1-(.28+.29+.19-.14-.12-.1+.08)

## [1] 0.52

.3-((1-.35)-.4)

## [1] 0.05

.9-(1-.7)

## [1] 0.6

4/10*16/(16+c(4,20,24,44,64))+6/10*c(4,20,24,44,64)/(16+c(4,20,24,44,64))

## [1] 0.4400000 0.5111111 0.5200000 0.5466667 0.5600000

3000-600-(1320-600)-(1400-600)

## [1] 880

(210-102)/(937-312)

## [1] 0.1728

.4*(.65-.15)+.6*(.5-.15)+.8*.15

## [1] 0.53

((1-.12)+.22-.14)/2

## [1] 0.48

1-(.7+(.2-.7*.15))

## [1] 0.205

1-(3*sqrt(.15/2)-.15)

## [1] 0.3284162

1-((.14-.14/3)+.14/3+(.15-.14/3-.64/8)+.64/8+(.64-.64/8))

## [1] 0.1966667

(1-(.1*3+.12*3+(.04*3/2)))/(1-(.1+.12*2+(.04*3/2)))

## [1] 0.4666667

1-(1/(1/(1-.2))+1/(1/(1-.2))*.2)

## [1] 0.04

1-(1/4+1/3+5/12)/2

## [1] 0.5

library(MASS)
fractions((1/2^8*1/2+1/2^7*1/2^2+1/2^6*1/2^3+1/2^5*1/2^4)*(2))

## [1] 1/64

.1/(1-(.85-(.25-(1-.85))))

## [1] 0.4

pnorm(.005/sqrt(.0056^2+.0044^2)*2,0,1)-
  (1-pnorm(.005/sqrt(.0056^2+.0044^2)*2,0,1))

## [1] 0.8397228

.06*.08/(.06*.08+.03*.15+.02*.49+.04*.28)

## [1] 0.1584158

.1*.001/(.5*.01+.4*.005+.1*.001)

## [1] 0.01408451

.3*.9/(.1*.6+.3*.9+.6*.99)

## [1] 0.2922078

4*.2/(4*.2+2*.3+.5)

## [1] 0.4210526

.16*.08/(.08*.15+.16*.08+.45*.04+.31*.05)

## [1] 0.219554

(1/6+1/12+1/20+1/30)/(1/2+1/6+1/12+1/20+1/30)

## [1] 0.4

.95*.01/(.95*.01+.005*.99)

## [1] 0.6574394

2*.25/(2*.25+.75)

## [1] 0.4

.16*.05/(.16*.05+.18*.02+.2*.03)

## [1] 0.4545455

dbinom(1,30,.1)*1/5/(dbinom(1,30,.1)*1/5+dbinom(1,30,.02)*4/5)

## [1] 0.0956576

pexp(80,-log(.7)/50)

## [1] 0.4348589

dpois(2,c(1,2,3,4))/dpois(4,c(1,2,3,4))

## [1] 12.000000  3.000000  1.333333  0.750000

120/qbinom(.99,20,.02)

## [1] 60

dmultinom(c(0,2,2),4,c(.5,.3,.2))+dmultinom(c(0,1,3),4,c(.5,.3,.2))+
  dmultinom(c(0,0,4),4,c(.5,.3,.2))+dmultinom(c(1,0,3),4,c(.5,.3,.2))

## [1] 0.0488

library(triangle)
fractions(ptriangle(20-16,0,20,20)/ptriangle(20-8,0,20,20))

## [1] 1/9

library(actuar)
5/4*ppareto(6,1,10)

## [1] 0.46875

pbeta(.4,1,5,lower.tail = FALSE)/pbeta(.1,1,5,lower.tail = FALSE)

## [1] 0.1316872

100*(200*pexp(1,1/2)+100*(pexp(2,1/2)-pexp(1,1/2)))

## [1] 10255.9

(ppareto1(2,3,1)-ppareto1(1.5,3,1))/ppareto1(1.5,3,1,FALSE)

## [1] 0.578125

pbinom(2,20,.05) 

## [1] 0.9245163

qbeta(.64,2,1)-.5

## [1] 0.3

pbinom(8,10,.8,FALSE)*pbinom(8,10,.8)*2 

## [1] 0.4691535

.6*.3+.4*.3*pnorm(-1000/sqrt(2000^2+2000^2),0,1)

## [1] 0.2234204

#mean(rcompound(1000000, rbinom(1,.3), rnorm(9000, 2000))-
#       rcompound(1000000, rbinom(1,.4), rnorm(10000, 2000))>0)

pnbinom(3,4,.4)

## [1] 0.289792

5/15*100+4/15*200+3/15*300+2/15*350+1/15*400

## [1] 220

\[E(X)=\int_{-2}^0x\frac{-x}{10} \,\ dx+\int_0^4x\frac{x}{10}=-\frac{x^3}{30}\Bigg|_{-2}^0+\frac{x^3}{30} \Bigg|_0^4=\]

fractions(-0^3/30+(-2)^3/30+4^3/30-0^3/30)

## [1] 28/15

pexp(2,1/3)*2+pexp(2,1/3,FALSE)*(2+mexp(1,1/3))

## [1] 3.540251

1000/(pexp(1,1/10)+.5*(pexp(3,1/10)-pexp(1,1/10)))

## [1] 5644.227

4000*.4+3000*.6*.4+2000*.6*.6*.4+1000*.6*.6*.6*.4 

## [1] 2694.4

10000*((1.5-1)-(0-1)*dpois(0,1.5))

## [1] 7231.302

levpareto1(limit = 2,2.5,.6,order = 1) 

## [1] 0.9342733

k=1/(1+1/2+1/3+1/4+1/5);.05*(k/3+k/4*2+k/5*3) 

## [1] 0.03138686

#per=rbinom(1000000,1,.05)*sample(1:5,1000000,TRUE,c(k,k/2,k/3,k/4,k/5))-2
#mean(per[1>per]=0)

levpareto1(10,2,1,1) 

## [1] 1.9

.02*(15-1)+.04*(.5003*2)*(mexp(1,1/2)-levexp(1,1/2,1)-(mexp(1,1/2)-levexp(15,1/2,1)))

## [1] 0.3285073

mpareto(1,3,1)

## [1] 0.5

(munif(1,0,1000)-levunif(c(250,375,500,625,750),0,1000,1))/munif(1,0,1000)

## [1] 0.562500 0.390625 0.250000 0.140625 0.062500

sqrt(mgamma(2,4,1/2500)-mgamma(1,4,1/2500)^2) 

## [1] 5000

mgamma(3,10,1/2) 

## [1] 10560

qpareto1(.7,2.5,200)-qpareto1(.3,2.5,200) 

## [1] 93.05909

\[Var(1.2C)=(1.2)^2Var(C)=1.44(260)=\]

(1.2)^2*260 

## [1] 374.4

fractions(.5*(1)^2+.5*((1^2+2^2+2^2-1*2-1*2-2*2)/18+((1+2+2)/3)^2)-
            (.5*(1)+.5*(1+2+2)/3)^2)

## [1] 5/36

levunif(4,0,5,2)-levunif(4,0,5,1)^2

## [1] 1.706667

(20*.15+30*.1+40*.05+50*.2+60*.1+70*.1+80*.3)+c(
  sqrt(20^2*.15+30^2*.1+40^2*.05+50^2*.2+60^2*.1+70^2*.1+80^2*.3-
         (20*.15+30*.1+40*.05+50*.2+60*.1+70*.1+80*.3)^2),
  -sqrt(20^2*.15+30^2*.1+40^2*.05+50^2*.2+60^2*.1+70^2*.1+80^2*.3-
         (20*.15+30*.1+40*.05+50*.2+60*.1+70*.1+80*.3)^2))

## [1] 76.79449 33.20551

Range includes the values \(40\), \(50\), \(60\) and \(70\) with probability \(0.05+0.2+0.1+0.1=0.45\)

sqrt(munif(2,0,1500)-levunif(250,0,1500,2)-
       2*250*munif(1,0,1500)+2*250*levunif(250,0,1500,1)-
       (munif(1,0,1500)-levunif(250,0,1500,1))^2)

## [1] 403.4358

1000*sqrt((dpois(1,.6)+ppois(1,.6,FALSE)*4)
          -((dpois(1,.6)+ppois(1,.6,FALSE)*2))^2) 

## [1] 698.8996

if (250>qexp(.5,.004)) {qexp(.5,.004)} else 250 

## [1] 173.2868

pexp(5,-log(.5)/4,FALSE) 

## [1] 0.4204482

By memorylessness,

qexp(.95,1/300)+100

## [1] 998.7197

\[F(y)=P(Y\leq y)=P\Big(T^2\leq y\Big)=P\Big(T\leq \sqrt{y}\Big)=F(\sqrt{y})\] \[\Rightarrow g(y)=F'(y)=F'\Big(\sqrt{y}\Big)=f\Big(\sqrt{y}\Big)\Big(\sqrt{y}\Big)'=\frac{8}{y^{3/2}}\cdot\frac{1}{2y^{1/2}}=\frac{4}{y^2}\]

\[F(v)=P(V\leq v)=P\Big(10000e^R\leq v\Big)=P\Big(R\leq \log{\frac{v}{10000}}\Big)\] \[\Rightarrow F\Big(\log{\frac{v}{10000}}\Big)= \frac{\log{\frac{v}{10000}}-.04}{.08-.04}=25\Big(\log{\frac{v}{10000}}-0.04\Big)\]

\[F(y)=P\Big(10X^{0.8}\leq y\Big)=P\Big[X\leq \Big(\frac{y}{10}\Big)^{1.25} \Big]\] \[\Rightarrow f(y)=F'(y)\Big(\Big(\frac{y}{10}\Big)^{1.25}\Big)'=.125\Big(\frac{y}{10}\Big)^{0.25}\exp{\Big[-\Big(\frac{y}{10}\Big)^{1.25}\Big]}\]

\[F(r)=P(R\leq r)=P(10/T\leq r)=P(10/r\leq T)=S_T\Big(\frac{10}{r}\Big)\] \[\Rightarrow f(r)=S_T'\Big(\frac{10}{r}\Big) \Big(\frac{10}{r}\Big)'=-\frac{1}{4}\Big(-\frac{10}{r^2}\Big)=\frac{5}{2r^2}\]

\[F_{II}(y)=P(2I\leq y)=P(I\leq y/2)=F_{I}\Big(\frac{y}{2}\Big)\] \[\Rightarrow g(y)=F_{II}'(y)=F_{I}'\Big(\frac{y}{2}\Big) \Big(\frac{y}{2}\Big)'=f_{I}\Big(\frac{y}{2}\Big) \Big(\frac{1}{2}\Big)\]

1000*(mpareto1(1,3,1)*3-3*mpareto1(1,6,1)+mpareto1(1,9,1))

## [1] 2025

\[1-P(T_{XY}\geq 1)=1-\int_1^2\int_1^2 \frac{x+y}{8} \,\ dxdy=1-\int_1^2 \Big[\frac{x^2}{16}+\frac{xy}{8} \Big]_1^2dy=\] \[1-\int_1^2 \Big(\frac{2^2}{16}-\frac{1^2}{16}+\frac{2y}{8}-\frac{y}{8}\Big)dy = 1-\Big[\frac{3}{16}y+\frac{1}{16}y^2\Big]_1^2=\]

1-(6/16-3/16+4/16-1/16)

## [1] 0.625

\[1-P(T_{ST}\geq 0.5)=1-\int_{0.5}^1\int_{0.5}^1 f(s,t) \,\ dsdt=\int_0^{0.5} \int_{0.5}^1 f(s,t) \,\ dsdt + \int_0^1\int_0^{0.5} f(s,t) \,\ dsdt\]

qnorm(.9,0,1)*250*sqrt(2025)+2025*3125

## [1] 6342542

pnorm((20000-19400)/5000*sqrt(25),0,1,FALSE) 

## [1] 0.2742531

pnorm((2600-2500)/sqrt(2)/sqrt(1250))-pnorm((2450-2500)/sqrt(2)/sqrt(1250))

## [1] 0.8185946

qnorm(.9772,0,1,FALSE)*sqrt(c(14,16,20,45,55)*1)+3*c(14,16,20,45,55)

## [1]  34.52014  40.00369  51.05985 121.58978 150.17445

pnorm((7100-(50+20)*100)/sqrt(100*(50+30+2*10)),0,1)

## [1] 0.8413447

pnorm((110000-100*1000)/sqrt(100*1000^2),0,1,FALSE)

## [1] 0.1586553

e1=.6*0+.4*.25*1+.4*.75*2;v1=.6*0^2+.4*.25*1^2+.4*.75*2^2-(e1)^2
pnorm((90+.5-100*e1)/sqrt(100*v1),0,1)

## [1] 0.9886301

pnorm(.25/sqrt(25/12/48),0,1)-pnorm(-.25/sqrt(25/12/48),0,1)

## [1] 0.7698607

\[P(G\leq 3)P(B\leq 2)=\Big(1-e^{-1/2}\Big)\Big(1-e^{-2/3}\Big)=1-e^{-1/2}-e^{-2/3}+e^{-7/6}\]

The region of integration goes from \(20\) to \(30\) in \(x\) and from \(20\) to \(50-x\) in \(y\), therefore the probability is

\[\frac{6}{125000}\int_{20}^{30}\int_{20}^{50-x}(50-x-y)dydx\]

\[P(B>D)=\int_0^{\infty}f_D(x)S_B(x)dx=\int_0^{\infty}\frac{e^{-x/3}}{3}e^{-x/2}dx=\int_0^{\infty}\frac{e^{-5x/6}}{3}dx=\] \[\frac{1}{3}\frac{6}{5}\int_0^{\infty}\frac{5e^{-5x/6}}{6}dx=\]

1/3*6/5*1

## [1] 0.4

\[P(T_{XY}\geq 1)=\int_0^1\int_{1-x}^2\frac{2x+2-y}{4}dydx=\int_0^1\Big[\frac{2xy+2y-0.5y^2}{4}\Big]_{1-x}^2dx=\] \[\int_0^1\Big(\frac{4x+4-2}{4}-\frac{2x(1-x)+2(1-x)-0.5(1-x)^2}{4}\Big)dx=\int_0^1\frac{3x+0.5+2.5x^2}{4}dx=\] \[\Big[\frac{1.5x^2+0.5x+2.5x^3/3}{4}\Big]_0^1=\]

(1.5+0.5+2.5/3)/4-0

## [1] 0.7083333

92. The region of integration in thousands is

The required probability is

1-(180^2/2+180^2/2)/200^2

## [1] 0.19

93. The region is

, with probability

8*8/2/10^2-(1*1/2+2*2/2)/10^2

## [1] 0.295

The region is

and by symmetry

\[E(T_1+T_2)=2E(T_1)=\frac{2}{(4)(6)+(2)(4)+(2)(2)/2}\Big(\int_0^4\int_0^6t_1dt_2dt_1+\int_4^6\int_0^{10-t_1}t_1dt_2dt_1\Big)=\] \[\frac{2}{34}\Big(\int_0^4t_1t_2\Big|_0^6dt_2+\int_4^6t_1t_2\Big|_0^{10-t_1}dt_1\Big)=\frac{2}{34}\Big(\int_0^46t_1 \,\ dt_1+\int_4^6t_1(10-t_1)dt_1\Big)=\] \[\frac{2}{34}\Big(6\Big[\frac{t_1^2}{2}\Big]_0^4+\Big[5t_1^2-\frac{t_1^3}{3}\Big]_4^6\Big)=\]

2/34*(6*4^2/2+5*6^2-6^3/3-5*4^2+4^3/3)

## [1] 5.72549

\[X,Y\sim N.-0,1 \Rightarrow W,Z\sim N.-0,2 \Rightarrow E\Big(e^{t_1W+t_2Z}\Big)=E\Big(e^{t_1W}\Big)E\Big(e^{t_2Z}\Big)=e^{t_1^2}e^{t_2^2}=e^{t_1^2+t_2^2}\]

50*21*pbinom(20,21,.98)+(21*50-1*100)*pbinom(20,21,.98,FALSE)

## [1] 984.5744

\[E(T_1^2+T_2^2)=\frac{2}{L^2}\int_0^L\int_0^{t_1}(t_1^2+t_2^2)dt_2dt_1=\frac{2}{L^2}\int_0^L\Big[(t_1^2t_2+\frac{t_2^3}{3}\Big]_0^{t_1}dt_1=\] \[\frac{2}{L^2}\int_0^L\Big(t_1^3+\frac{t_1^3}{3}\Big)dt_1=\frac{2}{L^2}\frac{t_1^4}{3}\Big|_0^{L}=\frac{2}{3}L^2\]

\[E\Big(e^{tY}\Big)=2^3/3^3e^t+1-2^3/3^3=19/27+8/27e^t\]

5000+1.1^2*10000+2*1.1*(17000-10000-5000)/2

## [1] 19300

0^2*1/6+1^2*3/12+2^2*7/12-(0^1*1/6+1^1*3/12+2^1*7/12)^2

## [1] 0.5763889

3^2*1+(-1)^2*2

## [1] 11

2*(mexp(2,1/10)-mexp(1,1/10)^2)

## [1] 200

1-pexp(3,1/1)*pexp(3,1/1.5)*pexp(3,1/2.4)

## [1] 0.4137812

Since te region is rectangular,

\[f(x,y)=f(x)f(y) \Rightarrow E(X,Y)=E(X)E(Y) \Rightarrow Cov(X,Y)=0\]

Since the region is not rectangular,

\[E(X)=\int_0^1\int_x^{2x}\frac{8}{3}x^2ydydx=\frac{8}{3}\int_0^1\frac{x^2y^2}{2}\Big|_x^{2x}dx=\frac{8}{3}\int_0^1\frac{4x^4-x^4}{2}dx=\frac{8}{3}\Big[\frac{3x^5}{10}\Big]_0^1=\frac{4}{5}\] \[E(Y)=\int_0^1\int_x^{2x}\frac{8}{3}xy^2dydx=\frac{8}{3}\int_0^1\frac{xy^3}{3}\Big|_x^{2x}dx=\frac{8}{3}\int_0^1\frac{8x^4-x^4}{3}dx=\frac{8}{3}\Big[\frac{7x^4}{15}\Big]_0^1=\frac{56}{45}\] \[E(XY)=\int_0^1\int_x^{2x}\frac{8}{3}x^2y^2dydx=\frac{8}{3}\int_0^1\frac{x^2y^3}{3}\Big|_x^{2x}dx=\frac{8}{3}\int_0^1\frac{8x^5-x^5}{3}dx=\frac{8}{3}\Big[\frac{7x^5}{18}\Big]_0^1=\frac{28}{27}\] \[\Rightarrow Cov(X,Y)=\frac{28}{27}-\frac{4}{5}\frac{56}{45}=0.04148148\]

\[E(X)=\int_0^{12}\int_0^xx\frac{1}{12}\frac{1}{x}dydx=\frac{1}{12}\int_0^{12}y\Big|_0^xdx=\frac{1}{12}\int_0^{12}xdx=\frac{1}{12}\frac{x^2}{2}\Big|_0^{12}=6\] \[E(Y)=\int_0^{12}\int_0^xy\frac{1}{12}\frac{1}{x}dydx=\frac{1}{12}\int_0^{12}\frac{y^2}{2x}\Big|_0^xdx=\frac{1}{12}\int_0^{12}\frac{x}{2}dx=\frac{1}{12}\frac{x^2}{4}\Big|_0^{12}=3\] \[E(XY)=\int_0^{12}\int_0^xxy\frac{1}{12}\frac{1}{x}dydx=\frac{1}{12}\int_0^{12}\frac{y^2}{2}\Big|_0^xdx=\frac{1}{12}\int_0^{12}\frac{x^2}{2}dx=\frac{1}{12}\frac{x^3}{6}\Big|_0^{12}=24\] \[\Rightarrow Cov(X,Y)=\]

24-6*3

## [1] 6

(27.4-5^2)+1.2*(51.4-7^2)+2.2*(8-(27.4-5^2)-(51.4-7^2))/2

## [1] 8.8

\[P(2T_1+T_2\leq x)=\int_0^x\int_0^{(x-t_2)/2}e^{-t_1}e^{-t_2}dt_1dt_2=\int_0^x\Big[-e^{-t_1}e^{-t_2}\Big]_0^{(x-t_2)/2}dt_2=\] \[\int_0^xe^{-t_2}(1-e^{-(x-t_2)/2})dt_2=\Big[-e^{-t_2}+2e^{-x/2}e^{-t_2/2}\Big]_0^x=e^{-x}+1-2e^{-x/2}\] \[\Rightarrow g(x)=(e^{-x}+1-2e^{-x/2})'=-e^{-x}+e^{-x/2}\]

\[F(x)=\int_0^{\infty}\int_0^{px}\frac{e^{-p/2}}{2}e^{-c}dcdp=\int_0^{\infty}\Big(\frac{e^{-p/2}}{2}-\frac{e^{-p/2}}{2}e^{-px}\Big)dp=\] \[\Bigg[-e^{-p/2}+\frac{e^{-p(x+1/2)}}{2x+1}\Bigg]_0^{\infty}=1-\frac{1}{2x+1} \Rightarrow f(x)=\frac{2}{(2x+1)^2}\]