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Solution Paper Of ISI Previous Years Sample Paper(PSB)

SOLUTION OF ISI SAMPLE PAPER

In ISI M.Stat exam  PSB/MS(Short Answer Type) paper contain 10 questions of which They will take the best 5 answers in concern, i.e. You need to attend 5 correctly at most.Here is an ISI Sample PSB paper with solutions.



Q1. (a) A be a n*n orthogonal matrix, where n is even and suppose |A| = -1,where |A| denotes the determinant of A. Show that | I - A |= 0,where I denotes the n*n identity matrix.
(b) Let A & B be two n*n real matrices such that A^2=A and B^2= B.Suppose that I-(A+B) is invertible. Show that rank(A)=rank(B).
Solution : (a) A' = A^(-1),as A is orthogonal.
Let k be the characteristic root  of A.
Then k = 1/k  , i.e. k=+-1.
Then |A| = -1 implying that at least one ki must be 1.
The characteristic equation is | 1.I - A |= 0,putting k=1.
Hence the proof.
(b) A[ I - A - B ]= - AB,since A^2=A
Similarly, [I - A - B]B= - AB
So,rank(A)=rank[ A(I - A - B)]=rank(-AB)=rank(B).

Q2. Let { Xn : n ≥ 0 } be a sequence of real numbers such that
Xn+1 = p*Xn + (1-p)*Xn-1 ,n≥0,for some 0<p<1.
(a) Show that Xn = Xo + (X1-Xo)∑(p-1)^k,where the sum is taken over k=0 to n-1.
(b) Hence, or otherwise, show that Xn converges & find the limit.
Solution : Xn+1 - Xn =pXn + (1-p)Xn-1 - pXn-1 + pXn-1 -Xn
=(p-1) [ Xn - Xn-1 ]=(p-1)^2 [ Xn-1 - Xn-2 ]= . . . . =(p-1)^n [ X1 - Xo ]
Proceeding in the similar way we have
Xn - Xn-1 =(p-1)^(n-1) [ X1 - Xo ]
.
.
.
X1 - Xo =(p-1)^0 [ X1 - Xo ]
Adding we get Xn - Xo =(X1 - Xo)∑(p-1)^k,where the sum is taken over k=0 to n-1.
Hence the first proof is complete.
(b) Hints: Take the limit in both sides for n tending to infinity. Put the value of infinte geometric series. The limiting value of Xn is Xo + (X1 - Xo)*{1/(2-p)}.

Q3. X1,X2, . . . ,Xn be a random sample drawn from a continuous distribution. The random samples are ranked in increasing order of magnitude. Ri be the rank of the ith sample. Find the correlation coefficient between R1 & R2.
Solution : Here X1,X2, . . . ,Xn are ranked in increasing order. Ri be the rank of Xi.
Ri be the random variable such that
P( Ri = ri )=1/n where ri=1,2, . . . , n.
∑ Ri = 1+2+3+ . . .+n =n(n+1)/2=constant.
And Ri are identically distributed RVs.
So, Cov( R1, ∑ Ri )=0
implying Var(R1)+(n-1)Cov(R1,R2) =0,since Cov(Ri,Rj)=Cov(Ri,Rj')
So, Cov(R1,R2)= - Var(R1)/(n-1)= -(n+1)/12, the variance of uniform distn is  (n^2 -1)/12.
So,the correlation coefficient between R1 & R2 is (use the formula cov/root(v1v2) & put the values from above) - 1/(n+1).

Q4. Suppose X is the number of heads in 10 tosses of a fair coin. Given X=5, what is the probability that the first head occured in the third toss?
Solution : Let X be a RV representing the number of heads in 10 tosses of a fair coin.
So, X following bin(10,1/2).
Now we are to calculate the following
P[ Head occured in the third toss| X=5 ]
=P[ Two tails occured in the first two tosses, a head occured in the third toss, 4 heads occured in 7 tosses ]/P( X = 5 ),trials are independent so express them in term of products of probabilities.
={(1/2)^3 (7C4)(1/2)^7}/{(10C5)(1/2)^10}=5/36.

Q5. Let Y1,Y2,Y3 be i.i.d. Continuous RVs for all  i=1,2. Define Ui as Ui takes
 the value 1 if Yi+1 > Yi & take 0 otherwise. Find the mean & variance of U1+U2.
Solution : E(Ui)=1.P[ Yi+1 > Yi ]=1/2 &
E(Ui^2)= 1^2.P[ Yi+1 > Yi ]=1/2.
V(Ui)=1/2 - 1/4 =1/4
E(U1+U2)=1.1.P[ Y2>Y1 , Y3>Y2 ]=P[Y3>Y2>Y1]=1/6.
Cov(U1,U2)= E(U1,U2)-E(U1)E(U2)=1/6 - 1/2*1/2 = -1/12.
V(U1+U2)= V(U1)+V(U2)+2Cov(U1,U2)=1/4 + 1/4 - 1/6 =1/3.

Q6.Let f(x) = exp{-(x-m)} if x ≥ m & f(x) = 0 otherwise. Find
(a) MLE of m .
(b) 95% C.I. for m .
Solution : (a) The likelihood function is
L(m|¥)=exp[- ∑(xi - m)],where i=1(1)n, ¥ denotes x curl.
The likelihood function will be maximum when ∑(xi - m) is minimum, i.e. , when m is maximum, MLE of m is X(1).
(b) Hints : Find the sufficient statistic by Neyman-Fisher Factorisation theorem,see that X(1) is sufficient. Then find the pdf of this above order statistic.
Then note that 2n(X(1) - m) following chi-square with degree of freedom 2.
Then Use the method of finding confidence interval.

Q7. Suppose X has a normal distn with mean 0 & variance 25. Let Y be an independent RV taking values -1 & +1 with equal probability. Define S=XY + X/Y and T= XY - X/Y .
(a) Find the probability distribution of S.
(b) Find the probability distribution of {(S+T)/10}^2.
Solution : (a) The distribution function of S is given by=P[ S ≤ s ]
= P[ S ≤ s | Y = -1 ] P[ Y = -1 ]+P[ S ≤ s | Y = 1 ] P[ Y = 1 ]
=1/2 *P[-2X ≤ s ] + 1/2*P[2X ≤ s ], since P[ Y = 1 ]=P[ Y = -1 ]=1/2.
=P[ X ≤ s/2 ], since X is symmetrically distributed about '0'.
=P[ (X - 0)/5 ≤ (s - 0)/10 ],since s.d. is equal to 5(given)
= ะค( s/10)
Now we have S following N(0,100)
(b)Hints: S+T = 2XY
(S+T)^2 = 4(XY)^2, here we know P[ Y^2 =1 ]=1.

Q8. Let Y1,Y2,Y3,Y4 be four uncorrelated random variables with E(Yi)= im, Var(Yi)= i^2 v^2 , i=1,2,3,4.,where m,v(>0) are unknown parameters. Find the values of c1,c2,c3,c4 for which ∑ciYi (i=1(1)4) is unbiased for m & has least variance.
Solution : E(∑ciYi)=(∑i*m*ci)m,where ∑ci=1
Again, V(∑ciYi)=∑ci^2*i^2*v^2
As ∑ciYi is unbiased.
=> 1 = (∑ ici )^2 ≤ (i^2ci^2)(∑i), i=1(1)4
=> 1 ≤ i^2*ci^2*4
=> i^2*ci^2 ≥ 1/4
Equality holds when ci=1/4i.