ssrs 2008 r2 barcode font SOLVED PROBLEMS in Software

Printer QR Code ISO/IEC18004 in Software SOLVED PROBLEMS

Expectation of random variables 3.1. In a lottery there are 200 prizes of $5, 20 prizes of $25, and 5 prizes of $100. Assuming that 10,000 tickets are to be issued and sold, what is a fair price to pay for a ticket

Let X be a random variable denoting the amount of money to be won on a ticket. The various values of X together with their probabilities are shown in Table 3-2. For example, the probability of getting one of the 20 tickets giving a $25 prize is 20 > 10,000 0.002. The expectation of X in dollars is thus E(X) (5)(0.02) (25)(0.002) (100)(0.0005) (0)(0.9775) 0.2

or 20 cents. Thus the fair price to pay for a ticket is 20 cents. However, since a lottery is usually designed to raise money, the price per ticket would be higher. Table 3-2 x (dollars) P(X x) 5 0.02 25 0.002 100 0.0005 0 0.9775

Let X and Y be the points showing on the two dice. We have E(X) Then, by Theorem 3-2, E(X E(Y) 1 1 6 Y) 1 2 6 c 1 6 6

E(X)

E(Y)

3.3. Find the expectation of a discrete random variable X whose probability function is given by f (x)

We have E(X)

1 2

1, 2, 3, c )

1 2 4 1 3 8

1 2 4 1 4 1 4

1 3 8 1 8

1 2 8

3.4. A continuous random variable X has probability density given by f (x) Find (a) E(X), (b) E(X2).

2e 0

E(X)

E(X2)

3.5. The joint density function of two random variables X and Y is given by f (x, y) e xy>96 0 0 x 4, 1 otherwise 3Y).

2 B(x2)

2 B (x)

30 x(2e (1)

R 2

0 ` 0

(2x)

(2)

R 2

(a) (b) (c)

E(X) E(Y) E(XY)

8 3 31 9 248 27

Another method (c) Since X and Y are independent, we have, using parts (a) and (b), E(XY) E(X)E(Y) 8 31 3 9 8 2 3 248 27 3 31 9

3y)

47 3

(d) By Theorems 3-1 and 3-2, pages 76 77, together with (a) and (b), E(2X 3Y) 2E(X) 3E(Y)

47 3

y) f (x, y) a a yf (x, y)

a a xf (x, y)

E(X)

E(Y)

If either variable is continuous, the proof goes through as before, with the appropriate summations replaced by integrations. Note that the theorem is true whether or not X and Y are independent.

Let f (x, y) be the joint probability function of X and Y, assumed discrete. If the variables X and Y are independent, we have f (x, y) f1 (x) f2 ( y). Therefore, E(XY) a a xyf (x, y)

a [(xf1(x)E( y)]

E(X)E(Y)

a a xyf1(x) f2 ( y)

If either variable is continuous, the proof goes through as before, with the appropriate summations replaced by integrations. Note that the validity of this theorem hinges on whether f (x, y) can be expressed as a function of x multiplied by a function of y, for all x and y, i.e., on whether X and Y are independent. For dependent variables it is not true in general.

Variance and standard deviation 3.8. Find (a) the variance, (b) the standard deviation of the sum obtained in tossing a pair of fair dice.

(a) Referring to Problem 3.2, we have E(X) E(X2) Then, by Theorem 3-4, Var (X) E(Y2) 1 12 6 E(Y) 1 22 6 91 6 1 > 2. Moreover, c 1 62 6 35 12

91 6

Var (Y)

7 2

35 6