10.3 Problems on functions of random variables

Applied probability Page 1 / 3

Suppose X is a nonnegative, absolutely continuous random variable. Let $Z = g (X) = C e^{- a X}$ , where $a > 0, C > 0$ . Then $0 < Z \leq C$ . Use properties of the exponential and natural log function to show that

F_{Z} (v) = 1 - F_{X} (- \frac{ln (v / C)}{a}) for 0 < v \leq C

$Z = C e^{- a X} \leq v$ iff $e^{- a X} \leq v / C$ iff $- a X \leq ln (v / C)$ iff $X \geq - ln (v / C) / a$ , so that

F_{Z} (v) = P (Z \leq v) = P (X \geq - ln (v / C) / a) = 1 - F_{X} (- \frac{ln (v / C)}{a})

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Use the result of [link] to show that if $X \sim$ exponential $(λ)$ , then

F_{Z} (v) = {(\frac{v}{C})}^{λ / a} 0 < v \leq C

F_{Z} (v) = 1 - [1 - e x p (- \frac{λ}{a} \cdot ln (v / C))] = {(\frac{v}{C})}^{λ / a}

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Present value of future costs. Suppose money may be invested at an annual rate a , compounded continually. Then one dollar in hand now, has a value $e^{a x}$ at the end of x years. Hence, one dollar spent x years in the future has a present value $e^{- a x}$ . Suppose a device put into operation has time to failure (in years) $X \sim$ exponential $(λ)$ . If the cost of replacement at failure is C dollars, then the present value of the replacement is $Z = C e^{- a X}$ . Suppose $λ = 1 / 10$ , $a = 0.07$ , and $C = $ 1000$ .

Use the result of [link] to determine the probability $Z \leq 700, 500, 200$ .
Use a discrete approximation for the exponential density to approximate the probabilities in part (a). Truncate X at 1000 and use 10,000 approximation points.

P (Z \leq v) = {(\frac{v}{1000})}^{10 / 7}

v = [700 500 200];P = (v/1000).^(10/7)
P =  0.6008    0.3715    0.1003tappr
Enter matrix [a b]of x-range endpoints  [0 1000]
Enter number of x approximation points  10000Enter density as a function of t  0.1*exp(-t/10)
Use row matrices X and PX as in the simple caseG = 1000*exp(-0.07*t);
PM1 = (G<=700)*PX'
PM1 =  0.6005PM2 = (G<=500)*PX'
PM2 =  0.3716PM3 = (G<=200)*PX'
PM3 =  0.1003

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Optimal stocking of merchandise. A merchant is planning for the Christmas season. He intends to stock m units of a certain item at a cost of c per unit. Experience indicates demand can be represented by a random variable $D \sim$ Poisson $(μ)$ . If units remain in stock at the end of the season, they may be returned with recovery of r per unit. If demand exceeds the number originally ordered, extra units may be ordered at a cost of s each. Units are sold at a price p per unit. If $Z = g (D)$ is the gain from the sales, then

For $t \leq m, g (t) = (p - c) t - (c - r) (m - t) = (p - r) t + (r - c) m$
For $t > m, g (t) = (p - c) m + (t - m) (p - s) = (p - s) t + (s - c) m$

Let $M = (- \infty, m]$ . Then

g (t) = I_{M} (t) [(p - r) t + (r - c) m] + I_{M} (t) [(p - s) t + (s - c) m]

= (p - s) t + (s - c) m + I_{M} (t) (s - r) (t - m)

Suppose $μ = 50 m = 50 c = 30 p = 50 r = 20 s = 40.$ .
Approximate the Poisson random variable D by truncating at 100. Determine $P (500 \leq Z \leq 1100)$ .

mu = 50;
D = 0:100;c = 30;
p = 50;r = 20;
s = 40;m = 50;
PD = ipoisson(mu,D);G = (p - s)*D + (s - c)*m +(s - r)*(D - m).*(D<= m);
M = (500<=G)&(G<=1100);
PM = M*PD'PM =  0.9209[Z,PZ] = csort(G,PD);         % Alternate: use dbn for Zm = (500<=Z)&(Z<=1100);
pm = m*PZ'pm =  0.9209

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(See Example 2 from "Functions of a Random Variable") The cultural committee of a student organization has arranged a special deal for tickets to a concert. The agreement is that the organization will purchase tentickets at $20 each (regardless of the number of individual buyers). Additional tickets are available according to the following schedule:

11-20, $18 each
21-30, $16 each
31-50, $15 each
51-100, $13 each

If the number of purchasers is a random variable X , the total cost (in dollars) is a random quantity $Z = g (X)$ described by

g (X) = 200 + 18 I_{M 1} (X) (X - 10) + (16 - 18) I_{M 2} (X) (X - 20) +

(15 - 16) I_{M 3} (X) (X - 30) + (13 - 15) I_{M 4} (X) (X - 50)

where M 1 = [10, \infty), M 2 = [20, \infty), M 3 = [30, \infty), M 4 = [50, \infty)

Suppose $X \sim$ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine $P (Z \geq 1000), P (Z \geq 1300)$ , and $P (900 \leq Z \leq 1400)$ .

X = 0:150;
PX = ipoisson(75,X);G = 200 + 18*(X - 10).*(X>=10) + (16 - 18)*(X - 20).*(X>=20) + ...
     (15 - 16)*(X- 30).*(X>=30) + (13 - 15)*(X - 50).*(X>=50);
P1 = (G>=1000)*PX'
P1 =  0.9288P2 = (G>=1300)*PX'
P2 =  0.1142P3 = ((900<=G)&(G<=1400))*PX'
P3 =  0.9742[Z,PZ] = csort(G,PX);         % Alternate: use dbn for Zp1 = (Z>=1000)*PZ'
p1 =  0.9288

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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