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**SOLUTION OF ISI SAMPLE PAPER **

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*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.