MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #10 SOLUTIONS: SECTIONS 4.7, 4.8, and 4.4 FALL 2003 4.76: If X1 and X2 are independent random variables with normal distributions N (6, 1) and N (7, 1) then the distribution of X1 - X2 is N (6 - 7, 1 + 1) = N (-
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #9 SOLUTIONS: SECTIONS 4.2 and 4.3 FALL 2003 4.17: If f (x) = 1/3 for x = 1, 2, 3 then the density function for Y = 2X + 1 is P (Y = y) = P (2X + 1 = y) = P (X = (y - 1)/2) = 1/3 for y = 3, 5, 7. 4.20: If f (x1
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #8 SOLUTIONS: SECTIONS 3.3, 4.1 FALL 2003 3.35: If X is 2 (5) and P (c < X < d) = 0.95 and P (X < c) = 0.025 then P (X < d) - P (X < c) = P (X < d) - 0.25 = 0.95 so P (X < d) = 0.975. From the tables in the back
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #7 SOLUTIONS: SECTIONS 2.1, 2.2, 2.3, and 2.4 FALL 2003 2.1: If f (x1 , x2 ) = 4x1 x2 for 0 < x1 < 1, 0 < x2 < 1 is the joint density function of X1 and X2 then
1/2 1
P (0 < X1 < 1/2, 1/4 < X2 < 1) =
0 1/4
4x1 x
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #6 SOLUTIONS: SECTION 3.4 FALL 2003 3.39: If X N (75, 100) then = 75 and = 10 so for part a) we have P (X < 60) = P X - 60 - 75 < 10
= P (Z < -1.5) = (-1.5) = 1 - (1.5) = 0.067 using the table in the back of the
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #5 SOLUTIONS: SECTIONS 3.1 and 3.2 FALL 2003 3.3: If X B(n, p) then E[X] = = np and V AR[X] = 2 = np(1 - p). Using that expectation is a linear function we have E[X/n] = (1/n)E[X] = (np)/n = p and E[(X/n - p)2 ]
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #4 SOLUTIONS: SECTIONS 1.8 AND 1.9 FALL 2003 1.80: If f (x) = (x + 2)/18 then E[X] =
4 -2
x(x + 2)/18 = 2 and
4 -2
E[(X + 2)3 ] =
(x + 2)3 (x + 2)/18 = 86.4.
For the third part, since expectation is linear we ha
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #3: SOLUTIONS: SECTIONS 1.5, 1.6 FALL 2003 2 x for x = 1, 2, 3, . . . then we need c x=1 2 = 1. Since in general 3 3 x rx = 1/(1 - r) then x=1 rx = 1/(1 - r) - 1 so that c x=1 2 = c(1/(1 - 2/3) - 1) - 1 = c 2 =
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #2: SOLUTIONS: SECTION 1.4 FALL 2003 1.34: For part a), notice that the event in which the colors alternate occurs only if we choose from the 3524 bowl either RBRB or BRBR in these orders. Since the probability
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #11 SOLUTIONS: SECTIONS 5.2 AND 5.3 FALL 2003 5.9: If Wn is a random variable with mean and variance b/np , to show that Wn converges in probability to we need to show that for arbitrary > 0 we have
n
lim P (|Wn
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #12 SOLUTIONS: SECTIONS 5.4 AND 6.1 FALL 2003 5.20: If X is the mean of a random sample of size 100 from a distribution that is 2 (50) then = r/2 = 25 and = 2 so that = = 50 and 2 = 2 = 100 so that P (49 < X < 5
MA 316: MATHEMATICAL PROBABILITY
ASSIGNMENT #1 SOLUTIONS: SECTIONS 1.2 AND 1.3 FALL 2003
5 1.2: For part a), the answer is cfw_x : 0 < x < 8 and for part b), the answer is cfw_(x, y, z) : x2 +y 2 +z 2 < 1. For part c), the complement should consist of fo
MA 316: MATHEMATICAL PROBABILITY FINAL EXAM REVIEW FALL 2003 The final exam will take place on Wednesday, December 10th from 8:30-11:00 am and will cover everything in the course. It is strongly recommended that you review your assignments and the first e
MA 316: MATHEMATICAL PROBABILITY ASSIGNMENT #13 SOLUTIONS: SECTIONS 6.2 and 6.3 (OPTIONAL) FALL 2003 6.14: If the observed value of X is 81.2 from a random sample of size n = 20 from N (, 80), then a 95% confidence interval for is - 1.96 , X + 1.96 = 81.2
Sample Problems
Statstics 517
1. This problem consists of a collection of unrelated short questions. (a) Let X1 , , Xn be a random sample from Poisson(). Show that M (t) = e(e
t -1)
n i=1 Xi
is dis-
tributed as Poisson(n). (Hint: The moment generating fun
Sample Problems
Statstics 517
1. In a contingency table, 1000 individuals are classified by gender and by whether they favor, oppose, or have no opinion on a complete ban on smoking in public places. The data is the following. Smoking in Public Places Fav
Null(H0): default state, we tend to believe it unless shown otherwise ex: the accused is innocent
Hypothesis Testing
Statistical Hypotheses: assertions about the distribution (parameters) ex. X N (, 2) = 0? = 0?
2 2 ex. X N (1, 1 ) Y N (2, 2 ) 1 = 2? 1 >
Confidence intervals for Means
Central Limit Theorem (CLT) X1, X2, , Xn: observations of a random sample from a distribution with mean and variance 2. Then n D i=1 Xi - n = n(Xn - ) - N (0, 1) Yn = n proof: Use MGF. (p247) X Exponential(), Yn = Another wa
Point Estimation
Likelihood function X1, X2, , Xn: a random sample, considered as outcomes of a random experiment X f (x; ), f (x1, , xn; ) is the joint pdf of X1, , Xn. Given that (X1, , Xn) = (x1, , xn) is observed, the function of defined by L(; x1, ,
Change of Variables
Bivariate case: Discrete: (X1, X2) f (x1, x2) on A y1 = u1(x1, x2), y2 = u2(x1, x2): 1-1 (x1 = w1(y1, y2), x2 = w2(y1, y2) What is the pdf of (Y1, Y2)? g(y1, y2) = f [w1(y1, y2), w2(y1, y2)], (y1, y2) B Continuous:
For (y1, y2) B, g(y1
Probability
Properties: Experiment: any action or process that generates observations. Sample Space C: set of all possible outcomes of the experiment. (or random phenomenon) Event: any collection of outcomes contained in the sample space C. Axioms Conditi
Bayesian Estimation
Bayes' theorem If A1, A2, , An is a partition of the sample space, and B is any set, then for each i, P (Ai|B) = = P (B|Ai)P (Ai) n j=1 P (B|Aj )P (Aj ) P (B|Ai)P (Ai) . P (B)
is a random variable! X f (x|), h() (prior) X : a rv x : o