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MTH 202 : Probability and Statistics Homework 3 13th January, 2017 (1) Any point in the interval [0, 1) can be represented by its decimal expansion 0.x1 x2 . . . . Suppose a point is chosen at random from the interval [0, 1). Let X be the first digit in the decimal expansion representing the point. Compute the density of X. [Ref : Exercise-1, Hoel, Port, Stone, Page-77] Solution : Note that every point x ∈ [0, 1) has a unique decimal expansion except the rational numbers : x = 0.x1 x2 . . . xN = 0.x1 x2 . . . xN −1 (xN − 1)99 . . . where x1 , x2 , . . . , xN ∈ {0, 1, . . . , 9} with xN ≥ 1. Hence the only points which does not have a unique first digit in its decimal expansion (considering the above ambiguous cases) are 1 2 9 , ,..., 10 10 10 corresponding to N = 1, as above. For these numbers we would disallow the second kind of expansion, which mean the allowed decimal expansion of i/10 is i = 0.i (and not 0.(i − 1)999 . . . ) 10 for i = 1, 2, . . . , 9. With this note, the random variable X (which take values 0, 1, . . . , 9) is well defined. Then X(ω) = i if ω ∈ [ 10i , i+1 ) and hence we can assign the probability P (X = 10 1 i) = 10 which is the length of the interval [ 10i , i+1 ). The density 10 function f : R → R is thus defined as ( 1 if x = 0, 1, . . . , 9, f (x) = 10 0 otherwise which come from the relation f (x) = P (X = x). (2) Let N be a positive integer and let ( c2x if x = 1, 2, . . . , N, f (x) = 0 otherwise 1 2 Find the value of c such that f is a probability density. [Ref : Exercise-4, Hoel, Port, Stone, Page-77] Solution : The values of x where f (x) > 0 are x = 1, 2, . . . , N . The PN necessary condition to make f a probability density is x=1 f (x) = 1. The value c thus satisfy N X c2x = 1 ⇒ 2c(2N − 1) = 1 x=1 and hence c = 1 . 2N +1 −2 (3) Suppose a box has 12 balls labelled 1, 2, . . . , 12.Two independent repetitions are made of the experiment of selecting a ball at random from the box. Let X denote the larger of the two numbers on the balls selected. Compute the density of X. [Ref : Exercise-8, Hoel, Port, Stone, Page-78] Solution : Denote by X1 and X2 , the random variables which represent the two independent experiments of selecting a ball at random from the box. Then X = max(X1 , X2 ). Both of the random variables X1 , X2 satisfy the uniform distribution on 12 discrete points given by : ( 1 if x = 1, 2, . . . , 12, P (Xi = x) = 12 0 otherwise for i = 1, 2. For x = 1, 2, . . . , 12, we have P (X ≤ x) = P (X1 ≤ x, X2 ≤ x) = P (X1 ≤ x)P (X2 ≤ x) since X1 and X2 are independent. Now P (Xi ≤ x) = x 2 hence P (X ≤ x) = 12 . Then, x , 12 and 2 x 2 P (X = x) = P (X ≤ x) − P (X ≤ x − 1) = 12 − x−1 = 12 2x−1 . The density function f corresponding to X is thus given 144 by : ( 2x−1 if x = 1, 2, . . . , 12, f (x) = 144 0 otherwise 3 (4) Let X and Y be independent random variables each having the uniform density on {0, 1, . . . , N }. Find the densities of : (a) min(X, Y ), (b) max(X, Y ), (c) |Y − X|. [Ref : Exercise-15, Hoel, Port, Stone, Page-78] Solution : Both of the independent random variables satisfy the density given by ( 1 if x = 0, 1, . . . , N, f (x) = N +1 0 otherwise For (a) let Z = min(X, Y ). Then, for x ∈ {0, 1, . . . , N } we have P (Z ≥ x) = P (X ≥ x, Y ≥ x) = P (X ≥ x)P (Y ≥ x). Hence 2 P (Z ≥ x) = N −x+1 . Then we have : N P (Z = x) = P (Z ≥ x) − P (Z ≥ x + 1) = 2(N − x) + 1 (N + 1)2 for x = 0, 1, . . . , N . For (b) let Z = max(X, Y ). Then, for x ∈ {0, 1, . . . , N } we have P (Z ≤ x) = P (X ≤ x, Y ≤ x) = P (X ≤ x)P (Y ≤ x). 2 Hence P (Z ≤ x) = x+1 . Then we have : N P (Z = x) = P (Z ≤ x) − P (Z ≤ x − 1) = 2x + 1 (N + 1)2 for x = 0, 1, . . . , N . For (c) let Z = |Y − X|. Notice that if x 6= 0, P (|Y − X| = x) = P (Y − X = x) + P (Y − X = −x) Hence for x 6= 0 P (Y − X = x) = N X P (Y − X = x, X = t) t=0 = N X P (Y = x + t, X = t) = t=0 N −x X P (Y = x + t)P (X = t) t=0 Hence P (Y − X = x) = (N − x + 1) (N + 1)2 Similarly, P (Y − X = −x) = (N − x + 1) (N + 1)2 4 Combining these we have 2(N − x + 1) (N + 1)2 while x 6= 0. While x = 0 we have P (|Y − X| = x) = N X P (|Y − X| = 0) = P (X = Y ) = P (X = Y, X = t) t=0 = N X P (Y = t, X = t) = t=0 N X P (Y = t)P (X = t) t=0 Hence, P (|Y − X| = 0) = 1 N +1 (5) Let X and Y be independent random variables having geometric densities with parameters p1 and p2 respectively. Find the density of (a) min(X, Y ), (b) X + Y . [Ref : Exercise-17, Hoel, Port, Stone, Page-79] Solution : (a) The densities of the random variables X and Y satisfy ( p1 (1 − p1 )x if x = 0, 1, . . . P (X = x) = 0 otherwise and ( p2 (1 − p2 )y P (Y = y) = 0 if y = 0, 1, . . . otherwise For an integer x ≥ 0 we have, ∞ ∞ X X P (X ≥ x) = P (X = t) = p1 (1 − p1 )t = (1 − p1 )x t=x t=x and similarly P (Y ≥ x) = (1 − p2 )x . Letting Z = min(X, Y ) we have P (Z ≥ x) = P (X ≥ x, Y ≥ x) = P (X ≥ x)P (Y ≥ x) and hence P (Z ≥ x) = (1 − p1 )x (1 − p2 )x . This implies, P (Z ≤ x − 1) = 1 − P (Z ≥ x) = 1 − (1 − p1 )x (1 − p2 )x Now, P (Z = x) = P (Z ≤ x) − P (Z ≤ x − 1) 5 = [1 − (1 − p1 )(1 − p2 )](1 − p1 )x (1 − p2 )x = q(1 − q)x for x ≥ 0, where q = p1 + p2 − p1 p2 . Note that while x = 0, P (Z = 0) = P (Z ≤ 0) = 1 − (1 − p1 )(1 − p2 ) = q Hence Z satisfies geometric density with parameter q. For (b) set Z = X + Y . Then P (Z = x) x x X X = P (X + Y = x, Y = t) = P (X = x − t)P (Y = t) t=0 t=0 = p1 p2 (1 − p1 )x = Px t=0 x+1 p1 p2 [(1−p2 ) p1 −p2 1−p2 1−p1 t −(1−p1 )x+1 ] using the finite geometric sum. (6) Suppose 2r balls are distributed at random into r boxes. Let Xi denote the number of balls in box i. (a) Find the joint density of X1 , . . . , Xr . (b) Find the probability that each box contains exactly 2 balls. [Ref : Exercise-21, Hoel, Port, Stone, Page-79] Solution : (a) Let I := {0, 1, . . . , 2r}, and A ⊆ I r defined by A = {(x1 , x2 , . . . , xr ) ∈ I r : x1 + x2 + . . . + xr = 2r} For a vector (x1 , x2 , . . . , xr ) ∈ A, the joint density is given by P (X1 = x1 , . . . , Xr = xr ) 2r 2r−x1 . . . 2r−(x1 +xx2 r+...+xr−1 ) x1 x2 = r2r The numerator is equal to C(2r; x1 , . . . , xr ) and calculated using induction in section 3.4.1, Page-68, Hoel, Port, Stone. This is given by (2r)! C(2r; x1 , . . . , xr ) = (x1 !) . . . (xr !) Hence we have (2r)! P (X1 = x1 , . . . , Xr = xr ) = (x1 !) . . . (xr !)r2r Also P (X1 = x1 , . . . , Xr = xr ) = 0 if (x1 , x2 , . . . , xr ) 6∈ A. For (b) setting (x1 , x2 , . . . , xr ) = (2, 2, . . . , 2) we have P (X1 = 2, . . . , Xr = 2) = (2r)! 2r r2r 6 (7) Use the Poisson approximation to calculate the probability that at most 2 out of 50 given people will have invalid driver’s licences if normally 5% of the people do. [Ref : Exercise-23, Hoel, Port, Stone, Page-80] Solution : The probability of the event with Binomial distribution with parameters (50, p), where p = 0.05 (probability of success) is given by 50 50 50 49 50 48 p + p (1 − p) + p (1 − p)2 50 49 48 The Poisson approximation as discussed in section 3.4.2, Page69, Hoel, Port, Stone, is given by 50 k (50p)k −50p p (1 − p)50−k ≈ e k k! Hence the Poisson approximation to above probability is given by e−5/2 [1 + 5/2 + 1/2 × (5/2)2 ] = (53/8)e−5/2 (8) Let X be a non-negative integer valued random variable whose 2 probability generating function is given by ΦX (t) = eλ(t −1) , where λ > 0. Find fX . [Ref : Exercise-33, Hoel, Port, Stone, Page-81] Solution : Recall that the probability generating function of a non-negative integer valued random variable X is given by ∞ X P (X = x)tx (−1 ≤ t ≤ 1) ΦX (t) = x=0 Here we have the Taylor series expansion (λt2 )k + ...] k! Now equating the coefficients of the Taylor expansions we get 2 −1) ΦX (t) = eλ(t = e−λ [1 + (λt2 ) + . . . + P (X = x) = e−λ λx/2 (x/2)! when x is a non-negative even integer, and P (X = x) = 0 otherwise. Hence the density function is given by ( λx/2 e−λ (x/2)! if x = 0, 2, 4, . . . P (X = x) = 0 otherwise 7 (9) Let X and Y be independent random variables having Poisson densities with parameters λ1 and λ2 respectively. Find P (Y = y|X + Y = z) for y = 0, . . . , z. [Hint : Use Theorem-1(iii), Page-75, Hoel, Port, Stone] [Ref : Exercise-35, Hoel, Port, Stone, Page-81] Solution : Using Theorem-1(iii), Page-75, Hoel, Port, Stone, we can see that X + Y satisfies a Poisson distribution with parameter λ1 + λ2 . Now, P (Y = y|X + Y = z) is P (Y = y, X + Y = z) P (X = z − y)P (Y = y) = = P (X + Y = z) P (X + Y = z) since X and Y are independent. Further plugging the parameter values we have this −(λ1 +λ2 ) −1 e−λ1 λ1 z−y e−λ2 λ2 y e (λ1 + λ2 )z = × × (z − y)! y! z! z−y y λ2 z λ1 for y = 0, . . . , z = λ1 + λ2 λ1 + λ2 y (10) Let X, Y and Z be independent random variables having Poisson densities with parameters λ1 , λ2 and λ3 respectively. Find P (X = x, Y = y, Z = z|X + Y + Z = x + y + z) for nonnegative integers x, y and z. [Hint : Use Theorem-1(iii), Page-75, Hoel, Port, Stone] [Ref : Exercise-36, Hoel, Port, Stone, Page-81] Proof : As in the previous exercise, using Theorem-1(iii), Page75, Hoel, Port, Stone, we can see that X + Y + Z satisfies a Poisson distribution with parameter λ1 + λ2 + λ3 . Now, P (X = x, Y = y, Z = z|X + Y + Z = x + y + z) is P (X = x, Y = y, Z = z, X + Y + Z = z) = P (X + Y + Z = x + y + z) P (X = x)P (Y = y)P (Z = z) = P (X + Y + Z = x + y + z) 8 since X, Y and Z are independent. parameter values we have this Further plugging the e−λ1 λ1 x e−λ2 λ2 y e−λ3 λ1 z e−(λ1 +λ2 +λ3 ) (λ1 + λ2 + λ3 )(x+y+z) = × × × x! y! z! (x + y + z)! (x + y + z)! λ1 x λ2 y λ3 z = × x!y!z! (λ1 + λ2 + λ3 )x+y+z !−1