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MATH254: Tutorial Exercise for Week 7 1. Let X and Y be independent random variables with common distribution function F and density function f . Let U = max{X,Y } and determine the distribution and density functions of V . Also, find the distribution and density functions of V = min{X,Y }. 2. Continuous random variables X and Y have joint pdf f(x, y) = x+ y, 0 x 1, 0 y 1, otherwise, it is zero. (a) Find the marginal pdfs of X and Y . (b) Find the two conditional densities. (c) By integrating over the appropriate regions, find (i) Pr(X > Y ), (ii) Pr(X + Y 1), (iii) Pr(X 0.5). 3. Suppose the joint density of X and Y is given by f(x, y) = 4y(x y)e(x+y), 0 x <1, 0 y x; otherwise, it is zero. Compute E[X | Y = y]. 1 MATH254: Solutions of Tutorial Exercise for Week 7 1. Note that max{X,Y } u if and only if X u and Y u. Hence, by independence, P (U u) = P (X u, Y u) = P (X u)P (Y u) = F (u)2. Di erentiate to obtain the density function fU (u) = 2F (u)f(u). In a similar way, min{X,Y } > v if and only if X > v and Y > v. Hence, the distribution function FV (v) is: P (V v) = 1 P (V > v) = 1 P (X > v)P (Y > v) = 1 [1 F (v)]2. Further, fV (v) = 2f(v)[1 F (v)]. 2. (a) Marginal pdfs: fX(x) = Z 1 y=0 (x+ y) dy = xy + y2 2 1 y=0 = x+ 0.5 (0 x 1) fY (y) = Z 1 x=0 (x+ y) dx = x2 2 + xy 1 x=0 = 0.5 + y (0 y 1) (b) Conditional densities: for y 2 (0, 1), we have fX|Y (x|y) = x+ y0.5 + y (0 x 1) and for x 2 (0, 1), we have fY |X(y|x) = x+ yx+ 0.5 (0 y 1) (c) Probabilities: (i) Pr(X > Y ) = Z 1 x=0 Z x y=0 (x+ y) dy dx = Z 1 x=0 xy + y2 2 x y=0 dx = Z 1 x=0 x2 + x2 2 ◆ dx = Z 1 x=0 3 2 x2 dx = x3 2 1 0 = 1 2 (Also follows by symmetry, since f(x, y) = f(y, x), so that Pr(X > Y ) = Pr(Y > X) = 0.5.) (ii) Pr(X + Y 1) = Z 1 x=0 Z 1x y=0 (x+ y) dy dx = Z 1 x=0 xy + y2 2 1x y=0 dx = Z 1 x=0 x(1 x) + (1 x) 2 2 ◆ dx = Z 1 x=0 x x2 + 0.5 x+ 0.5×2 dx = Z 1 x=0 0.5 0.5×2 dx = x 2 x 3 6 1 x=0 = 1 3 (iii) Pr (X 0.5) = Z 0.5 x=0 Z 1 y=0 (x+ y) dy dx = Z 0.5 x=0 xy + y2 2 1 y=0 dx = Z 0.5 x=0 (x+ 0.5) dx = x2 2 + x 2 0.5 x=0 = 3 8 (Or could use marginal pdf fX(x) to find Pr (X 0.5).) 1 3. If y < 0, it is clear that E[X |Y = y] = 0. We only consider the case y 0. The conditional density of X, given that Y = y, x > y, is given by fX|Y (x|y) = f(x, y)fY (y) = 4y(x y)e(x+y)R1 y 4y(x y)e(x+y)dx = (x y)e(x+y)R1 y (x y)e(x+y)dx = (x y)exR1 y (x y)exdx . Integrating by parts shows that the above gives fX|Y (x|y) = (x y)e x ey = (x y)e(xy), x > y. Therefore, E[X|Y = y] = Z 1 1 xfX|Y (x|y)dx = Z 1 y x(x y)e(xy)dx. Integration by parts yields E[X|Y = y] = x(x y)e(xy) 1 y + Z 1 y (2x y)e(xy)dx = Z 1 y (2x y)e(xy)dx = y + 2. 2 MATH254: Tutorial Exercise for Week 8 1. Discrete random variables X and Y have joint probability mass function given by Y = 1 Y = 2 Y = 3 Y = 4 X = 2 0.25 0.25 0 0 X = 4 0 0 0.25 0.25 Find the correlation Corr[X,Y ]. Comment on your answer. 2. Continuous random variables X and Y have joint pdf f(x, y) = 60x2y, 0 x, 0 y, x+ y 1, otherwise, it is zero. (a) Find the marginal pdfs of X and Y . (b) Find the conditional density of Y given X = x, and hence evaluate P (Y > 0.1 | X = 0.5). (c) Find the correlation Corr[X,Y ]. 3. Continuous random variables X and Y have joint pdf f(x, y) = (2/ ) exp x2 + y2 /2 , x, y > 0, otherwise, it is zero. (a) Are X and Y independent Justify your answer. (b) Random variables U and V are defined by U = X + 2Y, V = X/Y, Find the joint pdf of U and V . 1 MATH254: Solutions of Tutorial Exercise for Week 8 1. Marginals: P (X = 2) = 0.5, P (X = 4) = 0.5, P (Y = 1) = 0.25, P (Y = 2) = 0.25, P (Y = 3) = 0.25, P (Y = 4) = 0.25. So separate means and variances are E[X] = 0.5 2 + 0.5 4 = 3 E[Y ] = 0.25 1 + 0.25 2 + 0.25 3 + 0.25 4 = 2.5 Var[X] = 0.5 22 + 0.5 42 32 = 10 9 = 1 Var[Y ] = 0.25 12 + 0.25 22 + 0.25 32 + 0.25 42 2.52 = 7.5 6.25 = 1.25 and now Cov[X,Y ] = (0.25 2 1 + 0.25 2 2 + 0.25 4 3 + 0.25 4 4) 3 2.5 = 8.5 7.5 = 1, Corr[X,Y ] = 1/ p 1 1.25 = p 4/5 0.894. Comment: Correlation is large and positive, indicating that X and Y tend to be large together and small together. This can be seen from the joint probability mass function given in the question: when X is small (X = 2), then only the small Y values have non-zero probability (Y = 1, 2) whereas when X is large (X = 4) only the large Y values have non-zero probability (Y = 3, 4). 2. (a) Marginal: fX(x) = Z 1x y=0 60x2y dy = 30x2y2 1x y=0 = 30×2(1 x)2, 0 x 1, fY (y) = Z 1y x=0 60x2ydx = 20x3y 1y x=0 = 20(1 y)3y, 0 y 1, (b) Conditional density: For 0 x 1, fY |X(y|x) = f(x, y)fX(x) = 60x2y 30×2(1 x)2 = 2y (1 x)2 , 0 y 1 x, and hence P (Y > 0.1 | X = 0.5) = Z 10.5 y=0.1 2y (1 0.5)2 dy = Z 0.5 y=0.1 8ydy = 4y2 0.5 y=0.1 = 4 0.52 0.12 = 0.96. (c) From marginal, E[X] = Z 1 0 30×3(1 x)2 dx = Z 1 0 30 x3 2×4 + x5 dx = 30 x4 4 2x 5 5 + x6 6 1 0 = 7.5 12 + 5 = 0.5 E[Y ] = Z 1 0 20y2(1 y)3 dy = Z 1 0 20 y2 3y3 + 3y4 y5 dy = 20 y3 3 3y 4 4 + 3y5 5 y 6 6 1 0 = 20 3 15 + 12 10 3 = 1 3 Var[X] = Z 1 0 30×4(1 x)2 dx 0.52 = Z 1 0 30 x4 2×5 + x6 dx 0.25 = 30 x5 5 x 6 3 + x7 7 1 0 0.25 = (6 10 + (30/7)) 0.25 = 1/28 Var[Y ] = Z 1 0 20y3(1 y)3 dy (1/3)2 = Z 1 0 20 y3 3y4 + 3y5 y6 dy (1/9) = 20 y4 4 3y 5 5 + y6 2 y 7 7 1 0 (1/9) = (5 12 + 10 (20/7)) (1/9) = 2/63. 1 From joint density: E[XY ] = Z 1 x=0 Z 1x y=0 (xy) 60x2y dy dx = Z 1 x=0 Z 1x y=0 60x3y2 dy dx = Z 1 x=0 20x3y3 1x y=0 dx = Z 1 x=0 20×3(1 x)3 dx = Z 1 x=0 20 x3 3×4 + 3×5 x6 dx = 20 x4 4 3x 5 5 + x6 2 x 7 7 1 x=0 = 5 12 + 10 (20/7) = 1/7, and so Cov[X,Y ] = (1/7) (1/2) (1/3) = 1/42 Corr[X,Y ] = 1/42p (1/28) (2/63) = 1p 2 0.7071. 3. (a) X and Y are independent, since range of each variable is independent of the value of the other, and the density function is the product of two density functions of x and of y, respectively. (b) Inverting the relationship, V = X/Y ) X = Y V substituting for X, then U = X + 2Y = Y V + 2Y = Y (V + 2)) Y = U V + 2 and hence X = Y V = UV V + 2 Jacobian: @x @u = v v + 2 @x @v = (v + 2)u uv (v + 2)2 = 2u (v + 2)2 @y @u = 1 v + 2 @y @v = u (v + 2)2 and so J = @(x, y) @(u, v) = v v + 2 ◆ u (v + 2)2 ◆ 1 v + 2 ◆ 2u (v + 2)2 ◆ = u (v + 2)2 Joint pdf of (U, V ) is therefore given by fU,V (u, v) = fX,Y (x(u, v), y(u, v)) @(x, y)@(u, v) = 2 ◆ exp ( uv v + 2 ◆2 + u v + 2 ◆2!, 2 ) u (v + 2)2 = 2u (v + 2)2 exp ( u v + 2 ◆2 v2 + 1 , 2 ) for u > 0, v > 0. Otherwise, fU,V (u, v) = 0. 2 MATH254: Tutorial Exercise for Week 9 1. Consider a fair 3-sided die with faces 1, 2 and 4 each occuring with probability 1/3. Write down the PGF of the score for one throw of the die. Hence find the PGF of the total score for n independent throws. Using the PGF obtain the mean and variance of the total score when n = 2. 2. Find the PGF of the discrete random variable X with probability mass function P (X = k) = (2/3) (1/3)k1 for k = 1, 2, . . . Hence find the mean and variance of X. 3. Find the MGF of (a) the Uniform[0, 1] distribution; (b) the discrete random variable X with P (X = 4) = 1; (c) the continuous random variable Y with probability density function f(y) = 2y for 0 y 1, density zero elsewhere. 4. For the standard normal random variable Z N(0, 1), write down the MGF MZ(t) of Z. Defining X = μ+ Z for real numbers μ, with > 0, what is the distribution of X Use the MGF MZ(t) to find the MGF MX(t) of X. 1 MATH254: Solutions of Tutorial Exercise for Week 9 1. Denoting by X the score on one throw, the probability mass function of X is P (X = 1) = 1/3 P (X = 2) = 1/3 P (X = 4) = 1/3 and so the probability generating function of X is GX(s) = E sX = (1/3) s+ s2 + s4 Denoting by Sn the sum of the scores on n independent throws, and by Xi the score on throw i, then GSn(s) = E sSn = E sX1+···+Xn = E sX1 · · · sXn = E sX1 n = (1/3)n s+ s2 + s4 n When n = 2, then GS2(s) = (1/9) s+ s2 + s4 2 G0S2(s) = (2/9) s+ s2 + s4 1 + 2s+ 4s3 G00S2(s) = (2/9) 1 + 2s+ 4s3 1 + 2s+ 4s3 + s+ s2 + s4 2 + 12s2 hence G0S2(1) = (2/9) 3 7 = 14/3 G00S2(1) = (2/9)(7 7 + 3 14) = 182/9 and so E [S2] = G 0 S2(1) = 14/3 Var [S2] = G 00 S2(1) +G 0 S2(1) G0S2(1) 2 = 182 9 + 14 3 14 3 ◆2 = 28 9 2. PGF: GX(s) = E sX = 1X k=1 sk 2 3 ◆ 1 3 ◆k1 = 2s 3 1X k=0 s 3 k = 2s 3 1 1 (s/3) ◆ for 3 < s < 3 = 2s 3 s (Note range of validity 3 < s < 3.) To find mean and variance: G0X(s) = (3 s) 2 2s (1) (3 s)2 = 6 (3 s)2 G 0 X(1) = 3 2 G00X(s) = 12 (3 s)3 G 00 X(1) = 3 2 E [X] = G0X(1) = 3/2 Var [X] = G00X(1) +G 0 X(1) (G0X(1))2 = (3/2) + (3/2) (3/2)2 = 3/4 3. (a) For X Uniform[0, 1], pdf is f(x) = 1 for 0 x 1, density zero elsewhere, so MX(t) = E etX = Z 1 0 etx dx = etx t 1 x=0 = et 1 t 1 (b) With P (X = 4) = 1, then MX(t) = E etX = e4t. (c) Integrating by parts, MY (t) = E etY = Z 1 0 ety 2y dy = 2y ety t 1 y=0 Z 1 0 2ety t dy = 2et t 2ety t2 1 y=0 = 2et t 2e t t2 + 2 t2 = 2 ((t 1)et + 1) t2 4. For Z N(0, 1), MGF is MZ(t) = et2/2. Defining X = μ+ Z then X N μ,2. MGF of X is MX(t) = E h et(μ+Z) i = E etμetZ = etμMZ(t) = e tμe(t) 2/2 = exp tμ+ 2t2 2 2 MATH254: Tutorial Exercise for Week 10 1. A sweet maker produces mints whose weights are normally distributed with mean 21.37 and vari- ance 0.16. (a) Let X denote the weight of a single mint selected at random from the production line. Find P (X < 20.857). (b) Let Xˉ denote the mean weight of a sample of 100 mints. Find P (21.31 Xˉ 21.39). 2. The random variables X1, X2, . . . are independent and identically distributed with the Probability Mass Function Pr(X = 1) = Pr(X = 3) = 0.5. (a) Calculate μ = E(X) and 2 = V ar(X). (b) For the case n = 200 use the Central Limit Theorem to approximate the probability Pr{ln(Xˉ) > 0}, where Xˉ = (1/200) 200X i=1 Xi is the sample mean and ln denotes the natural logarithm (i.e., logarithm with base e). (c) Find the minimum n for which Pr n nX i=1 Xi > 190 o > 0.99. 3. For V Binomial(100, 0.2), compute P (V = 20) (i) exactly; (ii) using a Poisson approximation; (iii) using a normal approximation. Comment on your results. 4. Suppose that a measurement has mean μ and variance 2 = 25. Let Xˉ be the average of n such independent measurements. Use the Central Limit Theorem to estimate how large n should be so that P |Xˉ μ| < 1 = 0.95. 1 MATH254: Solutions of Tutorial Exercise Week 10 1. (a) It is told that X N(21.37, 0.16), so P (X < 20.857) = P Z < 20.857 21.37 0.4 ◆ where Z N(0, 1) = P (Z < 1.2825) = P (Z 1.2825) 1 (1.28) = 1 0.8997 = 0.1003. (b) With n = 100 then Xˉ N(21.37, 0.16/100) = N 21.37, 0.042, so P (21.31 Xˉ 21.39) = P 21.31 21.37 0.04 Z 21.39 21.37 0.04 ◆ = P (1.5 Z 0.5) = (0.5) (1.5) = (0.5) (1 (1.5)) = 0.6915 (1 0.9332) = 0.6247. 2. (a) μ = 1 · 12 + 3 · 12 = 1; EX2 = 12 + 92 = 5; 2 = V ar(X) = EX2 (EX)2 = 4. (b) Pr{ln(Xˉ) > 0} = Pr{Xˉ > 1} = Pr{ 200X i=1 Xi > 200} = Pr nP200 i=1Xi 200 · 1 2 p 200 > 200 200 · 1 2 p 200 o Pr{Z > 0} = 0.5 (here Z has standard normal distribution.) (c) Pr n nX i=1 Xi > 190 o Pr n Z > 190 n · 1 2 p n o > 0.99. The 99% critical value is 2.33, so we want to find n such that 190 n 2 p n < 2.33. Set n = x2 and solve equation 190 x2 = 4.66x, i.e., x2 4.66x 190 = 0. = 4.662 + 4 · 190 = 781.72; p = 27.96; x1 = 4.66 + 27.96 2 = 16.31; x2 = 4.66 27.96 2 = 11.65. Thus, n1 = 266.02; n2 = 135.72. Clearly, 190n2 > 0 so n2 does not satisfy our requirement. Thus the minimum n is 267. 3. (i) Exact: P (V = 20) = 100 20 0.220 0.880 = 0.099300. (ii) Poisson approximation: Have E[V ] = 100 0.2 = 20, so approximate by Poisson(20) distribu- tion, and then P (V = 20) 20 20 20! e20 = 0.088835. (iii) Normal approximation: Have Var[V ] = 100 0.2 0.8 = 16, so approximate by N(20, 16), and then P (V = 20) = P (19.5 V 20.5) P 19.5 20 4 Z 20.5 20 4 ◆ = P (0.125 Z 0.125) = (0.125) (0.125) = (0.125) (1 (0.125)) = 2(0.125) 1 (0.12) + (0.13) 1 = (0.5478 + 0.5517) 1 = 0.0995 Comment: Normal approximation rather more accurate (0.0995/0.099300 = 1.002, close to 1, whereas 0.088835/0.099300 = 0.8946, not so close to 1), as we would expect, since np = 20 > 5 and n(1 p) = 80 > 5, so conditions for normal approximation valid, whereas condition for Poisson approximation that np < 5 is not satisfied. 1 4. By the CLT, for large n we approximately have Xˉ N (μ, 25/n). Hence P |Xˉ μ| < 1 = P Xˉ μ5/pn < 15/pn ◆ P |Z| < p n 5 ◆ where Z N(0, 1) = P p n 5 < Z < p n 5 ◆ = p n 5 ◆ 1 p n 5 ◆◆ = 2 p n 5 ◆ 1, so we require 2 p n 5 ◆ 1 = 0.95 ) p n 5 ◆ = 0.975 ) p n 5 = 1.96 ) n = (1.96 5)2 = 96.04 So we require n = 96 (approximately). 2 MATH254: Tutorial Exercise for Week 11 1. Suppose U and V are independent chi-square random variables with m and n degrees of freedom, respectively. Show that the random variable W defined by W = U/m V/n has density fW (w) = (m+n2 ) (m2 )( n 2 ) m n m/2 w m 2 1 1 + m n w m+n2 , w > 0. Hint: Recall that the density of a chi-square distribution with n degrees of freedom is fV (v) = (1/2)n/2 (n2 ) v n 21 ev/2. Recall also that if A and B are independent random variables, then the density of the quotient C = B/A is fC(c) = Z |a|fA(a)fB(ca)da. Finally, recall that the gamma density is f ,(x) = ( ) x 1 ex, x > 0, , > 0. 2. Suppose that (X,Y ) are jointly Normal distributed. Identify the conditional density functions fY |X(y | x) and fX|Y (x | y). 3. Suppose that (X,Y ) has a standard jointly Normal distribution with parameter 2 (1, 1). Let V = (Y X)/ p 1 2. Verify that X and V are independent, standard Normal random variables. 1 MATH254: Solutions of Tutorial Exercise Week 11 1. We know that density of V is fV (v) = 1 2 n/2 n 2 vn/21ev/2. Consider A = V/n. Using transformation method fA(a) = fV (na)n = 1 2 n/2 n 2 (na)n/21ena/2n. Similarly, setting B = U/m, we obtain fB(b) = 1 2 m/2 m 2 (mb)m/21emb/2m. Since A and B are independent, we can use the formula for the density of the quotient W = B/A: fW (w) = Z 1 0 |a|fA(a)fB(wa)da = Z 1 0 a 1 2 n+m 2 n 2 m 2 (na)n/21n(mwa)m/21mena/2emwa/2da = 1 2 n+m 2 n 2 m 2 nn/21(mw)m/21nm Z 1 0 an/2+m/21e 1 2 (n+mw)ada. Note that Gamma density is f ,(x) = ( ) x 1ex. In our case = 12 (n+mw) and = n+m 2 and a plays the role of x. So fW (w) = 1 2 n+m 2 n 2 m 2 nn/2mm/2wm2 1 m+n2 1 2 (n+mw) n+m 2 = m+n 2 n 2 m 2 m n m/2 w m 2 1 1 + m n w n+m2 . 2. By a direct computation, we have fY |X(y | x) = fX,Y (x, y)fX(x) = 1 Y p 2 (1 2) · exp 1 2 h y μY YX (x μX) i2 2Y (1 2) . This is a normal density function with mean μY + (x μX)Y /X and variance 2Y (1 2). In a similar way, we can get fX|Y (x | y) = fX,Y (x, y)fY (y) = 1 X p 2 (1 2) · exp 1 2 h x μX XY (y μY ) i2 2X(1 2) . 3. By the last theorem in Chapter 9, since X and V are linear combinations of X and Y , it follows that any linear combination of X and V is a linear combination of X and Y , so is normally distributed (particularly, X or V is Normal) and so (X,V ) has a bivariate Normal distribution. Noting that E(V ) = 0 and V ar(V ) = (1 + 2 2 2)/(1 2) = 1, the random variable V is standard Normal distributed like the random variable X. To prove the independence of X and V , we only verify that Cov(X,V ) = 0. This is immediate from Cov(X,V ) = (Cov(X,Y ) · V ar(X))/ p 1 2 = ( )/ p 1 2 = 0. 1