Continuous r.v practice problems

Transcription

1 Continuous r.v practice problems SDS 321 Intro to Probability and Statistics 1. ( pts) The annual rainfall (in inches) in a certain region is normally distributed with mean 4 and standard deviation 4. (a) What is the probability that, starting with this year, it will take over 1 years before a year occurs having a rainfall of over 5 inches? Let R denote the amount of rainfall. P (R 5) P ( R ) 1 Φ(2.5).6. Let X be the number of years before we see over 5 inches of rainfall. P (X 1) P (None of the first 1 years have more than 5 inches of rain) (1.6) pt for getting P (R 5). 1 pt for correct calculation with geometric. (b) What is the probability that at least 4 out of the next 5 years will have a rainfall of over 5 inches? This is a binomial probability with n 5, p.6 and np.3. May be better to use X P oisson(3) where X denotes number of years with rainfall more than 5 inches. P (X 4) 1 ( 3 P (X i)) i 1 pt for understanding this is a binomial. 1 pt for the expression. OK if they dont evaluate it. (c) What is the expected number of years with over 5 inches of rainfall in the next 5 years? Expectation of a Poisson: pt for using the expectation of a binomial. (d) What assumptions are you making? The rainfall on each day is independent of each other. 1 pt for correct answer. 2. ( pts) The joint pdf of two random variables X and Y are given by: 24xy x, y [, 1], x + y 1 f X,Y (x, y) otherwise (a) Show that f X,Y (x, y) is a valid joint probability density function. x 24xydxdy 12 x(1 x) 2 dx 12 (x 2x 2 + x 3 ) 1 1 pt for correct limits on y (or x) and 1 pt for seeing it integrates to one. 1

2 (b) Find f X (x). f X (x) y f X,Y (x, y)dxdy x 24xydy 12x(1 x) 2 When x and zero otherwise. same as before. (c) Find E[X] and var(x). E[X] 12 xf X (x)dx 12 var(x) E[X 2 ] E[X] 2 12 x 2 (1 x) 2 dx (x 2 2x 3 + x 4 )dx 12(1/3 1/2 + 1/5) 2/5 x 2 f X (x)dx (2/5) 2 x 3 (1 2x + x 2 )dx 4/25 12 (x 3 2x 4 + x 5 )dx 4/25.4 Any consistent rubric would do as long as they know the definitions of E and Var and know how to integrate x i. (d) Find f Y (y). By symmetry, this is f Y (y) 12y(1 y) 2 when y and zero otherwise. Like the corresponding rubric for X. (e) Find E(Y ) and var(y ). Since the pdf s are the same, the expectation and variances are too. 3. ( pts) The random variables X and Y have joint density function cxy(1 x) < x < 1, < y < 1 f X,Y (x, y) otherwise (a) Find c. c xy(1 x)dxdy c c 12 x(1 x)dx ydy c 12 1 (b) Find E[X]. f X (x) f X,Y (x, y)dy 6x(1 x). So E[X] 6x2 (1 x)dx.5 (c) Find E[Y ]. f Y (y) f X,Y (x, y)dx 2y. So E[Y ] 2y2 dy 2/3. (d) Find V ar(x). var(x) E[X 2 ].25 6x3 (1 x)dx.25 3/1 1/4 1/2 (e) Find V ar(y ). var(y ) E[Y 2 ] 4/9 2y3 dy 4/9 1/2 4/9 1/18 2

3 4. The random variables X and Y have a joint density function given by: 2e 2x /x x <, y x f(x, y) otherwise (a) What are f X (x) and f Y (y)? (b) What are E[X] and E[Y ]? We have: f X (x) y f X,Y (x, y)dy x 2e 2x /xdy 2e 2x. So And So, E[Y ] f Y (y) E[X] yf Y (y)dy 2e 2x x ( x x x2e 2x dx 1/2 f X,Y (x, y)dx y( ydy)dx y 2e 2x x y dx)dy 2e 2x /xdx xe 2x dx 1/4 ve v dv 1/4 In the last step, we changed the order of the integrals. Originally the outer integral was over y < and inner was over y x <. But for ease of integration, the outer integral is now over x < and the inner is over y x. 5. f X (x) C(2x x 3 ) < x < 5/2 x Could f be a probability density function? If so find C. We want 1 5/2 C(2x x 3 )dx C(2.5 2 (2.5) 4 /4) C2.5 2 (1 25/16) 3.51C. Now say we use C 1/ If f(x) > for all x and integrates to one, then we call it a valid pdf. If you can find a x where it is negative then this is not. f(1) C(x 2 x 4 /4) 3C/4. If C < then this is negative and thats not okay. So no, this cannot be a probability density function. 6. Consider the density function f X (x) C(2 x) < x < 2 x Could f be a probability density function? If so find C. We know that 2 C(2 x)dx 1 and so C(4 (2) 2 /2) 2C 1 and so C 1/2. Also (2 x)/2 for < x < 2. So this is a valid pdf. (a) We know that 2 C(2 x)dx 1 and so C(4 (2)2 /2) 2C 1 and so C 1/2. (b) Also (2 x)/2 for < x < 2. (c) So this is a valid pdf. 3

4 7. Let U be an uniform [, 1] r.v and let a < b be constants. Show that: (a) If b > then bu Uniform([, b]). (b) a + U Uniform([a, a + 1]) (c) What function of U is distributed as Uniform([a, b]) (d) Show that min(u, 1 U) Uniform(, 1/2) (a) Let X bu. Fist note the values X bu can take. X [, b] When F X (x) P (X x) P (bu x) For x < F X (x) P (U x/b) x/b du u/b For x/b 1 1 For x/b > 1 So f X (x) 1/b when x [, b] and otherwise. But this is the pdf of a Uniform([, b]). (b) First note that a + U [a, a + 1]. Now, F a+u (x) P (a + U x) P (U x a) (x a), when x [a, a + 1], when x < a and 1 when x > a. So f a+u (x) 1 when x [a, a + 1] and otherwise. This is Uniform[a, a + 1]. (c) From the last two exercises we see that adding a constant shifts the Uniform distribution, and multiplying by a constant stretches it. To convert U nif orm([, 1]) to U nif orm([a, b]) we need both shifting and stretching. Let X αu + β Uniform([a, b]). X [β, α + β]. β a and α + β b and so α b a. So (b a)u + a Uniform([a, b]) (d) Let X min(u, 1 U). First note that X has to lie in [, 1/2]. F X (x) P (min(u, 1 U) x) 1 P (U x, 1 U x) 1 P (x U 1 x) 1 (1 2x) 2x If x 1/2 and otherwise And f X (x) 2 if x [, 1/2]. This is Uniform([, 1/2]). 8. The joint density of X and Y are given by: xe (x+y) x >, y > f X,Y (x, y) otherwise (a) Are X and Y independent? (b) What are f X (x) and f Y (y)? (c) What is P (X + Y 2)? (a) Check if the joint factorizes for all x, y? Yes. So independent. Always remember, check if the bounds on x involve y, that can lead to dependence. Here it does not. (b) f X (x) y xe (x+y) dy xe x (c) f Y (y) x xe (x+y) dy e y xe x dx e y. 4

5 (d) This is convolution, because the random variables are independent. P (X + Y 2) x 2 x 2 x 2 x y f X (x)f Y (y)dxdy 2 x xe x e y dydx y xe x (1 e (2 x) )dx 2 x 2 xe x dx e 2 xdx (1 3/e 2 ) 2/e 2 1 5/e 2 Note that after you plug in f Y (y), you basically say that 2 x so x 2 and that changes the limits on x. 9. The joint density function of X and Y is: x + y < x < 1, < y < 1 f X,Y (x, y) otherwise (a) Are X and Y independent? (b) Find f X (x) (c) Find P (X + Y < 1) (a) No, the joint density does not factorize into a product of functions of x and y. (b) f X (x) y (x + y)dy x + 1/2 and f Y (y) y + 1/2 (c) P (X + Y 1) f X,Y (x, Y )dxdy (x,y):x+y 1 x x y (x + y)dxdy (x(1 x) + (1 x) 2 /2)dx (x x 2 + 1/2 x + x 2 /2) (1/2 x 2 /2) 1/2 1/6 1/3 1. The running time in seconds of an algorithm on a medium sized data set is approximately normally distributed with mean 3 and variance 25. (a) What is the probability that the running time of a run selected at random will exceed 25 seconds? (b) What is the probability that the running time of at least one of four randomly selected runs will exceed 25 seconds? 5

6 (c) What is the probability that the running time of all runs will exceed 25 seconds given at least one of four randomly selected runs will exceed 25 seconds? (a).8413 (b) 1 (1.8413) 4 (c) /(1 (1.8413) 4 ) 11. ( pts) The joint density of X and Y is given by c < x < y, < y < 1 f X,Y (x, y) otherwise (a) What is c? (b) What is f Y (y)? (c) What is E[Y ]? (d) What is the conditional pdf f X Y (x y)? (e) Are X and Y independent? (a) The pdf has to normalize to one. xy f X,Y (x, y)dxdy c 2 x yx cdxdy c (1 x)dx c/2 1 x (b) f Y (y) x f X,Y (x, y)dx y x 2dx 2y When < y < 1 Otherwise (c) E[Y ] yf Y (y)dy 2y 2 dy 2/3 (d) f X Y (x y) f X,Y (x, y) f Y (y) 2 2y 1 y When < x < y, < y < 1 and otherwise. (e) f X (x) y f X,Y (x, y)dy yx 2dy 2(1 x) When < x < 1. otherwise No, since < x < y < 1, there are values such as x.5, y.1, where the pdf is zero but f X (x) and f Y (y) is nonzero. 6

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