10.3 Problems on functions of random variables  (Page 3/3)

${f}_{XY}\left(t,u\right)=\frac{3}{88}\left(2t+3{u}^{2}\right)$ for $0\le t\le 2$ , $0\le u\le 1+t$ (see Exercise 15 from "Problems on Random Vectors and Joint Distributions").

$Z={I}_{\left[0,1\right]}\left(X\right)4X+{I}_{\left(1,2\right]}\left(X\right)\left(X+Y\right)$

Determine $P\left(Z\le 2\right)$

$P\left(Z\le 2\right)=P\left(Z\in Q=Q1M1\bigvee Q2M2\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{where}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M1=\left\{\left(t,u\right):0\le t\le 1,\phantom{\rule{0.277778em}{0ex}}0\le u\le 1+t\right\}$
$M2=\left\{\left(t,u\right):1
$Q1=\left\{\left(t,u\right):0\le t\le 1/2\right\},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}Q2=\left\{\left(t,u\right):u\le 2-t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(see}\phantom{\rule{4.pt}{0ex}}\text{figure)}$
$P=\frac{3}{88}{\int }_{0}^{1/2}{\int }_{0}^{1+t}\left(2t+3{u}^{2}\right)\phantom{\rule{0.166667em}{0ex}}dudt+\frac{3}{88}{\int }_{1}^{2}{\int }_{0}^{2-t}\left(2t+3{u}^{2}\right)\phantom{\rule{0.166667em}{0ex}}dudt=\frac{563}{5632}$
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 3] Enter number of X approximation points 200Enter number of Y approximation points 300 Enter expression for joint density (3/88)*(2*t + 3*u.^2).*(u<=1+t) Use array operations on X, Y, PX, PY, t, u, and PG = 4*t.*(t<=1) + (t+u).*(t>1); [Z,PZ]= csort(G,P); PZ2 = (Z<=2)*PZ' PZ2 = 0.1010 % Theoretical = 563/5632 = 0.1000

${f}_{XY}\left(t,u\right)=\frac{24}{11}tu$ for $0\le t\le 2$ , $0\le u\le min\left\{1,2-t\right\}$ (see Exercise 17 from "Problems on Random Vectors and Joint Distributions").

$Z={I}_{M}\left(X,Y\right)\frac{1}{2}X+{I}_{{M}^{c}}\left(X,Y\right){Y}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M=\left\{\left(t,u\right):u>t\right\}$

Determine $P\left(Z\le 1/4\right)$ .

$P\left(Z\le 1/4\right)=P\left(\left(X,Y\right)\in {M}_{1}{Q}_{1}\bigvee {M}_{2}{Q}_{2}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{1}=\left\{\left(t,u\right):0\le t\le u\le 1\right\}$
${M}_{2}=\left\{\left(t,u\right):0\le t\le 2,\phantom{\rule{0.277778em}{0ex}}0\le t\le min\left(t,2-t\right)\right\}$
${Q}_{1}=\left\{\left(t,u\right):t\le 1/2\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{Q}_{2}=\left\{\left(t,u\right):u\le 1/2\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(see}\phantom{\rule{4.pt}{0ex}}\text{figure)}$
$P=\frac{24}{11}{\int }_{0}^{1/2}{\int }_{0}^{1}tu\phantom{\rule{0.166667em}{0ex}}dudt+\frac{24}{11}{\int }_{1/2}^{3/2}{\int }_{0}^{1/2}tu\phantom{\rule{0.166667em}{0ex}}dudt+\frac{24}{11}{\int }_{3/2}^{2}{\int }_{0}^{2-t}tu\phantom{\rule{0.166667em}{0ex}}dudt=\frac{85}{176}$
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 1] Enter number of X approximation points 400Enter number of Y approximation points 200 Enter expression for joint density (24/11)*t.*u.*(u<=min(1,2-t)) Use array operations on X, Y, PX, PY, t, u, and PG = 0.5*t.*(u>t) + u.^2.*(u<t); [Z,PZ]= csort(G,P); pp = (Z<=1/4)*PZ' pp = 0.4844 % Theoretical = 85/176 = 0.4830

${f}_{XY}\left(t,u\right)=\frac{3}{23}\left(t+2u\right)$ for $0\le t\le 2$ , $0\le u\le max\left\{2-t,t\right\}$ (see Exercise 18 from "Problems on Random Vectors and Joint Distributions").

$Z={I}_{M}\left(X,Y\right)\left(X+Y\right)+{I}_{{M}^{c}}\left(X,Y\right)2Y,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M=\left\{\left(t,u\right):max\left(t,u\right)\le 1\right\}$

Determine $P\left(Z\le 1\right)$ .

$P\left(Z\le 1\right)=P\left(\left(X,Y\right)\in {M}_{1}{Q}_{1}\bigvee {M}_{2}{Q}_{2}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{1}=\left\{\left(t,u\right):0\le t\le 1,\phantom{\rule{0.277778em}{0ex}}0\le u\le 1-t\right\}$
${M}_{2}=\left\{\left(t,u\right):1\le t\le 2,\phantom{\rule{0.277778em}{0ex}}0\le u\le t\right\}$
${Q}_{1}=\left\{\left(t,u\right):u\le 1-t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{Q}_{2}=\left\{\left(t,u\right):u\le 1/2\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(see}\phantom{\rule{4.pt}{0ex}}\text{figure)}$
$P=\frac{3}{23}{\int }_{0}^{1}{\int }_{0}^{1-t}\left(t+2u\right)\phantom{\rule{0.166667em}{0ex}}dudt+\frac{3}{23}{\int }_{1}^{2}{\int }_{0}^{1/2}\left(t+2u\right)\phantom{\rule{0.166667em}{0ex}}dudt=\frac{9}{46}$
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 300Enter number of Y approximation points 300 Enter expression for joint density (3/23)*(t + 2*u).*(u<=max(2-t,t)) Use array operations on X, Y, PX, PY, t, u, and PM = max(t,u)<= 1; G = M.*(t + u) + (1 - M)*2.*u;p = total((G<=1).*P) p = 0.1960 % Theoretical = 9/46 = 0.1957

${f}_{XY}\left(t,u\right)=\frac{12}{179}\left(3{t}^{2}+u\right)$ , for $0\le t\le 2$ , $0\le u\le min\left\{2,3-t\right\}$ (see Exercise 19 from "Problems on Random Vectors and Joint Distributions").

$Z={I}_{M}\left(X,Y\right)\left(X+Y\right)+{I}_{{M}^{c}}\left(X,Y\right)2{Y}^{2},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M=\left\{\left(t,u\right):t\le 1,\phantom{\rule{0.277778em}{0ex}}u\ge 1\right\}$

Determine $P\left(Z\le 2\right)$ .

$P\left(Z\le 2\right)=P\left(\left(,Y\right)\in {M}_{1}{Q}_{1}\bigvee \left({M}_{2}\bigvee {M}_{3}\right){Q}_{2}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{1}=\left\{\left(t,u\right):0\le t\le 1,\phantom{\rule{0.277778em}{0ex}}1\le u\le 2\right\}$
${M}_{2}=\left\{\left(t,u\right):0\le t\le 1,\phantom{\rule{0.277778em}{0ex}}0\le u\le 1\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{3}=\left\{\left(t,u\right):1\le t\le 2,\phantom{\rule{0.277778em}{0ex}}0\le u\le 3-t\right\}$
${Q}_{1}=\left\{\left(t,u\right):u\le 1-t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{Q}_{2}=\left\{\left(t,u\right):u\le 1/2\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(see}\phantom{\rule{4.pt}{0ex}}\text{figure)}$
$P=\frac{12}{179}{\int }_{0}^{1}{\int }_{0}^{2-t}\left(3{t}^{2}+u\right)\phantom{\rule{0.166667em}{0ex}}dudt+\frac{12}{179}{\int }_{1}^{2}{\int }_{0}^{1}\left(3{t}^{2}+u\right)\phantom{\rule{0.166667em}{0ex}}dudt=\frac{119}{179}$
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 300Enter number of Y approximation points 300 Enter expression for joint density (12/179)*(3*t.^2 + u).*(u<=min(2,3-t)) Use array operations on X, Y, PX, PY, t, u, and PM = (t<=1)&(u>=1); Z = M.*(t + u) + (1 - M)*2.*u.^2;G = M.*(t + u) + (1 - M)*2.*u.^2; p = total((G<=2).*P) p = 0.6662 % Theoretical = 119/179 = 0.6648

${f}_{XY}\left(t,u\right)=\frac{12}{227}\left(3t+2tu\right)$ , for $0\le t\le 2$ , $0\le u\le min\left\{1+t,2\right\}$ (see Exercise 20 from "Problems on Random Variables and Joint Distributions")

$Z={I}_{M}\left(X,Y\right)X+{I}_{{M}^{c}}\left(X,Y\right)\frac{Y}{X},\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}M=\left\{\left(t,u\right):u\le min\left(1,2-t\right)\right\}$

Detemine $P\left(Z\le 1\right)$ .

$P\left(Z\le 1\right)=P\left(\left(X,Y\right)\in {M}_{1}{Q}_{1}\bigvee {M}_{2}{Q}_{2}\right),\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{1}=M,\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{M}_{2}={M}^{c}$
${Q}_{1}=\left\{\left(t,u\right):0\le t\le 1\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}{Q}_{2}=\left\{\left(t,u\right):u\le t\right\}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\text{(see}\phantom{\rule{4.pt}{0ex}}\text{figure)}$
$P=\frac{12}{227}{\int }_{0}^{1}{\int }_{0}^{1}\left(3t+2tu\right)\phantom{\rule{0.166667em}{0ex}}dudt+\frac{12}{227}{\int }_{1}^{2}{\int }_{2-t}^{t}\left(3t+2tu\right)\phantom{\rule{0.166667em}{0ex}}dudt=\frac{124}{227}$
tuappr Enter matrix [a b]of X-range endpoints [0 2] Enter matrix [c d]of Y-range endpoints [0 2] Enter number of X approximation points 400Enter number of Y approximation points 400 Enter expression for joint density (12/227)*(3*t+2*t.*u).*(u<=min(1+t,2)) Use array operations on X, Y, PX, PY, t, u, and PQ = (u<=1).*(t<=1) + (t>1).*(u>=2-t).*(u<=t); P = total(Q.*P)P = 0.5478 % Theoretical = 124/227 = 0.5463

The class $\left\{X,\phantom{\rule{0.166667em}{0ex}}Y,\phantom{\rule{0.166667em}{0ex}}Z\right\}$ is independent.

$X=-2{I}_{A}+{I}_{B}+3{I}_{C}$ . Minterm probabilities are (in the usual order)

$0.255\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.025\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.375\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.045\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.108\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.012\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.162\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.018$

$Y={I}_{D}+3{I}_{E}+{I}_{F}-3$ . The class $\left\{D,\phantom{\rule{0.166667em}{0ex}}E,\phantom{\rule{0.166667em}{0ex}}F\right\}$ is independent with

$P\left(D\right)=0.32\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(E\right)=0.56\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}P\left(F\right)=0.40$

Z has distribution

 Value -1.3 1.2 2.7 3.4 5.8 Probability 0.12 0.24 0.43 0.13 0.08

Determine $P\left({X}^{2}+3X{Y}^{2}>3Z\right)$ .

% file npr10_16.m Data for [link] cx = [-2 1 3 0];pmx = 0.001*[255 25 375 45 108 12 162 18];cy = [1 3 1 -3];pmy = minprob(0.01*[32 56 40]);Z = [-1.3 1.2 2.7 3.4 5.8];PZ = 0.01*[12 24 43 13 8];disp('Data are in cx, pmx, cy, pmy, Z, PZ') npr10_16 % Call for dataData are in cx, pmx, cy, pmy, Z, PZ [X,PX]= canonicf(cx,pmx); [Y,PY]= canonicf(cy,pmy); icalc3Enter row matrix of X-values X Enter row matrix of Y-values YEnter row matrix of Z-values Z Enter X probabilities PXEnter Y probabilities PY Enter Z probabilities PZUse array operations on matrices X, Y, Z, PX, PY, PZ, t, u, v, and PM = t.^2 + 3*t.*u.^2>3*v; PM = total(M.*P)PM = 0.3587

The simple random variable X has distribution

$X=\left[-3.1\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}-0.5\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}1.2\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}2.4\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}3.7\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}4.9\right]\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}PX=\left[0.15\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.22\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.33\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.12\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.11\phantom{\rule{0.277778em}{0ex}}\phantom{\rule{0.277778em}{0ex}}0.07\right]$
1. Plot the distribution function F X and the quantile function Q X .
2. Take a random sample of size $n=10,000$ . Compare the relative frequency for each value with the probability that value is taken on.
X = [-3.1 -0.5 1.2 2.4 3.7 4.9];PX = 0.01*[15 22 33 12 11 7];ddbn Enter row matrix of VALUES XEnter row matrix of PROBABILITIES PX % Plot not reproduced here dquanplotEnter VALUES for X X Enter PROBABILITIES for X PX % Plot not reproduced hererand('seed',0) % Reset random number generator dsample % for comparison purposesEnter row matrix of VALUES X Enter row matrix of PROBABILITIES PXSample size n 10000 Value Prob Rel freq-3.1000 0.1500 0.1490 -0.5000 0.2200 0.21641.2000 0.3300 0.3340 2.4000 0.1200 0.11843.7000 0.1100 0.1070 4.9000 0.0700 0.0752Sample average ex = 0.8792 Population mean E[X]= 0.859 Sample variance vx = 5.146Population variance Var[X] = 5.112

what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
how did you get the value of 2000N.What calculations are needed to arrive at it
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A fair die is tossed 180 times. Find the probability P that the face 6 will appear between 29 and 32 times inclusive By By Sam Luong    By    