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?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
yes
Asali
I'm not good at math so would you help me
Samantha
what is the problem that i will help you to self with?
Asali
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive