IIT JAM Mathematical Statistics (MS) 2006 SECTION A

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1 IIT JAM Mathematical Statistics (MS) 6 SECTION A. If a > for ad lim a / L >, the which of the followig series is ot coverget? (a) (b) (c) (d) (d) = = a = a = a a + / a lim a a / + = lim a / a / + = lim L. L = L < lim a / = lim a = L < lim a / = > ot coverget L. Let E ad F be two mutually disjoit evets. Further, let E ad F be idepedet of G. If p = P E + P F ad q = P G, the P E F G is (a) pq (b) q + p (c) p + q (d) p + q pq (d) P E F G = P E + P F + P G P E F P E G P F G + P E F G = p + q pq as P E F = disjoit ad P E G = P E P G P F G = P F P G

2 3. Let X be a cotiuous radom variable with the probability desity fuctio symmetric about. If V X <, the which of the followig statemets is true? (a) E X = E X (b) V X = V X (c) V X < V X (d) V X > V X (d) Te E X = ; V X = E X E X = E X E X > V X = E X E X > V X 4. Let f x = x x + x, < x <. Which of the followig statemets is true? (a) f is ot differetiable at x = ad x = (b) f is differetiable at x = but ot differetiable at x = (c) f is ot differetiable at x = but differetiable at x = (d) f is differetiable at x = ad x = (b) f x = x x + ; x = x x + ; x = x + x ; x At x =, LHD = RHD = At x =, LHD ; RHD = So, f is differetiable at x = but ot at x = 5. Let A x = b be a o homogeeous system of liear equatios. The augmeted matrix [A: b] is give by Which of the followig statemets is true? (a) Rak of A is 3 (b) The system has o solutio 3 3

3 (c) The system has uique solutio. (d) The system has ifiite umber of solutio. (b) 3 3 ~ 3 3 By R R + R ~ 3 ; by R 3 R 3 R Thus the system is icosistet i.e. has o solutio. 6. A archer makes idepedet attempts at a target ad his probability of hittig the target at each attempt is 5/6, the the coditioal probability that his two attempts are successful give that he has a total of 7 successful attempts is (a) /5 5 (b) 7 5 (c) 5/36 (d) 8! 3!5! (b) P(last two successful attempts 7 success). = c = = Let f(x) = x x x 3 x 4 x 5, < x <. The umber of distict real roots of the equatio d f x = is exactly dx (a) (b) 3 (c) 4 (d) 5 (c) d dx f x has 4 distict roots betwee (, ), (, 3), (3, 4) & (4, 5). See the graph below:

4 8. Let f x = k x + x 4, < x <. The the value of k for which f(x) is a probability desity fuctio is (a) /6 (b) ½ (c) 3 (d) 6 (c) f x will be probability desity fuctio, if f x dx = k x + x 4 dx = kx + x 4 = kβ, = k 4 = k 6 = k = 3 9. If M X t = e 3t+8t is the momet geeratig fuctio of a radom variable X, the P < X 9. 6 is (a) Equal to.7 (b) Equal to.95 (c) Equal to.975 (d) Greater tha.999 (b) M X t = e 3t+8t E e tx = e 3t+8t d dt M X t = 3 + 6t e3t+8t

5 d dt M X t = 3 + 6t e 3t+8t + 6e 3t+8t Puttig t =, we ca get E X = d dt M X t = 3 Puttig t =, we ca get E X = d dt M X V X = 5 9 = 6 σ = 4 P( 4.84 < X < 9.6) = P( 7.84 < X 3 < 6.6) = P(.96 σ < X μ.65 σ) = = =.95 t = 5. Let X be a biomial radom variable with parameters ad p, where is a positive iteger ad p. If α = P X p, the which of the followig statemets hold true for all ad p? (a) α 4 (b) α 4 (c) α 3/4 (d) 3 α 4 (a) P X p kσ < k (By Chebyshev s iequality) As σ = pq = σ pq p X p < pq α 4

6 . Let X, X,, X be a radom sample from a Beroulli distributio with parameter p; p. The bias of the estimator For estimatig p ad is equal to + i= + X i (a) (b) (c) (d) p p + p p p (b) i= X i E + + = + p + = + p + Bias = E Estimator p = + p + p = p + = + p. Let the joit probability desity fuctio of X ad Y be The E(X) is (a).5 (b) (c) (d) 6 (c) f x = f x, y dy = e x dy x = xe x ; x < E X = xf x dx = x e x E(X) = x e x xe x e x = x f x, y = e x, y x <, otherwise

7 3. Let f be defied as f t = ta t t,, t t = The the value of lim x x x 3 x f t dt (a) Is equal to (b) Is equal to (c) Is equal to (d) Does ot exists (a) lim x x x 3 x f t dt = lim x x x 3 ta t dt x t It is case 3x ta x 3 x x = lim 3 x x ta x x, usig L HospitalRule. = lim x 3 ta x 3 ta x x case 9x sec x 3 4xsec x = lim = x 4x 4. Let X ad Y have the joit probability mass fuctio: P X = x, Y = y = y+ y + y + y + x The the margial distributio of Y is (a) Poisso with parameter λ = /4 (b) Poisso with parameter λ = / (c) Geometric with parameter p = /4 (d) Geometric with parameter p = / (d)

8 P Y = y = P X = x, Y = y x= = x= y+ y + y + y + x = y + y + y+ y+ = ; y =,,, y+ Which is geometric with parameter λ = / 5. Let X, X, ad X 3 be a radom sample from N(3, ) distributio. If X = i= X i, ad S = i= X i X Deote the sample mea ad the sample variace respectively, the (a).49 (b).5 (c).98 (d) Noe of the above (b) X = 3 3 i= X i E X = 3 V X = 4 σ = z = X + X + X 3 3 So, P(.65 < X 4.35) = P(.35 < X 3.35) = P(.675σ < X μ.675σ) =.5 P. 65 < X 4. 35,. < S is

9 6. (a) Let X, X,, X be a radom sample from a expoetial distributio with the probability desity fuctio; f x; θ = θe θx, if x >, otherwise Where θ >. Obtai the maximum likelihood estimator of P(X > ). (b) Let X, X,, X be a radom sample from a discrete distributio with the probability mass fuctio give by P X = θ ; P X = = ; P X = = θ θ Fid the method of momets estimator for θ. (a) f x, θ = θe θx, x > P X > = θe θx dx = θe θx θ = θe θ θ = e θ L x, θ = θ e θ x i log L = log θ θ x i L. L θ = θ x i = θ = x i θ = x i MLE of P X > is = e θ = e /( x i) (b) E X =. +. θ = θ + E X = θ = θ + V X = θ + θ + = θ + 4 θ = 4 θ + θ = θ

10 Which is maximum whe θ = / ad miimum whe θ = + or (ot possible). θ = 7. (a) Let A be a o sigular matrix of order (>), with A = k. If adj(a) deotes the adjoit of the matrix A, fid the value of the adj (A). (b) Determie the values a, b ad c so that (,, ) ad (,, ) are eigevectors of the matrix, a 3 3 b c (a) A. adj. A = A I adj A = A adj A = k (b) a 3 3 b c = a 3 c = λ λ =, a =, 3 c =, a =, c = b 4 b 4 = λ = λ λ = Ad b 4 = b = 3 a =, b = 3, c = 4.

11 8. (a) Usig Lagrage s mea value theorem, prove that Where < ta a < ta b < π/ b a + b < ta b ta b a a < + a (b) Fid the area of the regio i the first quadrat that is bouded by y = x, y = x ad the x axis. (a) f x = ta 4 f x = +x By lagrage s mea value theorem i [a, b], we have f c = f b f a b a ta b ta a b a As, = ; a c b + c + b < + c < + a + b < ta b ta a < b a + a b a + b < ta b ta b a a < + a (b) x = x x = x 4x + 4 x 5x + 4 = x x 4 = x =.4 y Required area = dxdy y= x=y = y y dy = y y y3 3 = = 4 3 Area = 4 3 = 4 3.

12 9. Let X ad Y have the joit probability desity fuctio: Evaluate the costat c & P X > Y. f x, y = cx ye x +y, if x >, y >, otherwise We kow that f(x,y) will be joit probability desity fuctio if f x, y dx dy = cx ye x +y dx dy = x= y= c xe x dx ye y dy = c e x 4 e y = c 4 = c = 8 P X > Y = P X > Y + P X < Y = As X, Y > so P X < Y = P X > Y oly = 8xye x e y dx dy y= x=y = 8y e y. e x dy = 4y e 3y dy = 4 6 e 3y = 4 6 = 3

13 . Let PQ be a lie segmet of legth β ad midpoit R. A poit S is chose at radom o PQ. Let X, the distace from S to P, be a radom variable havig the uiform distributio o the iterval (, β). Fid the probability that PS, QS ad PR form the sides of a triagle. Let PS = x P X = x = ; < x < β β PS, SQ ad PR will form a triagle if PS = x SQ = β X ad PR = β/ x + β > β x > β 4 Ad β x + β > x x < 3β 4 Tus for traigle β 4 < x < 3β 4 Required probability = 3β 4 β 4 β =. Let X, X,., X be a radom sample from a N μ, distributio. For testig H : μ = agaist H μ =, the most powerful critical regio is X k, where X = Fid k i terms of such that the size of this test is.5. Further determie the miimum sample size so that the power of this test is at least.95. Size of test, α =.5 P(rejects H is true) =.5 Power of test, β.95 β.95 β.5 E X = μ H is true E X = μ = i= X i.

14 X K K π e x dx =.5 i Also K π e x dx.5 (ii) I itegratio (i) Let x = y dx = dy (k )/ π e y dy =.5. iii Thus K is solutio of equatio (ii) & equatio (iii). Cosider the sequece S,, of positive real umbers satisfyig the recurrece relatio S + S = S + for all. (a) Show that S + S = S S for all (b) Prove that S is a coverget sequece. (a) S + S = S + S + = S + S S = S + S S + S = S S = S S + S S = S S + S S 3 + S S proceedig i similar way we get S + S = S S

15 (b) Also, lim S + S = lim S S = lim S + = lim S S is a coverget sequece. 3. The cumulative distributio fuctio of a radom variable X is give by F x =, if x > /5 + x 3 if x < /5 3 + x, if x > if x Fid P( < X < ), P X ad P X 3. P(<X<) = P(X < ) P X = = = 3 5 P X = P X P X < = P X < + P X = P X < = = 5 P X 3 = P X 3 P X < = P X 3 P X < = = = Let A ad B be two evets with P A B =. 3 ad P A B C =. 4. Fid P B A ad P B C A C i terms of P(B). If 4 P B A /3 ad 4 P BC A C 9 6, the determie the value of P(B). P(A B) = P A B P B =.3

16 P A B =.3P B P A B C = P A BC P B C =.4 P A B C =.4 P B P A = P A B + P A B C =.4.P B P B C A C = P B C P B C A = P B.4 +.4P B =.6 P B P B A = P B A P A =.3P B.4.P B P(B C A C ) = P BC A C P A C =.6 P B.6 +.P B As 4.3P B.4.P B 3.4.P B P B P B 4 3. i &.9P B.4.P B P B 5 ii.6 P B P B 9 6 P B 8 5 ; P B 5. iii From the requiremets, we have P B = 5

17 SECTION B 5. Solve the iitial value problem y y + y x + x + =, y = y y + y x + x + = ; y = dy dx y = y x + y dy dx + y = x + Let y = z y. dy dx = dz dx dz + z = x + dx ze x = e x x + dx + c = x + x + + e x + c z = x + + ce x y = x + + ce x y = = + c c = y = x + is te required solutio. 6. Let y x ad y x be liearly idepedet solutios of xy + y + xe x y = If W x = y x y x y x y x with W =, fid W(5). W 5 = W e 5 5 dx =. e 5 x dx =. e l x 5 =. e [ l 5] = = e l 5 e l 5 = 5

18 7. (a) Evaluate x e xy dx dy. y (b) Evaluate W z dx dy dz where W is the regio bouded by the plaes x =, y =, z =, z = ad the cylider x + y = with x, y. (a) y x e xy dx dy x = x e xy dy dx x= y= = xe xy x dx = xe x dx = e x = e (b) W zdx dy dz x = z dz dy dx x= y= z= = x dy dx = x dx = x x + si x = π = π 8

19 8. A liear trasformatio T R 3 R is give by T x, y, z = 3x + y + 5z, x + 8y + 3z. Determie the matrix represetatio of this trasformatio relative to the ordered bases: {(,, ), (,, ), (,, )}, {(,), (,)}. Also fid the dimesio of the ull space of this trasformatio. T x, y, z = 3x + y + 5z, x + 8y + 3z =, 3x + y + 5z = & x + 8y + 3z = 3y + 4z = y = 4 3 z x = T z 5z 3 = 7z 3 4 a, a, a =, 3 Null space = 7 3 a, 4 a, a a R 3 So, dimesio of ull space= Now T,, = 8,4 = 4, + 4, T,,, = 6, =, + 5, T,, = 3, =, +, Hece required trasformatio i matrix represetatio is T = or T = (a) Let f x, y = x + y x + y,, if x + y = if x + y = Determie if f is cotiuous at the poit (,). (b) Fid the miimum distace from the poit (,, ) to the coe z = x + y. (a)

20 lim x,y, x + y x f x, y = lim x x + y x y = lim x + x x + y x = Alog ay lie or curve. As f, = So, f(x,y) is cotiuous at (,) (b) Ay poit o the coe z = x + y is α, β, α + β Distace of (,, ) from poit i (i) is D = α + β + α + β D = α + β α 4β + 5 For D to be maximum, D will be maximum. D α = 4α = α = D β = 4β 4 = β = D αα = 4; D ββ = 4; D αβ = D αα D ββ D αβ > ad D αα < so,,, 5 is the poit whose distace is miimum. Miimum distace = = 7.

21 SECTION C 3. Let X, X,, X be a radom sample from a expoetial distributio with the probability desity fuctio; f x, θ = x θ e, θ, if x > otherwise Where θ >. Derive the Cramer Rao lower boud for the variace of ay ubiased estimator of θ. Hece prove that T = L x, θ = e θ x i θ i= X i is the uiformly miimum variace ubiased estimator of θ. log L = log θ x i θ L L θ = θ + x i θ = θ = x i θ = sample mea Hece E T = E X i i= = θ Hece T is miimum variace ubiased estimator. 3. Let X, X,, X be a radom sample from a expoetial distributio with the probability desity fuctio; f x = x e x, if x >, otherwise Show that lim P X + + X 3. E X = X = μ = = x 3 xf x dx e x dx = x3 3x 6x 6 e x = 3

22 E X = x f x dx = x 4 4 e x dx = x4 4x 3 x 4x 4 e x = V X = E X E X = 3 = 3 E X + X + + X = 3 V X + X + + X = 3 σ X + X + + X = 3 P X + X + + X As more tha ½ area is covered. P X + X + + X 3 3. Let X, X,, X be a radom sample from a N μ, σ distributio, where both m ad σ are ukow. Fid the value of b that miimizes the mea squared error of the estimator T b = b i= X i X For estimatig σ, where X = i= X i Mea squared error of the estimator T b = b i= X i X If we take b =, the T = i= X i X, is the ubiased estimator of sample variace, so b = is required for ubiased estimate of for estimatig σ.

23 33. Let X, X,, X 5 be a radom sample from a N, σ distributio, where σ is ukow. Derive the most powerful test of size α =. 5 for testig H : σ 4 = 4 Vs. H : σ =. Size, α =.5 P(rejects H is true) =.5 P(rejects σ = 4) =.5 For probability =.5 Area of acceptace = 95% μ.96σ < X < μ +.96σ.9, 5.9 is acceptace rage. Rest regio is rejectio regio β = P acceptace H is true = P 3.9σ, < X < + 3.9σ β =. (approx) power, β =.999 (approx). 34. Let X, X,, X be a radom sample from a cotiuous distributio with the probability desity fuctio; f(x; λ) = x λ e x /λ, if x >, otherwise where λ >. Fid the maximum likelihood estimator of λ ad show that it is sufficiet ad a ubiased estimator of λ. f x, λ = x λ e x /λ, if x >, x Likelihood fuctio L x, λ = x f x, λ = x i λ e x i λ log L = log log λ + log x i x i L. L λ = λ + λ x i λ

24 L λ = λ = λ x i λ = x i Hece maximum likelihood estimator of λ is λ = x i Also, E x i = λ Ad as values of x ad λ ca be writte separately so, x i is sufficiet ad ubiased estimate of λ.

2. The volume of the solid of revolution generated by revolving the area bounded by the

2. The volume of the solid of revolution generated by revolving the area bounded by the IIT JAM Mathematical Statistics (MS) Solved Paper. A eigevector of the matrix M= ( ) is (a) ( ) (b) ( ) (c) ( ) (d) ( ) Solutio: (a) Eigevalue of M = ( ) is. x So, let x = ( y) be the eigevector. z (M