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This is the Exam of Statistical Science which includes Stochastic Differential Equation, Brownian Motion, Solution, Measurable Function, Markov Process, Starting, Bounded Functions, Local Martingale, First Time etc. Key important points are: Sample of Families, Distributed, Components, Conditional, Heredity Study, Frets, Lengths, Sample of Families, Corresponding Quantities, Vector
Typology: Exams
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Tuesday 7 June, 2005 1:30 to 3:
Attempt THREE questions. There are FOUR questions in total.
The questions carry equal weight.
Cover sheet None Treasury Tag Script paper
1 Suppose the p-dimensional vector X is distributed as Np(μ, V ). Show that if we partition X into components X 1 , X 2 , so that XT^ = (XT 1 , X 2 T ), then the covariance matrix of X 1 conditional on X 2 = x 2 is V 11 − V 12 V 22 − 1 V 21 , where
In a classic heredity study, Frets (1921) measured the head lengths and head breadths on the first and second adult sons in a sample of families. Let (X 1 , X 2 ) be the head length and breadth of the first son and (Y 1 , Y 2 ) the corresponding quantities for the second son.
Considering these four measurements as the 4-vector ZT^ = (XT^ , Y T^ ) = (X 1 , X 2 , Y 1 , Y 2 ), a reasonable model for the population from which the sample has come is a Normal population with mean vector
μ = (μ 1 , μ 2 , μ 1 , μ 2 )T
and dispersion matrix
a b c c b a c c c c a b c c b a
for some positive a, b, c such that V is positive definite, a > c and b > c.
Obtain
(i) the joint distribution of X 1 − Y 1 and X 2 − Y 2 ;
(ii) the marginal distribution of X 1 − Y 1 ;
and
(iii) the conditional distribution of X 1 − Y 1 given that X 2 − Y 2 = 0.
Comment on the differences between (ii) and (iii)
3 Fisher’s classic “iris” data consists of a table 150 × 5, of which the first 3 rows are given in the Splus6 output below. There are 3 distinct species, denoted here by “setosa”, “versicolor” and “virginica”, and we wish to construct a classification tree to sort the 150 iris specimens into species according to the values of Sepal Length, Sepal Width, Petal Length and Petal Width. Explain carefully the construction of the Splus object “iris.tr” in the output below, and sketch the resulting classification tree.
iris[1:3,] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa iris.tr <- tree(Species~.,iris);summary(iris.tr)
Classification tree: tree(formula = Species ~ ., data = iris) Variables actually used in tree construction: [1] "Petal.Length" "Petal.Width" "Sepal.Length" Number of terminal nodes: 6 Residual mean deviance: 0.1253 = 18.05 / 144 Misclassification error rate: 0.02667 = 4 / 150
iris.tr node), split, n, deviance, yval, (yprob)
4 Write brief essays, which should include appropriate sketch graphs, on two of the following topics.
(i) Multivariate Analysis of Variance
(ii) Factor Analysis (iii) Clustering Algorithms.