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

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1 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 I)X = x ( ) ( y) = ( ) z y =, z = Hece ( ) is a eigevector.. The volume of the solid of revolutio geerated by revolvig the area bouded by the curve y = x ad the straight lies x = 4 ad y = about the x axis, is (a) π (b) 4π (c) 8π (d) π Solutio: (c) Volume of solid revolutio about x axis 4 v = π y dx = πxdx = πx 4 = 8π. 4

2 x x y y y y 3. Let I = xydxdy. The chage of order of itegratio i the itegral gives I as y y y y (a) I = xydxdy + xydxdy (b) I = xydxdy + xydxdy (c) I = xydxdy + xydxdy (d) I = xydxdy + xydxdy Solutio: (a) I the give itegratio, regio bouded is from y = x to y = x ad x = to x =. By chagig the order of itegratio, we get y y xydxdy + xydxdy y= x= y= x= 4. Let L = lim [f ( ) + f ( ) + + f (k ) kf()] where k is a positive iteger, if f(x)= si x, the L is equal to (a) (k+)(k+) 6 (b) (k+)(k+) (c) k(k+) (d) k(k + ) Solutio: (c) L = lim [f ( h ) + f ( h ) + + f (k ) kf()] = lim [si h + si h + + si k k si ]

3 L = lim si h h + si h L = k = + + si k k k(k + ) k 5. Let f(x, y) = { x + y si ( x + y ) if (x, y) (, ) if (x, y) = (, ) The at the poit (, ) (a) f is cotiuous ad f f ad exist. x y (b) f is cotiuous ad f f ad do ot exist. x y (c) f is ot cotiuous ad f f ad exist. x y (d) f is ot cotiuous ad f f ad do t exist. x y Solutio : (b) So, f(x,y) is cotiuous at (,) Which does ot exist. Lt f(x, y) = Lt f(r cos θ, r si θ) (x, y) (,) r = Lt r r si r = = f(,) t at (,) = Lt f( + h, ) f(,) x h h t x = Lt h h si h h f(,) + k f(,) at (,) = lim k k k si k = lim does ot exist. k k

4 6. Let {a } be a real sequece covergig to a, where a >. The (a) a coverges but a diverges (b) a diverges but a coverges (c) Both a ad a coverge (d) Both a ad a diverge Solutio: (d) As, lim a = a > So, = a diverges as lim a = is ecessary coditio for covergece. a Also = diverges as lim ( a ) = a > ad by Śleszyński Prigsheim theorem lim u = is ecessary for covergece of positive term series. 7. A four digit umber is chose at radom. The probability that there are exactly two zeroes i that umber is (a).73 (b).973 (c).7 (d).7 Solutio: (c) Number of 4 digit umbers = 9 3 = 9 Number of 4 digit umbers with exactly zeroes = 9C 3C 9 = 43 So, required probability = 43 9 =.7 8. A perso makes repeated attempts to destroy a target. Attempts are made idepedet of each other. The probability of destroyig the target i ay attempt is.8. Give that he fails to destroy the target i the first five attempts, the probability that the target is destroyed i the 8 th attempt is (a).8 (b).3 (c).6 (d).64 Solutio: (b) Probability that target will be hit i 8 th attempt is equal to product of probability of failure i the 6 th ad 7 th attempt ad success i the 8 th attempt. P = (.)(.)(.8) =.3

5 9. Let the radom variable X~B(5, p) such that P(X = ) = P(X = 3). The the variace of X is (a) /3 (b) /9 (c) 5/3 (d) 5/9 Solutio: (b) P(x = ) = P(x = 3) ( 5 ) p ( p) 3 = 5 c3 p 3 ( p) Variace of x = p( p) = = 3. p = p 3p = p = 3. Let X,., X 8 be i.i.d. M(, σ ) radom variables. Further, let U = X + X ad V = (a) /8 (b) ¼ (c) ¾ (d) ½ 8 i= X i. The correlatio coefficiet betwee U ad V is Solutio: (d) E(X i ) = ; V(X i ) = σ ; E(X i ) = σ ; E(UV) = E(X + X ) = σ E(U)E(V) = ; Var (U) = σ ; Var (V) = 8σ ; Cov (U, V) = E(UV) E(U)E(V) = σ Correlatio coefficiet U ad V is = COV (U, V) var (U), Var (V) = σ σ, 8σ =. Let X~F 8.5 ad Y~F 5.8. If P(X>4) =.ad P(Y k) =., the the value of k is (a).5 (b).5 (c) (d) 4 Solutio: (b) X~F 8, 5 ad Y~F 5, 8 Y = X ; P(X > 4) = P ( ) = P (Y )= P(Y.5). X 4 4

6 . Let X,, X be i.i.d. Exp () radom variables ad S = i= X i. Usig the cetral limit theorem, the value of lim P(S > ) is (a) (b) /3 (c) ½ (d) Solutio: (c) X i ~ exp() E(X i ) = ; V(X i ) = Where σ is v(x i ) S μ ξ, where ξ follows N(, σ ) lim P(S > ) = lim P ( S. > ) l N(,)P(x > ) = / 3. Let the radom variable X~U (5, 5 + θ). Based o a radom sample of size, say X, the ubiased estimator of θ is (a) 3(X 5) (b) X 5 (c) 3(X + 5) (d) X +5 Solutio: (a) 5+θ E(X ) = x. θ dx = (5 + θ) θ So, 3(X 5) is ubiased estimator of θ. 3θ = θ 3 + 5θ + 5 E(X) = x. θ dx = (5 + θ) 5 = θ θ E(X 5) = E(X ) E(X) + 5 = θ 3 E[3(X 5) ] = θ

7 4. Let X,, X be a radom sample of size from N(μ, 6) populatio. If a 95% cofidece iterval for μ is [X. 98, X +. 98], the the value of is (a) 4 (b) 6 (c) 3 (d) 64 Solutio: (d) 95% cofidece iterval for μ is [X.96σ, X +.96σ ].96σ =.98 = σ = (4) = 8 = A coi is tossed 4 times ad p is the probability of gettig head i a sigle trial. Let S be the umber of head(s) obtaied. It is decided to test H : p = agaist H : p Usig the decisio rule: Reject H if S is to 4. The probabilities of Type I error (α), ad Type II error (β) whe p = 3 4, are (a) α = 87, β = 4 8 (b) α = 87, β = 8 8 (c) α = 4, β = 8 56 (d) α = 4, β = 4 56 Solutio: (b) α = p (reject H /H is true) i.e., p = / & Rejectio H meas S =,4 α = P(S = or 4 P = /) = 4C ( 4 ) + 4C 4 ( 4 ) = 8 β = P(accept H H is true) = P(S =,,3 P = 3/4) = 4C ( 3 4 ) ( 4 ) 3 = + 4C ( 3 4 ) ( 4 ) + 4C 3 ( 3 4 ) 3 ( 4 ) = = 87 8.

8 6. (a) Fid the value(s) of λ for which the followig system of liear equatios λ x ( λ ) ( y) = ( ) λ z (i) (ii) (iii) Has a uique solutio Has ifiitely may solutios Has o solutio (b) Let a =, b = ad for, a + = a + b, b + = a b a + b Show that (i) b a, for all (ii) b + b for all, (iii) Solutio: (a) The sequeces {a }ad {b } coverge to the same limit. By R R R & R 3 R 3 λr By R 3 R 3 + R λ λ [A: B] = [ λ ] ~ [ λ λ ] λ λ λ λ λ ~ [ λ λ ] λ λ λ (i) For uique solutio (λ )( λ λ ) (λ ) (λ + ) λ or (ii) (iii) For λ =, we have ifiitely may solutios For λ =, we have o solutio. (b) As a + is arithmetic mea ad b + is harmoic mea betwee a ad b, so a + b +, as A. M. G. M. Hece, we have (i) b a, for all Also b b b 3 mootoically decreasig ad bouded below by b. So it is coverget ad similarly {b } is mootoically icreasig ad bouded above by a. So it is also coverget. Let Lt a = l ad Lt b = m

9 So, {a }ad {b } coverges to same limit. Let l be the commo limit of {a } ad {b } Lt a + = lim a + b l + m l = l = m. a + b + = a b = a b = = a b a b = a b = lim a b = l = l = Hece {a } ad {b } coverges to same limit =. 7. (a) Solve: (x y 3 + xy)dy = dx (b) Fid the geeral solutio of the differetial equatio (D 4D + 4)y = x si x, where D d dx Solutio: (a) (x y 3 + xy)dy = dx dx dy = xy + x y 3 x dx dy = x y + y3 ; let x = z x dx dy = dz dy dz dy = yz + y3 dz + yz = y3 dy ze y = e y ( y 3 )dy + e; Let y = t ydy = dt ze t = e t t dt + c = [tet e t ] + c z = [t ] + ce t x = [y ] + ce y is the solutio. (b) (D 4D + 4)y = x si x (D ) y = x si x Auxiliary equatio is (m ) =

10 m =, So, complemetary fuctio is y = (c + c x)e x Particular itegral is = y = x si x (D ) (D ) = [x (D ) ] si x (D ) = [x D ] D si x 4D + 4 = = [x D ] si x 4D = [x cos x ] D 8 x cos x 8 x cos x 8 = 4 D + D (cos x) 4 + ( si x + cos x) 3 x cos x 8 cos x si x + 6 = [( + x) cos x si x] 6 So, the complete solutio is y = (c + c x)e x + [( + x) cos x si x] 6 8. (a) Fid all the critical poits of the fuctio f(x, y) = x 3 + y 3 + 3xy ad examie those poits for local maxima ad local miima (b) If f is a cotiuous real valued fuctio o [,], show that there exists a poit c (, ) such that Solutio: (a) f(x, y) = x 3 + y 3 + 3xy xf(x)dx = f(x)dx. c

11 f x = 3x + 3y ; f y = 3y + 3x f xx = 6x ; f yy = 6y f xy = 3 For critical poits, we have f x = ad f y = x + y = ad y + x = y = x x 4 + x = x( + x 3 ) = x =, At x =, y = ; at x =, y = So, critical poits are (,) ad (, ) f xx f yy (f xy ) = 36xy 9 = 9(4xy ) At (,); f xx f yy (f xy ) = 9 <. So, (, ) is a saddle poit. At (, ); f xx f yy (f xy ) = 9(4 ) = 7 > Ad f xx at (, ) = 6 <. So(, ) is local maxima poit. (b) xf(x)dx lies betwee f(x)dx ad. f(x)dx Now as f(x) is a cotiuous fuctio i the iterval [,], so f(x)dx f(x)dx Now, as f(x) is cotiuous fuctio, so itegral Hece, f(x)dx exist for each c [,]. c f(x)dx = f(x)dx for a poit c (,) c

12 9. (a) Evaluate the triple itegral z=4 x= z y= 4z x dydxdz z= x= y= (b) Let M = ( ). If M = 5 I + km M, where I is the idetity matrix of order 3, fid the value of k. hece or otherwise, solve the system of equatios: x M ( y) = ( ) z Solutio: (a) z=4 x= z y= 4z x dydxdz = z= x= y= z=4 = z= z=4 x= z 4z x dxdz z= x= [ x 4z x + 4z x= z x si z ] dz = x= z=4 z= πzdz = 8π. so, (b) By Cayley Hamilto theorem, every square matrix satisfies it s characteristic equatio, M λi = is satisfied by M. λ M λi = λ = ( λ)[ + λ 3λ] + λ = λ 3 + 4λ 5λ + 4 Hece, M 3 + 4M 5M + 4I = M + 4M 5I + 4M = (By takig product by M ) M = 4 [M 4M + 5I] = 4 M M I K = 6 Now, M = ( ) M = ( 4 3)

13 M = 4 [M 4M + 5I] = 4 ( ) M = 4 ( ) x x M ( y) = ( ) ( y) = M ( ) z z 4 ( 4 4 x ( y) = z ( ) 4 ) 4 ( 4) x = ; y = 4 & z = 4. (a) Let N be a radom variable represetig the umber of fair dice throw with probability mass fuctio P(N = i) =, i =,,.. i Let S be the sum of umbers appearig o the faces of the dice. Give that S = 3, what is the probability that dice were throw? (b) Let X~ N(, )ad Y = X + X. Fid E(Y 3 ). Solutio: (a) Prob. ( dice were throw sum o the face = 3) = Prob. ( dice were throw sum o the face = 3) Prob. (sum o the face = 3) = P( s = 3) P( s = 3) + P( s = 3) + P(3 s = 3)

14 = = = = (b) X~N(,) Y = X + X = X; X > = ; X Y 3 = 8X 3 X > = ; X E(Y 3 ) = 8x 3. π e x / dx Let z = x dz = dx. Let Y ~N(μ, σ y ) ad Y = l X. E(Y 3 ) = π 6ze z dz = π 6 = 8 π (a) Fid the probability desity fuctio of the radom variable X ad the media of X. (b) Fid the maximum likelihood estimator of the media of the radom variable X based o a radom sample of size. Solutio: (a) Y~N(μ y, σ y ); Y = l X is log ormal distributio whose PDF, f(x) is Media of X = M, the we have f(x) = (l x μ) σ πx e σ ; x > M f(x)dx = M = eμ (b) Likelihood fuctio f L (x μ, σ) = ( x i ) f N (l x; μ, σ) i=

15 f L is pdf of log ormal ad f N is pdf of ormal l L = (μ, σ x, x,, x ) = l x k + l N (μ, σ l x, l x,, l x ) = costat + l N (μ, σ l x, l x,, l x ) So, by usig the cocept of Normal distributio, we have k μ = k l x k ad σ = k (l x k μ ). (a) A radom variable X has probability desity fuctio f(x) = axe β x, x >, α >, β > If E(X) = π, determie α ad β. (b) Let X ad Y be two radom variables with joi probability desity fuctio f(x, y) = { e y if x y otherwise (i) (ii) (iii) Fid the margial desity fuctios of X ad Y Examie whether X ad Y are idepedet Fid Cov (X, Y) Solutio: (a) Let y = (βx) dy = β xdx E(X) = xf(x)dx + αx e β x dx E(X) = α β. y/ β e y dy = α β 3 Γ (3 ) = α π 4β3 π = α = β 3 () Also, f(x)dx = αxe β x dx =

16 α β e y dy = α β = α = β. () From () & () we get β = α = (b) (i) Margial desity fuctio of X, f(x) = e y dy; x < f(x) = e x ; x < Margial desity fuctio of Y, y f(y) = e y dx = ye y = ye y ; y < (ii) As f(x)f(y) = e x (ye y ) = ye (x+y) ad f(x, y) = e y So, f(x, y) f(x)f(y) hece X ad Y are ot idepedet. (iii) E(X) = xe x dx = [ xe x e x ] = E(Y) = y e y dy = [( y y z)e y ] = y E(XY) = xye y dxdy = y3 e y dy y= x= y= = [ y3 3y 6y 6]e y = 3 Cov (X, Y) = E(XY) E(X)E(Y) = 3 ()() = 3 =.

17 3. (a) Let X,., X be a radom sample from Exp ( ) populatio. Obtai the Cramer Rao θ lower boud for the variace of a ubiased estimator of θ. σ. (b) Let X,, X (>4) be a radom sample from a populatio with mea μ ad variace Cosider the followig estimators of μ U = X i, i= V = 8 X 3 + 4( ) (X + + X ) + 8 X (i) (ii) (iii) Examie whether the estimates U ad V are ubiased Examie whether the estimates U ad V are cosistet Which of these two estimates is more efficiet? Justify your aswer. Solutio: (a) If T(X) is ubiased estimator of a fuctio Ψ(θ) of the parameter θ, the Cramer Rao boud of variace of Ψ(θ) is give by Where I(θ) is Fisher s iformatio. V(T(X)) [Ψ (θ)] L(x, θ) I(θ) = E [( ) θ I(θ) L(x, θ) = log f(x, θ) ] = E [ L(x; θ) θ ] Give Ψ(θ) = θ Ψ (θ) = θ f(x i= θ) = θ e x/θ ; θ >, x > f(x, θ) = θ e xi/θ = θ i= e xi/θ L(x, θ) = log f(x, θ) = log θ ( xi)/θ e ( xi) = log θ θ = I(θ) = E [( θ + x i θ ) ]

18 Var [T(x)] 4θ E [( θ + xi θ ) ] So, lower boud for variace is 4θ E [( θ + xi θ ) ] (b) U = X i i= V = 8 X 3 + 4( ) (X + + X ) + 8 X (i) E(U) = E(X i) = μ = μ i= E(X) = μ ( ) 4 ( ) μ + 8 μ = μ ( ) = μ Both U ad V are ubiased estimator of μ as E(U) = E(V) = μ (ii) lim U = μ ad lim V = μ So, both U ad V are cosistet estimator of μ. (iii) As, Var(U) < Var (V) Var(U) = Var ( X i) = [Var X + Var X + + Var X ] i= = [σ + σ + + σ ] = σ Var (V) = 9 64 σ + 6( ) ( )σ + 64 σ = 9 3 σ + 6( ) σ

19 Here U is more efficiet estimator tha V. 4. Let X,, X be a radom sample from a Beroulli populatio with parameter p, (a) (i) Fid a sufficiet statistic for p (ii) Cosider a estimator U(X, X ) of P( p) give by U(X, X ) = { if X + X = otherwise Examie whether U(X, X ) is the a ubiased estimator (b) Usig the results obtaied i (a) above ad Rao Blackwell theorem, fid the uiformity miimum variace ubiased estimator (UMVUE) of Solutio: (a) p( p). (i) T(X) = X + X + + X is sufficiet statistics for p. As P r = (X = x) = Pr{X = x, X = x,., X = x } = P x ( p) x p x ( p) x P x ( p) x = P (x +x + +x )( p) (x +x+ +x ) = p x i( p) x i = P T(X) ( P) T(X) Hece T(x) = x i (ii) X + X = Evaluate U(X, X ) ad if E(U(X, X )) = P( p) the U is ubiased E(U) = (X =, X = ) + (X =, X = ) = p( p) p( P) p( p) + = Hece U is a ubiased estimator of p( p). (b) Now by Rao Blackwell theorem, we obtai UMVUE, which requires the statistics to be sufficiet.

20 Also, mea square error of Rao Blackwell estimator does ot exceed that of the origial estimator. Rest of the calculatios are left for you as a exercise. 5. (a) Let X,, X be a radom sample from the populatio havig probability desity fuctio x f(x, θ) = { e θ θ if x > otherwise Obtai the most powerful test for testig H : θ = θ agaist H : θ = θ (θ < θ ) x (b) Let X,, X be a radom sample of size from N(μ, ) populatio. To test H μ = 5 agaist H : μ = 4, the decisio rule is : Reject H if x c. If α =. 5 ad β =., determie (rouded off to a iteger) ad hece c. Solutio: (a) Likelihood fuctio, L(x, θ) = x i θ e x i /θ i= Cosider H : θ = θ & H : θ = θ The best critical regio, usig Neyma Pearso Lemma is give by = θ x ie x i /θi k i= θ i= x i e x i /θ K ( θ ) e x i ( θ θ ) θ i= log K (log θ log θ ) + ( θ θ ) x i Now P ( ( θ θ ) x i log k (log θ log θ ) H is true) = α i= So it is used for most powerful test. (b) Same as above. Do yourself. i=

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