4653: Math Elementary Probability: Spring 2007
Homework #6. Problems and Solutions
1. Sec. 4.2: #6: A Geiger counter is recording background radiation at an average rate of one hit per minute. Let T3 be the time in minutes when the third hit occurs after the counter is switched on. Find P (2 T3 4). Solution. The random variable T3 has a gamma distribution with parameter r = 3 and = 1, which gives us a density function f (t) = Therefore, P (2 T3 4) =
2 4 2 -t t e
r tr-1 r-1 -t 1 2 -t t e = t e . (r) 2
2
-4
dt = -e-t
t2 +t+1 2
4 2
= -e
(8 + 4 + 1) + e-2 (2 + 2 + 1) = 5e-2 - 13e-4 0.4386.
2. Sec. 4.4: #4: Suppose X has uniform distribution on (-1, 1). Find the density of Y = X 2 . Solution. It is easy to show that |X| has uniform distribution on [0, 1). Then Y has density fY (y) =
f|X| |dy/d|x| |
=
1 2|x|
=
1 2 y
for 0 < y < 1, otherwise.
0
3. Sec. 4.5: #2: Find the cumulative distribution functions of: a) the binomial (3, 1/2) distribution; b) the geometric (1/2) distribution on {1, 2, . . .}. Solution. 0 for x < 0, 1/8 for 0 x < 1, a) P (X x) = 1/2 for 1 x < 2, 7/8 for 2 x < 3, 1 for x 3. b) P (X x) = 0
2k -1 2k
for x < 1, for k x < k + 1,
k = 1, 2, . . . .
4. Sec. 4.5: #4: Let X be a random variable with c.d.f. F (x). Find the c.d.f. of aX + b first for a > 0, then for a < 0. Solution. Let Y = aX + b. Then the c.d.f. for Y is FY (y) = P (Y y) = P (aX + b y) + P (aX y - b). For a > 0, we have FY (y) = P X For a < 0, we have FY (y) = P X y-b a =1-P X < 1 y-b a =1-F y-b - . a y-b a =F y-b . a
5. Sec. 4.5: #6: Let X be a random variable with c.d.f. F (x) = x3 for 0 x 1. Find: 1 a) P (X 2 ); b) the density function f (x); c) E(X). d) Let Y1 , Y2 , Y3 be three points chosen independently and uniformly on the unit interval, and let X be the rightmost point. Show that X has the distribution described above. Solution. a) P (X 1/2) = 1 - F (1/2) = 1 - (1/2)3 = 7/8. b) Since f (x) = F (x), we have f (x) = c)
3x2 0
for 0 x 1, for x < 0 or x > 1.
1
E(X) =
-
xf (x)dx =
-
3 x 3x2 dx = x4 4
0
3 = . 4
d) Let Y be a random variable uniformly distributed on the unit interval (0, 1), let FY (y) denote its c.d.f., and let FX (x) denote the c.d.f. for X = max(Y1 , Y2 , Y3 ). Then 0 for x < 0, 3 3 for 0 x 1, FX (x) = [FY (x)] = x 1 for x > 1, which is the c.d.f. described above. 6. Sec. 4.6: #2: Let X have beta (r, s) distribution. a) Find E(X 2 ), and use the formula for E(X) given in this section to find Var (X). b) Find a formula for E(X k ), for integers k 1. Solution. a) We have
1 1 B(r + 2, s) xr+1 (1 - x)s-1 dx = B(r, s) 0 B(r, s) (r + 2)(s) (r + s) r(r + 1) = = , (r + s + 2) (r)(s) (r + s)(r + s + 1) r2 rs r(r + 1) Var (X) = E(X 2 ) - [E(X)]2 = - = . 2 2 (r + s + 1) (r + s)(r + s + 1) (r + s) (r + s)
E(X 2 ) =
b) We have E(X k ) =
1 1 B(r + k, s) xr+k-1 (1 - x)s-1 dx = B(r, 0 s) B(r, s) (r + k)(s) (r + s) r(r + 1) (r + k - 1) = = . (r + s + k) (r)(s) (r + s)(r + s + 1) (r + s + k - 10
7. Sec. 5.1: #2: A metal rod is l inches long. Measurements on the length of this rod are equal to l plus a random error. Assume that the errors are uniformly distributed over the range -0.1 inch to +0.1 inch, and are independent of each other. a) Find the chance that a measurement is less than 1/100 of an inch away from l. b) Find the chance that two measurements are less than 1/100 of an inch away from each other.
2
Solution. a) Let X be the measurement error. Then P (|X| < 0.01) = 0.02 = 0.1. 0.2
b) Let Y be the second measurement error, and let the set of possible values of (X, Y ) be S = {(x, y) : |x| < 0.1 and |y| < 0.1}. Then the region where the two measurement are less than 1/100 of an inch away from each other is B = {(x, y) S : |x - y| < 0.01}. Then P (|X - Y | < 0.01) = area (B) 0.22 - 0.192 39 = = = 0.0975. area (S) 0.22 400
8. Sec. 5.1: #6a): A group of 10 people agree to meet for lunch at a cafe between 12 noon and 12 : 15 P.M. Assume that each person arrives at the cafe at a time uniformly distributed between noon and 12 : 15 P.M., and that the arrival times are independent of each other. Jack and Jill are two members of the group. Find the probability that Jack arrives at least two minutes before Jill. Solution. Let X be Jacks arrival time, measured in minutes after 12 noon, and let Y be Jills arrival time, measured on the same scale. Then P (X < Y - 2) = 169 132 /2 = 0.3756. 152 450
9. Sec.5.2: #4: For random variables X and Y with joint density function f (x, y) = 6 e-2x-3y (x, y > 0)
and f (x, y) = 0 otherwise, find a) P (X x, Y y); b) fX (x); c) fY (y). d) Are X and Y independent? Give a reason for your answer. Solution. a)
y x y x
P (X x, Y y) =
- - y
f (t, s) dtds =
0 x 0 y
6e-2t e-3s dtds
y x 0
=
0
3e
-3s 0
2e
y
-2t
dtds =
0
3e
-3s
-e
-2t
ds =
0
3e-3s (1 - e-2x )ds
= (1 - e-2x )
0
3e-3s ds = (1 - e-2x )(1 - e-3y ).
b)
fX (x) =
-
f (x, y)dy =
0
6e
-2x -3y
e
dy = 2e
-2x 0
3e-3y dy = 2e-2x .
c)
fY (y) =
-
f (x, y)dx =
0
6e-2x e-3y dx = 3e-3y
0
2e-2x dx = 3e-3y .
d) X and Y are independent because f (x, y) = fX (x)fY (y). 3
10. Sec. 5.2: #8a): Random variables X and Y have joint density fX,Y (x, y) = c(y 2 - x2 )e-y 0 for - y x y, y > 0; otherwise.
Here c is a constant. Show that Y has a gamma density, and hence deduce that c = 1/8. Solution. For y > 0, the marginal density for Y is
y
fY (y) =
-
fX,Y (x, y)dx =
-y
c(y 2 - x2 )e-y dx = ce-y y 2 x - (-y)3 3 4c 3 -y y e . 3
x3 3
x=y x=-y
ce-y
y3 -
y3 3
- (-y)3 -
=
Recall that, for y > 0, the density function for the gamma distribution with parameters r and is fr, (y) = r y r-1 -y e . (r)
We see that Y has gamma distribution with parameters r = 4 and = 1 and therefore 1 1 4c = = , 3 (4) 3! 1 c= . 8
4