) + b = b + 8. etting these equal
More informationProbability and Distributions
Probability and Distributions What is a statistical model? A statistical model is a set of assumptions by which the hypothetical population distribution of data is inferred. It is typically postulated
More information1 Solution to Problem 2.1
Solution to Problem 2. I incorrectly worked this exercise instead of 2.2, so I decided to include the solution anyway. a) We have X Y /3, which is a - function. It maps the interval, ) where X lives) onto
More informationConstructing Taylor Series
Constructing Taylor Series 8-8-200 The Taylor series for fx at x = c is fc + f cx c + f c 2! x c 2 + f c x c 3 + = 3! f n c x c n. By convention, f 0 = f. When c = 0, the series is called a Maclaurin series.
More informationMultivariate Distributions (Hogg Chapter Two)
Multivariate Distributions (Hogg Chapter Two) STAT 45-1: Mathematical Statistics I Fall Semester 15 Contents 1 Multivariate Distributions 1 11 Random Vectors 111 Two Discrete Random Variables 11 Two Continuous
More informationMath 229 Mock Final Exam Solution
Name: Math 229 Mock Final Exam Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and that it
More information3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y:
3 Algebraic Methods b The first appearance of the equation E Mc 2 in Einstein s handwritten notes. So far, the only general class of differential equations that we know how to solve are directly integrable
More information7 Random samples and sampling distributions
7 Random samples and sampling distributions 7.1 Introduction - random samples We will use the term experiment in a very general way to refer to some process, procedure or natural phenomena that produces
More informationMath 142, Final Exam, Fall 2006, Solutions
Math 4, Final Exam, Fall 6, Solutions There are problems. Each problem is worth points. SHOW your wor. Mae your wor be coherent and clear. Write in complete sentences whenever this is possible. CIRCLE
More informationReview for the Final Exam
Math 171 Review for the Final Exam 1 Find the limits (4 points each) (a) lim 4x 2 3; x x (b) lim ( x 2 x x 1 )x ; (c) lim( 1 1 ); x 1 ln x x 1 sin (x 2) (d) lim x 2 x 2 4 Solutions (a) The limit lim 4x
More information2 Functions of random variables
2 Functions of random variables A basic statistical model for sample data is a collection of random variables X 1,..., X n. The data are summarised in terms of certain sample statistics, calculated as
More informationContinuous Distributions
A normal distribution and other density functions involving exponential forms play the most important role in probability and statistics. They are related in a certain way, as summarized in a diagram later
More information(a) x cos 3x dx We apply integration by parts. Take u = x, so that dv = cos 3x dx, v = 1 sin 3x, du = dx. Thus
Math 128 Midterm Examination 2 October 21, 28 Name 6 problems, 112 (oops) points. Instructions: Show all work partial credit will be given, and Answers without work are worth credit without points. You
More informationLecture 11: Probability, Order Statistics and Sampling
5-75: Graduate Algorithms February, 7 Lecture : Probability, Order tatistics and ampling Lecturer: David Whitmer cribes: Ilai Deutel, C.J. Argue Exponential Distributions Definition.. Given sample space
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More information5.9 Representations of Functions as a Power Series
5.9 Representations of Functions as a Power Series Example 5.58. The following geometric series x n + x + x 2 + x 3 + x 4 +... will converge when < x
More informationTopic 4: Continuous random variables
Topic 4: Continuous random variables Course 3, 216 Page Continuous random variables Definition (Continuous random variable): An r.v. X has a continuous distribution if there exists a non-negative function
More informationX. Numerical Methods
X. Numerical Methods. Taylor Approximation Suppose that f is a function defined in a neighborhood of a point c, and suppose that f has derivatives of all orders near c. In section 5 of chapter 9 we introduced
More informationMATH141: Calculus II Exam #4 review solutions 7/20/2017 Page 1
MATH4: Calculus II Exam #4 review solutions 7/0/07 Page. The limaçon r = + sin θ came up on Quiz. Find the area inside the loop of it. Solution. The loop is the section of the graph in between its two
More information3 rd class Mech. Eng. Dept. hamdiahmed.weebly.com Fourier Series
Definition 1 Fourier Series A function f is said to be piecewise continuous on [a, b] if there exists finitely many points a = x 1 < x 2
More informationSolutions to Math 41 Final Exam December 10, 2012
Solutions to Math 4 Final Exam December,. ( points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is or. x ln(t + ) dt (a) lim x x (5 points)
More informationMa 530 Power Series II
Ma 530 Power Series II Please note that there is material on power series at Visual Calculus. Some of this material was used as part of the presentation of the topics that follow. Operations on Power Series
More informationCh 4 Differentiation
Ch 1 Partial fractions Ch 6 Integration Ch 2 Coordinate geometry C4 Ch 5 Vectors Ch 3 The binomial expansion Ch 4 Differentiation Chapter 1 Partial fractions We can add (or take away) two fractions only
More informationSampling Distributions
Sampling Distributions In statistics, a random sample is a collection of independent and identically distributed (iid) random variables, and a sampling distribution is the distribution of a function of
More informationChapter 5: Integrals
Chapter 5: Integrals Section 5.5 The Substitution Rule (u-substitution) Sec. 5.5: The Substitution Rule We know how to find the derivative of any combination of functions Sum rule Difference rule Constant
More informationJoint Distributions. (a) Scalar multiplication: k = c d. (b) Product of two matrices: c d. (c) The transpose of a matrix:
Joint Distributions Joint Distributions A bivariate normal distribution generalizes the concept of normal distribution to bivariate random variables It requires a matrix formulation of quadratic forms,
More informationMath Review for Exam Answer each of the following questions as either True or False. Circle the correct answer.
Math 22 - Review for Exam 3. Answer each of the following questions as either True or False. Circle the correct answer. (a) True/False: If a n > 0 and a n 0, the series a n converges. Soln: False: Let
More informationPart IA Probability. Definitions. Based on lectures by R. Weber Notes taken by Dexter Chua. Lent 2015
Part IA Probability Definitions Based on lectures by R. Weber Notes taken by Dexter Chua Lent 2015 These notes are not endorsed by the lecturers, and I have modified them (often significantly) after lectures.
More information5.5. The Substitution Rule
INTEGRALS 5 INTEGRALS 5.5 The Substitution Rule In this section, we will learn: To substitute a new variable in place of an existing expression in a function, making integration easier. INTRODUCTION Due
More informationMath 230 Mock Final Exam Detailed Solution
Name: Math 30 Mock Final Exam Detailed Solution Disclaimer: This mock exam is for practice purposes only. No graphing calulators TI-89 is allowed on this test. Be sure that all of your work is shown and
More informationMA1131 Lecture 15 (2 & 3/12/2010) 77. dx dx v + udv dx. (uv) = v du dx dx + dx dx dx
MA3 Lecture 5 ( & 3//00) 77 0.3. Integration by parts If we integrate both sides of the proct rule we get d (uv) dx = dx or uv = d (uv) = dx dx v + udv dx v dx dx + v dx dx + u dv dx dx u dv dx dx This
More informationP (x). all other X j =x j. If X is a continuous random vector (see p.172), then the marginal distributions of X i are: f(x)dx 1 dx n
JOINT DENSITIES - RANDOM VECTORS - REVIEW Joint densities describe probability distributions of a random vector X: an n-dimensional vector of random variables, ie, X = (X 1,, X n ), where all X is are
More informationRepresentation of Functions as Power Series
Representation of Functions as Power Series Philippe B. Laval KSU Today Philippe B. Laval (KSU) Functions as Power Series Today / Introduction In this section and the next, we develop several techniques
More informationMath 162 Review of Series
Math 62 Review of Series. Explain what is meant by f(x) dx. What analogy (analogies) exists between such an improper integral and an infinite series a n? An improper integral with infinite interval of
More informationSTEP Support Programme. Pure STEP 1 Questions
STEP Support Programme Pure STEP 1 Questions 2012 S1 Q4 1 Preparation Find the equation of the tangent to the curve y = x at the point where x = 4. Recall that x means the positive square root. Solve the
More informationDistributions of Functions of Random Variables. 5.1 Functions of One Random Variable
Distributions of Functions of Random Variables 5.1 Functions of One Random Variable 5.2 Transformations of Two Random Variables 5.3 Several Random Variables 5.4 The Moment-Generating Function Technique
More informationChapter 6. Techniques of Integration. 6.1 Differential notation
Chapter 6 Techniques of Integration In this chapter, we expand our repertoire for antiderivatives beyond the elementary functions discussed so far. A review of the table of elementary antiderivatives (found
More informationDIFFERENTIAL EQUATIONS
DIFFERENTIAL EQUATIONS Basic Concepts Paul Dawkins Table of Contents Preface... Basic Concepts... 1 Introduction... 1 Definitions... Direction Fields... 8 Final Thoughts...19 007 Paul Dawkins i http://tutorial.math.lamar.edu/terms.aspx
More informationFinding local extrema and intervals of increase/decrease
Finding local extrema and intervals of increase/decrease Example 1 Find the relative extrema of f(x) = increasing and decreasing. ln x x. Also, find where f(x) is STEP 1: Find the domain of the function
More informationMTH739U/P: Topics in Scientific Computing Autumn 2016 Week 6
MTH739U/P: Topics in Scientific Computing Autumn 16 Week 6 4.5 Generic algorithms for non-uniform variates We have seen that sampling from a uniform distribution in [, 1] is a relatively straightforward
More information11.6: Ratio and Root Tests Page 1. absolutely convergent, conditionally convergent, or divergent?
.6: Ratio and Root Tests Page Questions ( 3) n n 3 ( 3) n ( ) n 5 + n ( ) n e n ( ) n+ n2 2 n Example Show that ( ) n n ln n ( n 2 ) n + 2n 2 + converges for all x. Deduce that = 0 for all x. Solutions
More information4. CONTINUOUS RANDOM VARIABLES
IA Probability Lent Term 4 CONTINUOUS RANDOM VARIABLES 4 Introduction Up to now we have restricted consideration to sample spaces Ω which are finite, or countable; we will now relax that assumption We
More informationSOLUTIONS TO EXAM II, MATH f(x)dx where a table of values for the function f(x) is given below.
SOLUTIONS TO EXAM II, MATH 56 Use Simpson s rule with n = 6 to approximate the integral f(x)dx where a table of values for the function f(x) is given below x 5 5 75 5 5 75 5 5 f(x) - - x 75 5 5 75 5 5
More informationSubsequences and Limsups. Some sequences of numbers converge to limits, and some do not. For instance,
Subsequences and Limsups Some sequences of numbers converge to limits, and some do not. For instance,,, 3, 4, 5,,... converges to 0 3, 3., 3.4, 3.4, 3.45, 3.459,... converges to π, 3,, 3.,, 3.4,... does
More informationSample Spaces, Random Variables
Sample Spaces, Random Variables Moulinath Banerjee University of Michigan August 3, 22 Probabilities In talking about probabilities, the fundamental object is Ω, the sample space. (elements) in Ω are denoted
More informationMATH Solutions to Probability Exercises
MATH 5 9 MATH 5 9 Problem. Suppose we flip a fair coin once and observe either T for tails or H for heads. Let X denote the random variable that equals when we observe tails and equals when we observe
More informationEssex County College Division of Mathematics MTH-122 Assessments. Honor Code
Essex County College Division of Mathematics MTH-22 Assessments Last Name: First Name: Phone or email: Honor Code The Honor Code is a statement on academic integrity, it articulates reasonable expectations
More informationPartial Fractions. (Do you see how to work it out? Substitute u = ax + b, so du = a dx.) For example, 1 dx = ln x 7 + C, x x (x 3)(x + 1) = a
Partial Fractions 7-9-005 Partial fractions is the opposite of adding fractions over a common denominator. It applies to integrals of the form P(x) dx, wherep(x) and Q(x) are polynomials. Q(x) The idea
More informationVII. Techniques of Integration
VII. Techniques of Integration Integration, unlike differentiation, is more of an art-form than a collection of algorithms. Many problems in applied mathematics involve the integration of functions given
More informationFirst Midterm Examination
Çankaya University Department of Mathematics 016-017 Fall Semester MATH 155 - Calculus for Engineering I First Midterm Eamination 1) Find the domain and range of the following functions. Eplain your solution.
More informationIn the real world, objects don t just move back and forth in 1-D! Projectile
Phys 1110, 3-1 CH. 3: Vectors In the real world, objects don t just move back and forth in 1-D In principle, the world is really 3-dimensional (3-D), but in practice, lots of realistic motion is 2-D (like
More informationn! (k 1)!(n k)! = F (X) U(0, 1). (x, y) = n(n 1) ( F (y) F (x) ) n 2
Order statistics Ex. 4. (*. Let independent variables X,..., X n have U(0, distribution. Show that for every x (0,, we have P ( X ( < x and P ( X (n > x as n. Ex. 4.2 (**. By using induction or otherwise,
More informationSuppose we have the set of all real numbers, R, and two operations, +, and *. Then the following are assumed to be true.
Algebra Review In this appendix, a review of algebra skills will be provided. Students sometimes think that there are tricks needed to do algebra. Rather, algebra is a set of rules about what one may and
More informationHomework Solutions: , plus Substitutions
Homework Solutions: 2.-2.2, plus Substitutions Section 2. I have not included any drawings/direction fields. We can see them using Maple or by hand, so we ll be focusing on getting the analytic solutions
More informationPerhaps the simplest way of modeling two (discrete) random variables is by means of a joint PMF, defined as follows.
Chapter 5 Two Random Variables In a practical engineering problem, there is almost always causal relationship between different events. Some relationships are determined by physical laws, e.g., voltage
More information4. Distributions of Functions of Random Variables
4. Distributions of Functions of Random Variables Setup: Consider as given the joint distribution of X 1,..., X n (i.e. consider as given f X1,...,X n and F X1,...,X n ) Consider k functions g 1 : R n
More informationChapter 6: Functions of Random Variables
Chapter 6: Functions of Random Variables We are often interested in a function of one or several random variables, U(Y 1,..., Y n ). We will study three methods for determining the distribution of a function
More informationS6880 #7. Generate Non-uniform Random Number #1
S6880 #7 Generate Non-uniform Random Number #1 Outline 1 Inversion Method Inversion Method Examples Application to Discrete Distributions Using Inversion Method 2 Composition Method Composition Method
More informationIntegrals. D. DeTurck. January 1, University of Pennsylvania. D. DeTurck Math A: Integrals 1 / 61
Integrals D. DeTurck University of Pennsylvania January 1, 2018 D. DeTurck Math 104 002 2018A: Integrals 1 / 61 Integrals Start with dx this means a little bit of x or a little change in x If we add up
More informationMathematics 1 Lecture Notes Chapter 1 Algebra Review
Mathematics 1 Lecture Notes Chapter 1 Algebra Review c Trinity College 1 A note to the students from the lecturer: This course will be moving rather quickly, and it will be in your own best interests to
More informationSUMMATION TECHNIQUES
SUMMATION TECHNIQUES MATH 53, SECTION 55 (VIPUL NAIK) Corresponding material in the book: Scattered around, but the most cutting-edge parts are in Sections 2.8 and 2.9. What students should definitely
More informationLast/Family Name First/Given Name Seat #
Math 2, Fall 27 Schaeffer/Kemeny Final Exam (December th, 27) Last/Family Name First/Given Name Seat # Failure to follow the instructions below will constitute a breach of the Stanford Honor Code: You
More informationReview Problems for the Final
Review Problems for the Final Math -3 5 7 These problems are provided to help you study. The presence of a problem on this handout does not imply that there will be a similar problem on the test. And the
More information5 Integrals reviewed Basic facts U-substitution... 4
Contents 5 Integrals reviewed 5. Basic facts............................... 5.5 U-substitution............................. 4 6 Integral Applications 0 6. Area between two curves.......................
More informationLecture 5: Moment generating functions
Lecture 5: Moment generating functions Definition 2.3.6. The moment generating function (mgf) of a random variable X is { x e tx f M X (t) = E(e tx X (x) if X has a pmf ) = etx f X (x)dx if X has a pdf
More informationStatistics 3657 : Moment Approximations
Statistics 3657 : Moment Approximations Preliminaries Suppose that we have a r.v. and that we wish to calculate the expectation of g) for some function g. Of course we could calculate it as Eg)) by the
More information