Feb 28 Homework Solutions Math 151, Winter Chapter 6 Problems (pages )

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1 Feb 8 Homework Solutions Math 5, Winter Chapter 6 Problems (pages 87-9) Problem 6 bin of 5 transistors is known to contain that are defective. The transistors are to be tested, one at a time, until the defective ones are identified. Denote by N the number of test made until the first defective is identified and by N the number of additional tests until the second defective is identified. Find the joint probability mass function of N and N. Note N and N take positive integer values such that N + N 5, i.e. N,..., and N,..., 5 N. Each pair of values for N and N are equally likely, and there are such pairs. So P {N i, N j} for i,..., and j,..., 5 i Problem 8 The joint probability density function of X and Y is given by f(x, y) c(y x )e y, y x y, < y <. (a) Find c. We want We compute f(x, y)dxdy y y c(y x )e y dxdy. y y (y x )e y dxdy 3 y 8 8. Thus c /8. (b) Find the marginal densities of X and Y. y (y x )e y dxdy ) ] xy [(xy x3 e y y 3 e y dy e y dy 3 x y dy using integration by parts

2 The marginal probability density function of X is f X (x) x x f(x, y)dy x e x + 8 (y x )e y dy ye y dy using integration by parts x x e x + e x e x ( x + ) e y dy using integration by parts Let f Y be the marginal probability density function of Y. For y < we have f Y (y), and for y we have f Y (y) 8 8 y y f(x, y)dx (y x )e y dx [(xy x3 3 6 y3 e y ) e y ] xy x y (c) Find E[X]. E[X] xf(x, y)dxdy 8 since x(y x )e y is an odd function of x. y Problem The joint probability density function of X and Y is given by y x(y x )e y dxdy, f(x, y) e x y, x <, y < (a) Find P {X < Y }. P {X < Y } x f(x, y)dydx e x dx [ e x] x x. x e x y dydx [ ] e x y y dx yx

3 (b) Find P {X < a}. P {X < a} a a f(x, y)dydx a e x dx [ e x] xa x e a. e x y dydx a [ e x y ] y y dx Problem 5 The random vector (X, Y ) is said to be uniformly distributed over a region R in the plane if, for some constant c, its joint density is { c if (x, y) R, f(x, y) otherwise. (a) Show that /c area of region R. Let R be the -dimensional plane. Observe that for E R, P {(X, Y ) E} f(x, y)dxdy c c area(e R). Since f(x, y) is a joint density function, we have So the area of R is /c. E E R P {(X, Y ) R } c area(r R) c area(r). (b) Suppose that (X, Y ) is uniformly distributed over the square centered at (, ) and with sides of length. Show that X and Y are independent, with each being distributed uniformly over (, ). Let R be the real line. For sets, R, we have P {X, Y } f(x, y)dxdy [,] [,] length( [, ])length( [, ]) dxdy and P {X } and P {Y } Thus f(x, y)dxdy f(x, y)dxdy [,] [,] dxdy length( [, ]) dxdy length( [, ]). P {X, Y } length( [, ])length( [, ]) P {X }P {Y }, so X and Y are independent. lso, our formulas for P {X } and P {Y } show that each of X and Y are uniformly distributed over (, ). 3

4 (c) What is the probability that (X, Y ) lies in the circle of radius centered at the origin? That is, find P {X + Y }. Let C denote the interior of the circle of radius centered at the origin. We want to compute P {(X, Y ) C}. We have P {(X, Y ) C} area(c R) area(c) π π. Problem 6 Suppose that n points are independently chosen at random on the circumference of a circle, and we want the probability that they all lie in some semicircle. That is, we want the probability that there is a line passing through the center of the circle such that all the points are on one side of that line. Let P,..., P n denote the n points. Let denote the event that all the points are contained in some semicircle, and let i be the event that all the points lie in the semicircle beginning at the point P i and going clockwise for 8, i,..., n. (a) Express in terms of the i. It s clear that i for each i. Hence n i i. Now we want to show the reverse inclusion. If occurs, then we can rotate the semicircle clockwise until it starts at one of the points P i, showing that i and thus n i i occurs. Hence n i i. So we ve shown that n i i. (b) re the i mutually exclusive? lmost, but not quite. Strictly speaking the i and j are not mutually exclusive since if P i P j, the other P k can lie in the semicircle beginning at P i P j and going clockwise for 8 degrees in which case both i and j occur. However, if P i P j, then i and j cannot both occur since if P j is in the semicircle starting at P i and going clockwise 8 degrees, then P i is not in the semicircle starting at P j and going clockwise 8 degrees. So i j {P i P j }. Moreover P {P i P j }, so P ( i j ). (c) Find P (). We argued in part (b) that P ( i j ). It follows that P ( i j k ), and the same is true for all higher order intersections. We have n P () P ( i ) i i n P ( i ) P ( i j ) + P ( i j k )... i<j i<j<k by the inclusion-exclusion identity n P ( i ) since all other terms are zero. i

5 Fix i {,..., n} and fix a randomly chosen P i. For j i, the probability that the randomly chosen point P j is in the semicircle starting at P i and going clockwise 8 degrees is /. So P ( i ) (/) n. Therefore P () n P ( i ) n(/) n. i Problem 8 Two points are selected randomly on a line of length L so as to be on opposite sides of the midpoint of the line. [In other words, the two points X and Y are independent random variables such that X is uniformly distributed over (, L/) and Y is uniformly distributed over (L/, L).] Find the probability that the distance between the two points is greater than L/3. Note that X and Y are jointly distributed with joint probability density function given by { x L L f(x, y) and L y L otherwise. Hence the probability that the distance between the two points is greater than L/3 is equal to P (Y X > L/3) f(x, y)dydx y x>l/3 L/ L L/6 L/ L dydx + L dydx max{l/,l/3+x} L/6 L L/ L L/6 L (L L/)dx + L dx + L/ L/6 L/6 L/3+x L/ L/6 8 3L L x dx [ 8 L (L/6 ) + 3L x ] L/ L x L/6 ( ) ( 9 8) 7 9. L dydx L ( L (L/3 + x) ) dx Note: it is much easier to find the limits for the integrals above if you first draw a picture of the region in the plane where x L, where L y L, and where y x > L/3. lternatively, after drawing this region in the plane, you can compare the area of this region to the area of the larger rectangle given by x L and L y L to see that the desired probability is 7/9. 5

6 Problem 3 The random variables X and Y have joint density function and equal to zero otherwise. (a) re X and Y are independent? f(x, y) xy( x), < x <, < y <. Let R be the real line. Let, R. Without loss of generality, we may assume that, (, ) since f is zero outside this range. We have P {X } f(x, y)dxdy xy( x)dxdy and P {Y } x( x)dx 6 x( x)dx f(x, y)dxdy x( x)dx ydy ydy 6 x( x)dx xy( x)dxdy ydy and P {X, Y } f(x, y)dxdy xy( x)dxdy 6 x( x)dx ydy P {X } P {Y }. Since P {X, Y } P {X } P {Y }, this means that X and Y are independent. Note that we ve also shown that f X (x) 6x( x) and f Y (y) y, and we will use this information in the remaining parts of this problem. (b) Find E[X]. E[X] (c) Find E[Y ]. (d) Find V ar(x). E[Y ] xf X (x)dx yf Y (y)dy 6x ( x)dx [ x 3 3x / ] /. y dx [ y 3 /3 ] /3. ydy so E[X ] x f X (x)dx 6x 3 ( x)dx [ 3x / 6x 5 /5 ] 3/, Var(X) E[X ] E[X] 3/ (/) /. 6

7 (e) Find V ar(y ). so E[Y ] y f Y (y)dy y 3 dx [ y / ] /. Var(Y ) E[Y ] E[Y ] / (/3) /8. Problem 6 Suppose that,, C are independent random variables, each being uniformly distributed over (, ). (a) What is the joint cumulative distribution function of,, C? The joint cumulative distribution function is f(a, b, c) f (a)f (b)f C (c) { if < x, y, z < otherwise. (b) What is the probability that all the roots of the equation x + x + C are real? Recall that the roots of x + x + C are real if C, which occurs with probability min{, ac} dbdcda min{,/(a)} / / / / ac dbdcda ac dbdcda + ( a / c / )dcda + [ c 3 a/ c 3/] c c da + 3 a/ da + / /(a) / /(a) / a da / [ a 8a3/] a/ + [ log(a)] a 9 a a/ 5 + log(). 36 Section 6 Theoretical Exercises (page 9-93) Problem Solve uffon s needle problem when L > D. ac dbdcda ( a / c / )dcda [ c 3 a/ c 3/] c/(a) c da See page 3 for the version of the problem where L D. What is new for L > D is that now X < min{d/, (L/) cos θ}. This does not reduce to X < (L/) cos θ as it did 7

8 before. Let φ be the angle such that D/ (L/) cos φ, that is, cos φ D/L (note we are using slightly different notation from the text). Then P {X < L } cos θ f X (x)f θ (y)dxdy x<(l/) cos y π/ min{d/,(l/) cos y φ D/ φd + dxdy + π/ φ π + L ( sin φ), where φ is the angle such that cos φ D/L. φ dxdy π/ (L/) cos y φ L cos y dy dxdy Problem 5 If X and Y are independent continuous positive random variables, express the density function of (a) Z X/Y and (b) Z XY in terms of the density functions of X and Y. Evaluate the density functions in the special case where X and Y are both exponential random variables. (a) Z X/Y. Note f Z (z) for z, since X and Y are both positive. For z >, we compute so (b) Z XY. P {Z < z} P {X/Y < z} P {X < zy } f Z (z) d dz P {Z < z} zy d zy f X (x)f Y (y)dxdy dz f X (x)f Y (y)dxdy, yf X (zy)f Y (y)dy. Note f Z (z) for z, since X and Y are both positive. For z >, we compute so P {Z < z} P {XY < z} P {X < z/y } z/y f X (x)f Y (y)dxdy, f Z (z) d dz P {Z < z} d z/y f X (x)f Y (y)dxdy dz y f X(z/y)f Y (y)dy. Now, let X and Y be exponential random variables with parameters λ and µ, respectively. We can now evaluate the density functions a bit more explicitly. 8

9 (a) Z X/Y. f Z (z) yf X (zy)f Y (y)dy λµ λz + µ λµ (λz + µ) e (λz+µ)y dy λµye λzy e µy dy after integrating by parts λµye (λz+µ)y dy (b) Z XY. f Z (z) y f λµ X(z/y)f Y (y)dy y e λz/y e µy λµ dy y e λz/y µy dy. 9