Math 151. Rumbos Spring Solutions to Review Problems for Exam 2

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1 Math 5. Rumbos Spring 22 Solutions to Review Problems for Exam 2. Let X and Y be independent Normal(, ) random variables. Put Z = Y X. Compute the distribution functions (z) and (z). Solution: Since X, Y Normal(, ), their pdfs are given by f X (x) = 2π e x2 /2, for x R, and f Y (y) = 2π e y2 /2, for y R, respectively. The joint pdf of (X, Y ) is then We compute the cdf o, (x, y) = 2π e (x2 +y2 )/2, for (x, y) R 2. () (z) = Pr(Z z) = Pr ( y x z ), or (z) = (x, y) dxdy, (2) y x z where the integrand in (2) is given in () and the integration is done over the region R = {(x, y) R 2 y } x z. Make the change variables u = x v = y x, so that in the integral in (2) to obtain where (z) = z x = u y = uv, (3) (u, uv) (x, y) (u, v) dudv, (4) ( ) (x, y) (u, v) = det = u, (5) v u

2 Math 5. Rumbos Spring 22 2 is the Jacobian determinant of the transformation in (3). It then follows from (4) and (5) that (z) = z (u, uv) u dudv, (6) Differentiating with respect to z and using the definition of the joint pdf of (X, Y ) in () we obtain from (6) that (z) = 2π u e (+z2 )u 2 /2 du, (7) where we have also used the Fundamental Theorem of Calculus. Since the integrand in (7) is an even function of u, we can rewrite the expression for in (7) as (z) = π u e (+z2 )u 2 /2 du. (8) Integrating the right hand side of equation in (8) we obtain (z) = π The cdf o is then obtained by integrating (9) to get (z) = z, for z R. (9) + z2 (z) dz = 2 + arctan(z), for z R. π 2. A random point (X, Y ) is distributed uniformly on the square with vertices (, ), (, ), (, ) and (, ). (a) Give the joint pdf for X and Y. (b) Compute the following probabilities: (i) Pr(X 2 + Y 2 < ), (ii) Pr(2X Y > ), (iii) Pr( X + Y < 2). Solution: The square is pictured in Figure and has area 4.

3 Math 5. Rumbos Spring 22 3 y x Figure : Sketch of square in Problem 2 (a) Consequently, the joint pdf of (X, Y ) is given by, for < x <, < y < ; 4 (x, y) = elsewhere. () (b) Denoting the square in Figure by R, it follows from () that, for any subset A of R 2, Pr[(x, y) A] = (x, y) dxdy = area(a R); () 4 A that is, Pr[(x, y) A] is one fourth the area of the portion of A in R. We will use the formula in () to compute each of the probabilities in (i), (ii) and (iii). (i) In this case, A is the circle of radius around the origin in R 2 and pictured in Figure 2. Note that the circle A in Figure 2 is entirely contained in the square R so that, by the formula in (), Pr(X 2 + Y 2 < ) = area(a) 4 = π 4.

4 Math 5. Rumbos Spring 22 4 y A x Figure 2: Sketch of A in Problem 2(i) (ii) The set A in this case is pictured in Figure 3 on page 5. Thus, in 2 this case, A R is a trapezoid of area = 2, so that, by the 2 formula in (), Pr(2X Y > ) = 4 area(a R) = 2. (iii) In this case, A is the region in the xy plane between the lines x+y = 2 and x + y = 2 (see Figure 4 on page 6). Thus, A R is R so that, by the formula in (), Pr( X + Y < 2) = area(r) 4 =. 3. Prove that if the joint cdf of X and Y satisfies F (X,Y ) (x, y) = F X (x)f Y (y), then for any pair of intervals (a, b) and (c, d), Pr(a < X b, c < Y d) = Pr(a < X b)pr(c < Y d).

5 Math 5. Rumbos Spring 22 5 y A x Figure 3: Sketch of A in Problem 2(ii) Solution: First show that Pr(a < X b, c < Y d) = F (X,Y ) (b, d) F (X,Y ) (b, c) F (X,Y ) (a, d) + F (X,Y ) (a, c) (see Problem in Assignment #5). Then, Pr(a < X b, c < Y d) = F X (b)f Y (d) F X (b)f Y (c) F X (a)f Y (d) + F X (a)f Y (c) = (F X (b) F X (a))f Y (d) (F X (b) F X (a))f Y (c) = (F X (b) F X (a))(f Y (d) F Y (c)) = Pr(a < X b)pr(c < Y d), which was to be shown. 4. The random pair (X, Y ) has the joint distribution shown in Table on page 6.

6 Math 5. Rumbos Spring 22 6 y A x Figure 4: Sketch of A in Problem 2(iii) X\Y Table : Joint Probability Distribution for X and Y, p (X,Y ) (a) Show that X and Y are not independent. Solution: Table 2 shows the marginal distributions of X and Y on the margins on page 7. Observe from Table 2 that p (X,Y ) (, 4) =, while p X () = 4 and p Y (4) = 3. Thus, p X () p Y (4) = 2 ;

7 Math 5. Rumbos Spring 22 7 X\Y p X p Y Table 2: Joint pdf for X and Y and marginal distributions p X and p Y so that p (X,Y ) (, 4) p X () p Y (4), and, therefore, X and Y are not independent. (b) Give a probability table for random variables U and V that have the same marginal distributions as X and Y, respectively, but are independent. Solution: Table 3 on page 7 shows the joint pmf of (U, V ) and the marginal distributions, p U and p V. U\V p U p V Table 3: Joint pdf for U and V and their marginal distributions. 5. Let X denote the number of trials needed to obtain the first head, and let Y be the number of trials needed to get two heads in repeated tosses of a fair coin. Are X and Y independent random variables? Solution: X has a geometric distribution with parameter p =, so that 2 p X (k) =, for k =, 2, 3,... (2) 2k On the other hand, Pr[Y = 2] = 4, (3) since, in two repeated tosses of a coin, the events are HH, HT, T H and T T, and these events are equally likely.

8 Math 5. Rumbos Spring 22 8 Next, consider the joint event (X = 2, Y = 2). Note that (X = 2, Y = 2) = [X = 2] [Y = 2] =, since [X = 2] corresponds to the event T H, while [Y = 2] to the event HH. Thus, Pr(X = 2, Y = 2) =, while by (2) and (3). Thus, Hence, X and Y are not independent. p X (2) p Y (2) = 4 4 = 6, p (X,Y ) (2, 2) p X (2) p X (2). 6. Let X Normal(, ) and put Y = X 2. Find the pdf for Y. Solution: The pdf of X is given by f X (x) = 2π e x2 /2, for < x <. We compute the pdf for Y by first determining its cdf: Thus, Pr(Y y) = P (X 2 y) for y = Pr( y X y) = Pr( y < X y), since X is continuous. Pr(Y y) = Pr(X y) Pr(X y) We then have that the cdf of Y is = F X ( y) F X ( y) for y >. F Y (y) = F X ( y) F X ( y) for y >, from which we get, after differentiation with respect to y, f Y (y) = F ( y) X 2 y + F ( y) X 2 y = f X ( y) 2 y + f ( y) X 2 y { = 2 e y/2 + } e y/2 y 2π 2π = e y/2, 2π y

9 Math 5. Rumbos Spring 22 9 for y >, where we have applied the Chain Rule. Hence, e y/2, for y > ; 2π y f Y (y) = for y. 7. Let X and Y be independent Normal(, ) random variables. Compute P (X 2 + Y 2 < ). Solution: Since X, Y Normal(, ), their pdfs are given by f X (x) = 2π e x2 /2, for x R, and f Y (y) = 2π e y2 /2, for y R, respectively. The joint pdf of (X, Y ) is then Thus, (x, y) = 2π e (x2 +y2 )/2, for (x, y) R 2. (4) P (X 2 + Y 2 < ) = x 2 +y 2 < (x, y) dxdy, (5) where the integrand is given in (4) and the integral in (5) is evaluated over the disc of radius centered around the origin in R 2. We evaluate the integral in (5) by changing to polar coordinates to get P (X 2 + Y 2 < ) = 2π 2π e r2 /2 rdrdθ = = e r2 /2 rdr [ ] e r2 /2 = e /2,

10 Math 5. Rumbos Spring 22 or Pr(X 2 + Y 2 < ) = e. 8. Suppose that X and Y are independent random variables such that X Uniform(, ) and Y Exponential(). (a) Let Z = X + Y. Find and. Solution: Since X Uniform(, ) and Y Exponential(), their pdfs are given by { if < x < ; f X (x) = elsewhere, and f Y (y) = { e y if y > ; if y, respectively. The joint pdf of (X, Y ) is then { e y if < x <, y > ; (x, y) = elsewhere. We compute the cdf o, or (z) = Pr(X u) = Pr (X + Y z), F U (u) = x+y z (6) (x, y) dxdy, (7) where the integrand in (7) is given in (6) and the integration is done over the region R = { (x, y) R 2 x + y z }. Make the change variables u = x v = x + y, so that x = u y = v u, (8)

11 Math 5. Rumbos Spring 22 in the integral in (7) to obtain z (z) = (u, v u) (x, y) (u, v) dudv, (9) where ( ) (x, y) (u, v) = det =, (2) is the Jacobian determinant of the transformation in (8). It then follows from (9) and (2) that (z) = z (u, v u) dudv. (2) Differentiating with respect to z and using the definition of the joint pdf of (X, Y ) in (6) we obtain from (2) that (z) = (u, z u) du. (22) where we have also used the Fundamental Theorem of Calculus. Next, use the definition of in (6) to rewrite (22) as (z) = (u, z u) du, for z >, (23) We consider two cases, (i) < z, and (ii) z >. (i) For < z, use (6) to obtain from (23) that (z) = z e u z du z = e z e u du = e z, so that (z) = e z, for < z. (24)

12 Math 5. Rumbos Spring 22 2 (ii) For z >, use (6) to obtain from (23) that (z) = e u z du so that = e z e u du = (e )e z, (z) = (e )e z, for z >. (25) Combining (24) and (25) we obtain the cdf for z ; (z) = e z, for < z ; (e )e z, for z >. Finally, integrating (26) yields the cdf for z ; (z) = z + e z, for < z ; e + (e )(e e z ), for z >. (26) (b) Let U = Y/X. Find F U and f U. Solution: Since X Uniform(, ) and Y Exponential(), their pdfs are given by { if < x < ; f X (x) = elsewhere, and f Y (y) = { e y if y > ; if y, respectively. The joint pdf of (X, Y ) is then { e y if < x <, y > ; (x, y) = elsewhere. (27)

13 Math 5. Rumbos Spring 22 3 We compute the cdf of U, F U (u) = Pr(U u) = Pr ( ) Y X u, or F U (u) = (x, y) dxdy, (28) y x u where the integrand in (28) is given in (27) and the integration is done over the region R = {(x, y) R 2 y } x u. Make the change variables w = x v = y x, so that x = w y = wv, (29) in the integral in (28) to obtain u F U (u) = (w, wv) (x, y) (w, v) dwdv, (3) where ( ) (x, y) (w, v) = det = w, (3) v w is the Jacobian determinant of the transformation in (29). It then follows from (3) and (3) that F U (u) = u (w, wv) w dwdv. (32) Differentiating with respect to u and using the definition of the joint pdf of (X, Y ) in (27) we obtain from (32) that f U (u) = (w, wu) w dw. (33) where we have also used the Fundamental Theorem of Calculus. Next, use the definition of in (27) to rewrite (33) as f U (u) = e uw w dw, for u >, (34)

14 Math 5. Rumbos Spring 22 4 We evaluate the integral in (34) by integration by parts to get f U (u) = [ w u e uw u 2 e uw ] (35) = u 2 u e u u 2 e u, for u >. In order to compute the cdf, F U, we can integrate (28) in Cartesian coordinates to get F U (u) = ux e y dydx = [ e ux ] dx = + u [e u ], so that + u [e u ], for u > ; F U (u) = (36) for u. Note that differentiating F U (u) in (36) with respect to u, for u >, leads to (35). We then have that u ( f U (u) = 2 e u ) u e u, for u > ; for u. 9. Let X Exponential(), and define Y to be the integer part of X + ; that is, Y = i + if and only if i X < i +, for i =,, 2,... Find the pmf of Y, and deduce that Y Geometric(p) for some < p <. What is the value of p? Solution: Compute Pr[Y = i + ] = Pr[i X < i + ] = Pr[i < X i + ],

15 Math 5. Rumbos Spring 22 5 since X is continuous; so that where since X Exponential(). Pr[Y = i + ] = f X (x) = i+ i { e x if x > ; if x, Evaluating the integral in (37), for i and f X f X (x) dx, (37) as given in (38), yields (38) Pr[Y = i + ] = i+ i e x dx = [ e x] i+ i = e i e i, so that Pr[Y = i + ] = ( ) i ( ) e e (39) It follows from (39) that Y Geometric(p) with p = e.. The expected number of typographical errors on a page of a certain magazine is.2. What is the probability that an article of pages contains (a) no typographical errors, and (b) two or more typographical errors. Explain your reasoning. Solution: Let X denote the number of typographical errors in one page. Then, X may be modeled by a Poisson random variable with parameter λ, where E(X) = λ, and λ =.2 in this problem. We then have that the pmf of X is p X (k) = λk k! e λ, for k =,, 2, 3,... (4) The moment generating function of X is ψ X (t) = e λ(et ), for t R. (4)

16 Math 5. Rumbos Spring 22 6 Let X, X 2,..., X n, where n =, denote the number of typographical errors in pages through n, respectively. We may assume that X, X 2,... X n are independent and that they all have a Poisson(λ) distribution. Put Y = X + X X n. Then Y gives the number of typographical errors in the n pages of the article. The moment generating function of Y is then ψ Y (t) = ψ X (t) ψ X2 (t) ψ Xn (t), for t R, (42) by the independence of the X i s. It then follows from (42) and (4) that or ψ Y (t) = [e λ(et ) ] n for t R, ψ Y (t) = e nλ(et ) for t R. (43) Comparing (43) with (4), we see that Y has a Poisson distribution with parameter nλ; that is, Y Poisson(nλ), so that, in view of (4), p Y (k) = (nλ)k k! In this problem n = and λ =.2. e nλ, for k =,, 2, 3,... (44) (a) The probability that the article contains no typographical errors is Pr[Y = ] = e nλ = e %. (b) The probability that the article contains two or more typographical errors is Pr[Y 2] = Pr[Y ] = Pr[Y = ] Pr[Y = ] = e nλ nλe nλ = e 2 2e %.