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
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
(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! 5 6 7 6 3 (b) P(last two successful attempts 7 success). = 8 5 5 6 5 6 c7 5 6 3 5 7 6 3 6 = 8 7 6 9 8 = 7 5 7. 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:
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 4. 84 < 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
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 +.9 =.475 +.45 =.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
. 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
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)
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 = 3 3 3 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 55. 6 is
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 =. + 4. θ = θ + V X = θ + θ + = θ + 4 θ = 4 θ + θ = θ
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 = 4 3 3 b 4 b 4 = λ = λ λ = Ad b 4 = b = 3 a =, b = 3, c = 4.
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 8 3 = 4 3 Area = 4 3 = 4 3.
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
. 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.
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
(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 = 5 3 + 5 + 3 = 4 5 5 = 3 5 P X = P X P X < = P X < + P X = P X < = 5 + 3 + 5 3 + 3 5 = 5 P X 3 = P X 3 P X < = P X 3 P X < = 5 3 + 3 5 + 3 3 5 = 3 9 4 3 = 3 4 4. 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
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 4.6 +.P B 9 6 P B 8 5 ; P B 5. iii From the requiremets, we have P B = 5
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
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
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 7 3 44 3 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 = 4 4 5 or T = 4 4 5 9. (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)
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 = + + 5 = 7.
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
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 3 + 3 3 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.
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
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.