Statistics Examples. Cathal Ormond

Transcription

1 Statistics Examples Cathal Ormond

2 Contents Probability. Odds: Betting Combinatorics: kdm Hypergeometric: Card Games Conditional Probability and Bayes Theorem: Urns Decision Trees: Urns Functions and Estimation 6. MGF and Cumulants: Exponential Distribution PGF: Geometric Distribution Cumulants: Sample Average PDF: 007 Q Multivariate Statistics 9. Joint Distributions: 008 Q Multivariate Normal: 007 Q

3 Chapter Probability. Odds: Betting Let s say that Ireland are playing a match against Cyprus, and the bookmaker makes the following offers: Odds Ireland Win Draw Cyprus Win Implied Probability Note, however, that the sum of the probabilities isn t, but.07. Thus, a model for probabibility would be: 7 9 Ireland Win Draw Cyprus Win By offering, the bookmakers are saying that P Ireland win < By offering the other two odds, we have that P Ireland don t win < , and so P Ireland win > , so we have bounds on the probability of a win for Ireland Combinatorics: kdm If we consider a 5D6 experiment, we have outcomes. Let Y be the maximum value on the 5 dice. Thus, we have P Y The number of outcomes where 7776 Y 5 is the same as an experiment where 6 is not included, i.e. a 5D5 experiment. Thus, P Y From this, using the fact that P y k P y k P y k, and 7776 so we can evaluate P y k for all valuse in the experiment.

4 4. Hypergeometric: Card Games Given a random 5-card hand of cards, we will work out the probability of getting certain hands based on the hypergeometric distribution. Note that : 4 of a Kind AAAAB: There are different cards we can choose for A, and there 48 are 5 4 cards we can choose for B. Thus P AAAAB Full House AAABB: We can choose A in ways and B in ways. However, we can also have 4 orderings for A and orderings for B. Thus P AAABB of a Kind AAABC: A can be chosed from cards with ordering 4 and B and C can be chosed in, both with ordering This P AAABC Pair AABBC: We can choose both the two A s and the two B s in ways, both with orderings 4, and we can choose the C in ways with ordering 4. Thus P AABBC Pair AABCD: We can choose the pair A in ways with ordering 4, and we can choose the other cards in ways with orderings. Thus P No Pair ABCDE: As there are no restrictions, we can choose all the cards in 5 ways with ordering each, so P ABCDE Conditional Probability and Bayes Theorem: Urns Let s say that we have urns filled with a certain amount of balls, and we have the following information: Urn Urn Urn 8 red, black 5 red, 5 black red, 4 black From this, we can evaluate the following probabilities: P R U 8 P R U P R U 5 P B U P B U P B U 4 5 where R or B means that the ball drawn was red or black respectively, and U, U, U mean that the ball was chosen from Urn, or. The question we ask is, if we draw a random ball, from a random urn, what is the probability that it is red?

5 5 We assume that the probability of picking an urn is equally likely, i.e. P U P U P U. Thus, we have: P R P R U P U + P R U P U + P R U P U Now, let s look at a different question: if draw a red ball, what is the probability that the ball we have drawn is from each of the urns? We evaluate this by noting, by Bayes Theorem, that P U n R P R U n P U n. Thus, P R by simple calculation, we have:.5 Decision Trees: Urns P U R 8 5 P U R P U R 5 Let s say we have an orn containing 0 balls, 8 red and black. You win a prize if you draw red balls before you draw black balls. If we take the notation n, k to mean that there are n red balls and k black balls left in the urn, the decision tree is as follows: 8, Choose a Red p0.8 7, Choose a Black p0. 8, Choose a Red p0.77 Win Choose a Black p0. 7, Choose a Red p0.88 7, Choose a Black 0.5 Lose Choose a Red p0.875 Win Chose a Black p0.5 Lose Choose a Red p0.857 Win Chose a Black p0. Lose

6 6 Note that in the above tree, p is the probability that the event occurs. Thus, the probability of a win may be derived by adding up all the cases that give us a win, so, we have P Win Now, let s say that if we draw a red on the first draw, we have the choice of quitting and winning 00 or continue as normal, and win 500. The pruned decision tree is as follows: 8, R 7, B 8, Keep Going Stop. R Win 500 B 7, Win 00 R Win 500 B Lose The question we ask is, if we draw a red first, which is better: to keep going or to stop? Note that if we draw a red first, the only way we can lose is if we draw two blacks balls afterwards. Thus, we have P RBB. Thus, the probability of a win is P Win The expected gain if we continue, given that we draw a red ball first is given by EGain R P WinGainWin + P LoseGainLose As this value is positive, 6 6 we should continue.

7 Chapter Functions and Estimation. MGF and Cumulants: Exponential Distribution Consider the following continuous distribution defined by: { λe λx x 0 f X x 0 x < 0 Where λ > 0. We wish to evaluate the MGF, ψ X t. If we let t < λ, we have: ψ X t e tx f X x dx λ e t λx dx [ ] e t λx λ t λ 0 λ t λ Thus, we see that the cumulants are given by κ n ψ n t EX n n! λ n.. PGF: Geometric Distribution We define a geometric discrete random variable by P X x pq x q p. We wish to evaluate the PGF: G X θ θ x P X x x0 θ x pq x x pθ θq x x pθ qθ for θq < for x N, where

8 8 We see that G X θ gives VarX q p. p pθ, and thus we have EX G X p. Computing G X. Cumulants: Sample Average n Let X X n i, where X i are i.i.d. random variables. Then the CFG X is given by i [ n ] C X t nc X t for any X X n i. Thus, we have κ X n C X tn κ X. For greater cumulants, we can see that κ k X κ n k k X. the average is less than the skewness of the original set, i.e. γx κ X.4 PDF: 007 Q The PDF of a random variabel X is given by: t0 We note that the skewness of κ X n γx. Compute E[X]. f X f + b x x b x [0, ], b b If b, compute P x If b, compute E[X X < ] A single observation x is observed. Show how you would obtain a MLE of b..4. Part a We know that E[X] xf X x dx. Thus: E[X] 0 +b b + b x x b+ dx [ b ] x 4 4 xb+4 b+4 +b b 4 b+4 0

9 9.4. Part b We know that P X c.4. Part c c We know that EX X c.4.4 Part d f X x dx. Thus: P X x x + dx [ x x E[X] P X c. Thus: ] EX X E[X] P X +b b 4 b We have Lx; b f X + b b b + b 4 b b. Thus, we have lx; b b loglx; b log + b logb + log + 4 logb b log. Now, if we differentiate l with respec to b, we have l b b + b + b log log 0. Solving this for b would give the b MLE of b.

10 Chapter Multivariate Statistics. Joint Distributions: 008 Q7 Consider the following joint distribution: f XY x, y 8 x xy + y x, y [, ]. Compute the marginal distribution of Y and show that E[Y ] 0. Argue that E[X] 0. Compute the covariance of X and Y. Are they independent?. Obtain an expression for the conditional distribution of Y X x 4. Y is to be predicated from X. A standard approach is to use EY X x as a predicator. Obtain a predication for Y when X 0... Part a We have f Y y We also have E[Y ] f XY x, y dx. And so: f Y Y 8 x xy + y dx [ ] x 8 x y + xy [ y + 8 y y y ] 8 + y yf Y y dy. And so: E[Y ] 8 8 y + y dx [ y + y4] 8 [ + + ] 0

11 as required. By symmetry, we must have that f X x + 8 x, and thus E[X] 0, as above... Partb We know that CovX, Y E[XY ] E[X]E[Y ], and thus we need only compute E[XY ]. Thus: E[XY ] xyf XY x, y dx dy x y x y + xy dx dy [ 4 x4 y x y + ] x y dy 4 y dy [ 4 y] 9 Thus, CovX, Y 0, and so X and Y are not independent... Partc We have f Y X y, x f XY x,y f X x. Thus, we have f Y Xx, y for y [, ]...4 Partd We have E[Y X x] yf Y X y, x dy, and so: E[Y X 0] 0 yf Y X y, 0 dy y y dy [ y4] 4 8 x xy + y + 8 x x xy + y + x

12 . Multivariate Normal: 007 Q8 We have the vector X X X X X 4 T has a multivariate normal distribution with the following parameters: 9 6 µ What is the distribution of X 4?. Derive the distribution of X, X 4 X 0.. Derive the distribution of X + X. 4. Derive the distribution of X + X X + X Compute P X 4 > X X 0.. Part a Quite simply, we know that X 4 is normally distributed with mean µ µ 4 and variance σ σ 44.. Part b We partition µ and Σ into the variables that we know and the variables we wish to include. Thus, we have: 9 6 µ and Σ where we have rearranges the rows/columns of the above matrices so that the variables with known values i.e. X are last. We evaluate µ 4 and Σ 4 µ 4 µ A + Σ AB Σ BB α µ B Σ 4 Σ AA Σ AB Σ BB Σ BA

13 .. Part c We have Y X + X 0 0 X X X Thus, we have µ Y and Σ Y..4 Part d 0 0 X Letting Y X + X and Y X + X 4, we wish to derive the conditional probability of Y given that Y 6. Thus, we have X Y X + X 0 0 X Y Y X + X X Thus, we have: 0 0 µ Y Σ Y X Now, we wish to work out the conditional distribution of Y given Y. This is a normaly distribution with the following parameters: µ µ A + Σ AB Σ BB α µ B Σ Σ AA Σ AB Σ BB Σ BA

14 4..5 Part e If we let Z X 4 X and Z X. We have that Thus, we have: Z Z Z X 4 X 0 0 X µ Z Σ Y As before, we wish to find the conditional distribution of Y given that Y 0. This will be a normal distribution with parameters: µ µ A + Σ AB Σ BB α µ B X X X X 4 Σ 5 Σ AA Σ AB Σ BB Σ BA Y µ Thus, we rescale Y to the normal 0, distribution to see that P Y > 0 P σ P Z > 0.496, according to the Standard Normal Tables. 94 >