Due Wednesday, February 1, 11:59 pm.

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Name:  SOLUTIONS

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Submit your computational work both as an R markdown (*.Rmd) document and as a pdf, along with any files needed to run the code.  Embed your answers to each problem in the document below after the question statement. If you have hand-written work, please scan or take pictures of it and include in a pdf file, ideally combined with your pdf output file from R Markdown. Be sure to show your work.

Problem 1 (10 pts)

Genotypes AA, Aa, and aa occur in the population with probabilities (π1 , π2 , π3 ) respectively. A random sample of n from the population gives genotype counts (n1 , n2 , n3 ), where n = n1 + n2 + n3 .

(a) (2 pts) If n = 3, show all the possible observation vectors (n1 , n2 , n3 ).

Answer:

(3,0,0), (2,1,0), (2,0,1), (1,1,1),

(0,3,0), (1,2,0), (0,2,1),

(0,0,3), (1,0,2), (0,1,2)

(b) (2 pts) If (π1 , π2 , π3 ) = (0.25, 0.50, 0.25) and n = 3, compute the multinomial probability that (n1 , n2 , n3 ) = (0, 2, 1).

Answer:

 * 0.250 * 0.502 * 0.251  =  * 0.25 * 0.25 = 0.1875

(c) (2 pts) If (π1 , π2 , π3 ) = (0.25, 0.50, 0.25) and n = 3, compute P (n3  = 1).

Answer:

n3  ~ Binomial(3, 0.25) so

P (n3  = 1) = ╱  \1(3) * 0.251 * 0.752  = 3 * 0.25 * 0.752  = 0.753  = 0.421875 ≈ 0.422

Alternative method:

dbinom(1 ,  size=3 , prob=0.25)

##  [1]  0 .421875

(d)  (2  pts)  If  (n1 , n2 , n3 )   =   (0, 2, 1),  compute  the  log  likelihood  for  (π1 , π2 , π3 )  = (0.25, 0.50, 0.25) using only the kernel of the likelihood.

Answer:

Ignoring the multiplicative constant in the multinomial probability density, the kernel of the likelihood is

y(T) = π1(n)1 π2(n)2 π3(n)3

so the log-likelihood has the form

L(T) = n1 log(π1 ) + n2 log(π2 ) + n3 log(π3 ).

Therefore,

L({0.25, 0.50, 0.25}) = 0 * log(0.25) + 2 * log(0.50) + 1 * log(0.25) = 2 * log(0.50) + log(0.25)

2*log(0.50)+log(0.25)

##  [1]  -2 .772589

(e) (2 pts) If (n1 , n2 , n3 ) = (0, 2, 1), compute (i) the sample proportions pj  = nj /n, j = 1, 2, 3, and (ii) the log likelihood ratio statistic 一2(L({0.25, 0.50, 0.25}) 一 L({p1 , p2 , p3 })).

Answer:

(i)  (p1 , p2 , p3 ) = (0, 2/3, 1/3)

(ii)  L({p1 , p2 , p3 }) = 2 * log(2/3) + log(1/3)

Lp  =  2*log(2/3)+log(1/3)    #  log - likelihood  for  sample  proportions

L0  =  -2.772589                         #  From  part   (d)

-2* (L0  -  Lp)

##  [1]  1 .726093

Problem 2 (10 pts)

For diagnostic testing, let X = true status (1 = disease, 2 = no disease), and let Y = diagnosis (1 = positive, 2 = negative). Let Tij  = P (X = i, Y = j) for i = 1, 2 and j = 1, 2. Also recall our notation that Ti+  = P (X = i) and T+j  = P (Y = j).

(a) (2 pts) The sensitivity of a diagnostic test is the conditional probability that the test is positive given true status = disease. Give a formula for the sensitivity in terms of the Tij .

Answer:

Sensitivity = P (Y = 1|X = 1) =  =

(b) (2 pts) The specificity of a diagnostic test is the conditional probability that the test is negative given true status = no disease. Give the formula for specificity in terms of the Tij .

Answer:

Specificity = P (Y = 2|X = 2) =  =

(c)  (2 pts) Suppose the probability of disease is P (X  =  1)  = 0.01.   Also suppose the sensitivity = 0.86, and the specificity = 0.88.   Determine all the cell probabilities and marginal probabilities for the following table:

Y = 2

T1+

T2+

T+2

Answer:

We are given

 = 0.86

so T11  = (0.01)(0.86) = 0.0086, p12  = T1+ 一 T11  = 0.01 一 0.0086 = 0.0014, and T2+  = 1 一 T1+  = 0.99. We are also given

 = 0.88

so T22  = (0.99)(0.88) = 0.8712 and pi21  = T2+ 一 T22  = 0.99 一 0.8712 = 0.1188.

Finally, T+1  = T11 + T21  = 0.0086 + 0.1188 = 0.1274 and T+2  = 1 一 T+1 = 1 一 0.1274 = 0.8726. Summarizing, we have the table: