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Suppose X is a nonnegative, absolutely continuous random variable. Let $Z=g\left(X\right)=C{e}^{-aX}$ , where $a>0,\phantom{\rule{0.277778em}{0ex}}C>0$ . Then $0<Z\le C$ . Use properties of the exponential and natural log function to show that
$Z=C{e}^{-aX}\le v$ iff ${e}^{-aX}\le v/C$ iff $-aX\le ln(v/C)$ iff $X\ge -ln(v/C)/a$ , so that
Use the result of [link] to show that if $X\sim $ exponential $\left(\lambda \right)$ , then
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 ${e}^{ax}$ at the end of x years. Hence, one dollar spent x years in the future has a present value ${e}^{-ax}$ . Suppose a device put into operation has time to failure (in years) $X\sim $ exponential $\left(\lambda \right)$ . If the cost of replacement at failure is C dollars, then the present value of the replacement is $Z=C{e}^{-aX}$ . Suppose $\lambda =1/10$ , $a=0.07$ , and $C=\$1000$ .
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
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 $\left(\mu \right)$ . 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 $Z=g\left(D\right)$ is the gain from the sales, then
Let $M=(-\infty ,m]$ . Then
Suppose
$\mu =50\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}m=50\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}c=30\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}p=50\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}r=20\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}s=40.$ .
Approximate
the Poisson random variable
D by truncating at 100. Determine
$P(500\le Z\le 1100)$ .
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
(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:
If the number of purchasers is a random variable X , the total cost (in dollars) is a random quantity $Z=g\left(X\right)$ described by
Suppose $X\sim $ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine $P(Z\ge 1000),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P(Z\ge 1300)$ , and $P(900\le Z\le 1400)$ .
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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