# 10.3 Problems on functions of random variables

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 . Use properties of the exponential and natural log function to show that

${F}_{Z}\left(v\right)=1-{F}_{X}\left(-,\frac{ln\left(v/C\right)}{a}\right)\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{for}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0

$Z=C{e}^{-aX}\le v$ iff ${e}^{-aX}\le v/C$ iff $-aX\le ln\left(v/C\right)$ iff $X\ge -ln\left(v/C\right)/a$ , so that

${F}_{Z}\left(v\right)=P\left(Z\le v\right)=P\left(X\ge -ln\left(v/C\right)/a\right)=1-{F}_{X}\left(-,\frac{ln\left(v/C\right)}{a}\right)$

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

${F}_{Z}\left(v\right)={\left(\frac{v}{C}\right)}^{\lambda /a}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0
${F}_{Z}\left(v\right)=1-\left[1,-,e,x,p,\left(-,\frac{\lambda }{a},·,ln,\left(v/C\right)\right)\right]={\left(\frac{v}{C}\right)}^{\lambda /a}$

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 $C=1000$ .

1. Use the result of [link] to determine the probability $Z\le 700,\phantom{\rule{0.277778em}{0ex}}500,\phantom{\rule{0.277778em}{0ex}}200$ .
2. 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.
$P\left(Z\le v\right)={\left(\frac{v}{1000}\right)}^{10/7}$
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

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

• For $t\le m,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}g\left(t\right)=\left(p-c\right)t-\left(c-r\right)\left(m-t\right)=\left(p-r\right)t+\left(r-c\right)m$
• For $t>m,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}g\left(t\right)=\left(p-c\right)m+\left(t-m\right)\left(p-s\right)=\left(p-s\right)t+\left(s-c\right)m$

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

$g\left(t\right)={I}_{M}\left(t\right)\left[\left(p-r\right)t+\left(r-c\right)m\right]+{I}_{M}\left(t\right)\left[\left(p-s\right)t+\left(s-c\right)m\right]$
$=\left(p-s\right)t+\left(s-c\right)m+{I}_{M}\left(t\right)\left(s-r\right)\left(t-m\right)$

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 $P\left(500\le Z\le 1100\right)$ .

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: • 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 $Z=g\left(X\right)$ described by

$g\left(X\right)=200+18{I}_{M1}\left(X\right)\left(X-10\right)+\left(16-18\right){I}_{M2}\left(X\right)\left(X-20\right)+\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}$
$\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\left(15-16\right){I}_{M3}\left(X\right)\left(X-30\right)+\left(13-15\right){I}_{M4}\left(X\right)\left(X-50\right)$
$\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M1=\left[10,\phantom{\rule{0.166667em}{0ex}}\infty \right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M2=\left[20,\phantom{\rule{0.166667em}{0ex}}\infty \right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M3=\left[30,\phantom{\rule{0.166667em}{0ex}}\infty \right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M4=\left[50,\phantom{\rule{0.166667em}{0ex}}\infty \right)$

Suppose $X\sim$ Poisson (75). Approximate the Poisson distribution by truncating at 150. Determine $P\left(Z\ge 1000\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(Z\ge 1300\right)$ , and $P\left(900\le Z\le 1400\right)$ .

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

can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
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