MTH 202 : Probability and Statistics

Transcription

1 MTH 202 : Probability and Statistics Homework 3 3th January, 207 ( Any point in the interval [0, can be represented by its decimal expansion 0.x x Suppose a point is chosen at random from the interval [0,. Let X be the first digit in the decimal expansion representing the point. Compute the density of X. [Ref : Exercise-, Hoel, Port, Stone, Page-77] Solution : Note that every point x [0, has a unique decimal expansion except the rational numbers : x 0.x x 2... x N 0.x x 2... x N (x N where x, x 2,..., x N 0,,..., 9} with x N. Hence the only points which does not have a unique first digit in its decimal expansion (considering the above ambiguous cases are 0, 2 0,..., 9 0 corresponding to N, as above. For these numbers we would disallow the second kind of expansion, which mean the allowed decimal expansion of i/0 is i 0.i (and not 0.(i for i, 2,..., 9. With this note, the random variable X (which take values 0,,..., 9 is well defined. Then X(ω i if ω [ i, i+ and hence we can assign the probability P (X 0 0 i i which is the length of the interval [, i+. The density function f : R R is thus defined as if x 0,,..., 9, f(x 0 which come from the relation f(x P (X x. (2 Let N be a positive integer and let c2 x if x, 2,..., N, f(x

2 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, 2,..., N. The necessary condition to make f a probability density is N x f(x. The value c thus satisfy c2 x 2c(2 N x and hence c 2 N+ 2. (3 Suppose a box has 2 balls labelled, 2,..., 2.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 X and X 2, the random variables which represent the two independent experiments of selecting a ball at random from the box. Then X max(x, X 2. Both of the random variables X, X 2 satisfy the uniform distribution on 2 discrete points given by : if x, 2,..., 2, P (X i x 2 for i, 2. For x, 2,..., 2, we have P (X x P (X x, X 2 x P (X xp (X 2 x since X and X 2 are independent. Now P (X i x x 2, and hence P (X x ( x 2 2. Then, P (X x P (X x P (X x ( x 2 ( 2 x 2 2 2x. The density function f corresponding to X is thus given 44 by : 2x if x, 2,..., 2, f(x 44

3 (4 Let X and Y be independent random variables each having the uniform density on 0,,..., N}. Find the densities of : (a min(x, Y, (b max(x, Y, (c Y X. [Ref : Exercise-5, Hoel, Port, Stone, Page-78] Solution : Both of the independent random variables satisfy the density given by if x 0,,..., N, f(x N+ For (a let Z min(x, Y. Then, for x 0,,..., N} we have P (Z x P (X x, Y x P (X xp (Y x. Hence P (Z x ( N x+ 2. N Then we have : 2(N x + P (Z x P (Z x P (Z x + (N + 2 for x 0,,..., N. For (b let Z max(x, Y. Then, for x 0,,..., N} we have P (Z x P (X x, Y x P (X xp (Y x. Hence P (Z x ( x+ 2. N Then we have : P (Z x P (Z x P (Z x 2x + (N + 2 for x 0,,..., N. For (c let Z Y X. Notice that if x 0, P ( Y X x P (Y X x + P (Y X x Hence for x 0 P (Y X x P (Y X x, X t P (Y x + t, X t Hence Similarly, P (Y X x P (Y X x N x P (Y x + tp (X t (N x + (N + 2 (N x + (N + 2 3

4 4 Combining these we have P ( Y X x 2(N x + (N + 2 while x 0. While x 0 we have P ( Y X 0 P (X Y P (X Y, X t Hence, P (Y t, X t P (Y tp (X t P ( Y X 0 N + (5 Let X and Y be independent random variables having geometric densities with parameters p and p 2 respectively. Find the density of (a min(x, Y, (b X + Y. [Ref : Exercise-7, Hoel, Port, Stone, Page-79] Solution : (a The densities of the random variables X and Y satisfy p ( p x if x 0,,... P (X x and P (Y y p 2 ( p 2 y if y 0,,... For an integer x 0 we have, P (X x P (X t p ( p t ( p x tx tx and similarly P (Y x ( p 2 x. Letting Z min(x, Y we have P (Z x P (X x, Y x P (X xp (Y x and hence P (Z x ( p x ( p 2 x. This implies, P (Z x P (Z x ( p x ( p 2 x Now, P (Z x P (Z x P (Z x

5 [ ( p ( p 2 ]( p x ( p 2 x q( q x for x 0, where q p + p 2 p p 2. Note that while x 0, P (Z 0 P (Z 0 ( p ( p 2 q Hence Z satisfies geometric density with parameter q. For (b set Z X + Y. Then P (Z x x x P (X + Y x, Y t P (X x tp (Y t p p 2 ( p ( t x x p 2 p p p 2 p p 2 [( p 2 x+ ( p x+ ] using the finite geometric sum. (6 Suppose 2r balls are distributed at random into r boxes. Let X i denote the number of balls in box i. (a Find the joint density of X,..., X r. (b Find the probability that each box contains exactly 2 balls. [Ref : Exercise-2, Hoel, Port, Stone, Page-79] Solution : (a Let I : 0,,..., 2r}, and A I r defined by A (x, x 2,..., x r I r : x + x x r 2r} For a vector (x, x 2,..., x r A, the joint density is given by P (X x,..., X r x r ( 2r ( 2r x ( x x r (x +x x r x r r 2r The numerator is equal to C(2r; x,..., x r and calculated using induction in section 3.4., Page-68, Hoel, Port, Stone. This is given by Hence we have C(2r; x,..., x r P (X x,..., X r x r (2r! (x!... (x r! (2r! (x!... (x r!r 2r Also P (X x,..., X r x r 0 if (x, x 2,..., x r A. For (b setting (x, x 2,..., x r (2, 2,..., 2 we have P (X 2,..., X r 2 (2r! 2 r r 2r 5

6 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 ( p 50 + ( p 49 ( p + ( 50 p 48 ( p 2 48 The Poisson approximation as discussed in section 3.4.2, Page- 69, Hoel, Port, Stone, is given by ( 50 p k ( p 50 k (50pk e 50p k k! Hence the Poisson approximation to above probability is given by e 5/2 [ + 5/2 + /2 (5/2 2 ] (53/8e 5/2 (8 Let X be a non-negative integer valued random variable whose probability generating function is given by Φ X (t e λ(t2, where λ > 0. Find f X. [Ref : Exercise-33, Hoel, Port, Stone, Page-8] Solution : Recall that the probability generating function of a non-negative integer valued random variable X is given by Φ X (t P (X xt x ( t x0 Here we have the Taylor series expansion Φ X (t e λ(t2 e λ [ + (λt (λt2 k +... ] k! Now equating the coefficients of the Taylor expansions we get λx/2 P (X x e λ (x/2! when x is a non-negative even integer, and P (X x 0 otherwise. Hence the density function is given by λ λx/2 e if x 0, 2, 4,... (x/2! P (X x

7 (9 Let X and Y be independent random variables having Poisson densities with parameters λ and λ 2 respectively. Find P (Y y X + Y z for y 0,..., z. [Hint : Use Theorem-(iii, Page-75, Hoel, Port, Stone] [Ref : Exercise-35, Hoel, Port, Stone, Page-8] Solution : Using Theorem-(iii, Page-75, Hoel, Port, Stone, we can see that X + Y satisfies a Poisson distribution with parameter λ + λ 2. Now, P (Y y X + Y z is P (Y y, X + Y z P (X + Y z P (X z yp (Y y P (X + Y z since X and Y are independent. Further plugging the parameter values we have this e λ z y λ e λ 2 y ( λ 2 e (λ +λ 2 (λ + λ 2 z (z y! y! z! ( ( z y ( y z λ λ2 for y 0,..., z y λ + λ 2 λ + λ 2 (0 Let X, Y and Z be independent random variables having Poisson densities with parameters λ, λ 2 and λ 3 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-(iii, Page-75, Hoel, Port, Stone] [Ref : Exercise-36, Hoel, Port, Stone, Page-8] Proof : As in the previous exercise, using Theorem-(iii, Page- 75, Hoel, Port, Stone, we can see that X + Y + Z satisfies a Poisson distribution with parameter λ + λ 2 + λ 3. 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 xp (Y yp (Z z P (X + Y + Z x + y + z 7

8 8 since X, Y and Z are independent. Further plugging the parameter values we have this ( e λ x λ e λ 2 y λ 2 e λ 3 z λ e (λ +λ 2 +λ 3 (λ + λ 2 + λ 3 (x+y+z x! y! z! (x + y + z! (x + y + z! x!y!z! λ x λ 2 y λ 3 z (λ + λ 2 + λ 3 x+y+z