Data 602 Week 5 Homework
Brian K. Liles
September 30, 2018
Library
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## df, qqplot, sdChoose indepenently two numbers B and C from the interval [0,1] with uniform density. Prove that B and c are proper probability distributions. Note that point (B,C) is then chosen at random in the unit square.
(A) Find the probability that B+C < 1/2
First, we create 10,000 random numbers from 0 to 1 and assign them to variables BA & CA respectively; the A suffix corresponds to question A.
BA <- runif(10000, min = 0, max = 1)
CA <- runif(10000, min = 0, max = 1)Next, we create a variable ZA which will host the sum of the variables BA and CA
ZA <- BA + CATo provide a visual we will create a tibble that will showcase the variables created
(ZAT <- tibble(BA,CA,ZA))## # A tibble: 10,000 x 3
## BA CA ZA
## <dbl> <dbl> <dbl>
## 1 0.728 0.462 1.19
## 2 0.170 0.839 1.01
## 3 0.660 0.766 1.43
## 4 0.688 0.692 1.38
## 5 0.189 0.640 0.829
## 6 0.556 0.852 1.41
## 7 0.372 0.752 1.12
## 8 0.136 0.803 0.939
## 9 0.876 0.273 1.15
## 10 0.112 0.0375 0.150
## # ... with 9,990 more rowsWe can also see that there is no variables in ZAT that are below 0 by running a summary.
ZAT %>% summary()## BA CA ZA
## Min. :0.0000804 Min. :0.0000112 Min. :0.005683
## 1st Qu.:0.2484283 1st Qu.:0.2500285 1st Qu.:0.707749
## Median :0.4987503 Median :0.5057283 Median :1.002737
## Mean :0.4995242 Mean :0.5021684 Mean :1.001693
## 3rd Qu.:0.7508814 3rd Qu.:0.7569676 3rd Qu.:1.293915
## Max. :0.9999382 Max. :0.9999994 Max. :1.997156In order to visualize the ZA variable we will plot its findings on a histogram.
hist(ZA, xlab="ZA",col="beige",breaks=10,probability = TRUE,ylim=c(0,1.0))
Using the integrate function we set the upper limits to 0.5 to obtain the probability of A
f <- function(x){x}
PA <- integrate(f, lower = 0, upper= 0.5)
PA## 0.125 with absolute error < 1.4e-15(B) Find the probability that BC < 1/2
First, we create 10,000 random numbers from 0 to 1 and assign them to variables BB & CB respectively.
BB <- runif(10000, min = 0, max = 1)
CB <- runif(10000, min = 0, max = 1)Next, we create a variable ZB which will host the product of the variables BB and CB
ZB <- BB * CBTo provide a visual we will create a tibble that will showcase the variables created
(ZBT <- tibble(BB,CB,ZB))## # A tibble: 10,000 x 3
## BB CB ZB
## <dbl> <dbl> <dbl>
## 1 0.383 0.949 0.364
## 2 0.799 0.825 0.659
## 3 0.585 0.0307 0.0179
## 4 0.309 0.923 0.286
## 5 0.515 0.372 0.191
## 6 0.0794 0.679 0.0539
## 7 0.0911 0.237 0.0216
## 8 0.908 0.385 0.349
## 9 0.593 0.811 0.481
## 10 0.651 0.258 0.168
## # ... with 9,990 more rowsWe can also see that there is no variables in ZBT that are below 0 by running a summary.
ZBT %>% summary()## BB CB ZB
## Min. :0.0000421 Min. :0.0001521 Min. :0.0000025
## 1st Qu.:0.2535308 1st Qu.:0.2475493 1st Qu.:0.0696264
## Median :0.5065082 Median :0.4964271 Median :0.1894975
## Mean :0.5019821 Mean :0.4987519 Mean :0.2507308
## 3rd Qu.:0.7508229 3rd Qu.:0.7515797 3rd Qu.:0.3849211
## Max. :0.9997904 Max. :0.9998306 Max. :0.9812452In order to visualize the ZB we will plot its findings on a histogram as we did with ZA
hist(ZB, xlab="ZB",col="purple",breaks=10,probability = TRUE,ylim=c(0,3.5))
As we can see the output is skewed to the left and don’t posses a symmetric look as ZA
Calling the lower limit of intergration of 0 and the upper limit as 0.5 based off the anonymous function.
f1 <- function(x){x}
PB <- integrate(f1, lower = 0, upper = 0.5)
PB## 0.125 with absolute error < 1.4e-15(C) Find the probability that |B-C| < 1/2
First, we create 10,000 random numbers from 0 to 1 and assign them to variables BC & CC respectively.
BC <- runif(10000, min = 0, max = 1)
CC <- runif(10000, min = 0, max = 1)Next, we create a variable ZC which will host the absolute difference of the variables BC and CC
ZC <- abs(BC * CC)To provide a visual we will create a tibble that will showcase the variables created
(ZCT <- tibble(BC,CC,ZC))## # A tibble: 10,000 x 3
## BC CC ZC
## <dbl> <dbl> <dbl>
## 1 0.216 0.229 0.0494
## 2 0.0116 0.306 0.00356
## 3 0.335 0.554 0.186
## 4 0.454 0.225 0.102
## 5 0.571 0.256 0.146
## 6 0.466 0.0412 0.0192
## 7 0.0375 0.867 0.0326
## 8 0.535 0.825 0.441
## 9 0.969 0.378 0.366
## 10 0.407 0.716 0.292
## # ... with 9,990 more rowsWe can also see that there is no variables in ZCT that are below 0 by running a summary.
ZCT %>% summary()## BC CC ZC
## Min. :0.0000565 Min. :0.0000532 Min. :0.0000152
## 1st Qu.:0.2519704 1st Qu.:0.2541732 1st Qu.:0.0683726
## Median :0.5051390 Median :0.5005201 Median :0.1921402
## Mean :0.5023966 Mean :0.5000019 Mean :0.2509183
## 3rd Qu.:0.7523445 3rd Qu.:0.7418004 3rd Qu.:0.3815652
## Max. :0.9998818 Max. :0.9999990 Max. :0.9851663In order to visualize the ZC we will plot its findings on a histogram.
hist(ZC, xlab="ZC",col="green",breaks=10,probability = TRUE,ylim=c(0,3.5))
f3 <- function(x){2-2*x}
PC <- integrate(f3, lower = 0, upper = 0.5)
PC## 0.75 with absolute error < 8.3e-15(D) Find the probability that max{B,C} < 1/2
First, we create 10,000 random numbers from 0 to 1 and assign them to variables BD & CD respectively.
BD <- runif(10000, min = 0, max = 1)
CD <- runif(10000, min = 0, max = 1)Next, we create a variable ZD which will host the maximum values of the variables BD and CD
ZD <- pmax(BD,CD)To provide a visual we will create a tibble that will showcase the variables created
(ZDT <- tibble(BD,CD,ZD))## # A tibble: 10,000 x 3
## BD CD ZD
## <dbl> <dbl> <dbl>
## 1 0.329 0.633 0.633
## 2 0.550 0.413 0.550
## 3 0.157 0.958 0.958
## 4 0.0391 0.860 0.860
## 5 0.384 0.228 0.384
## 6 0.990 0.910 0.990
## 7 0.0567 0.661 0.661
## 8 0.222 0.625 0.625
## 9 0.996 0.602 0.996
## 10 0.170 0.245 0.245
## # ... with 9,990 more rowsIn order to visualize the ZD we will plot its findings on a histogram.
hist(ZD, xlab="ZD",col="red",breaks=10,probability = TRUE,ylim=c(0,2.5))
f4 <- function(x){2*x}
PD <- integrate(f4, lower = 0, upper = 0.5)
PD## 0.25 with absolute error < 2.8e-15(E) Find the probability that min{B,C} < 1/2
First, we create 10,000 random numbers from 0 to 1 and assign them to variables BE & CE respectively.
BE <- runif(10000, min = 0, max = 1)
CE <- runif(10000, min = 0, max = 1)Next, we create a variable ZE which will host the maximum values of the variables BE and CE
ZE <- pmin(BE,CE)To provide a visual we will create a tibble that will showcase the variables created
(ZET <- tibble(BE,CE,ZE))## # A tibble: 10,000 x 3
## BE CE ZE
## <dbl> <dbl> <dbl>
## 1 0.765 0.745 0.745
## 2 0.883 0.851 0.851
## 3 0.0838 0.265 0.0838
## 4 0.584 0.352 0.352
## 5 0.978 0.625 0.625
## 6 0.725 0.886 0.725
## 7 0.720 0.170 0.170
## 8 0.856 0.151 0.151
## 9 0.985 0.648 0.648
## 10 0.950 0.567 0.567
## # ... with 9,990 more rowsIn order to visualize the ZE we will plot its findings on a histogram.
hist(ZE, xlab="ZE",col="yellow",breaks=10,probability = TRUE,ylim=c(0,2.5))
f5 <- function(x){2-2*x}
PE <- integrate(f5, lower = 0, upper = 0.5)
PE## 0.75 with absolute error < 8.3e-15