Data 605 Assignment 5

library(knitr)
library(rmdformats)

## Global options
options(max.print="31")
opts_chunk$set(echo=FALSE,
                 cache=TRUE,
               prompt=FALSE,
               tidy=TRUE,
               comment=NA,
               message=FALSE,
               warning=FALSE)
opts_knit$set(width=31)

Question 0

Choose independently two numbers B and C at random from the interval [0, 1] with uniform density. Prove that B and C are proper probability distributions.

Ans:
To prove B is a proper probability distribution, it suffies to show that its (empirical) CDF is 1 and comprises of individual pdf’s that are between 0 and 1. Here goes the checks for B. 

Empirical CDF:    100000 unique values with summary
     Min.   1st Qu.    Median      Mean   3rd Qu.      Max. 
0.0000166 0.2494526 0.4988924 0.4990694 0.7485272 0.9999714

Same checks go for C. 

Empirical CDF:    99999 unique values with summary
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.000004 0.249426 0.499106 0.500159 0.749656 0.999999

Both results checked out.

Find the probability that

(a) B + C < 1/2

Let B = x and C = y.

f(B,C) = f(x,y)

P(B+C > 1/2) = P(x + y < 1/2)

y < 1/2 - x

Now we integrate the function \[\begin{multline*} \begin{split} \int_{0}^{1/2}\int_{0}^{(1/2)-x} f(x,y) dxdy \\ &= \int_{0}^{1/2}\int_{0}^{(1/2)-x} 1 dxdy \\ &= \int_{0}^{1/2} ((\frac{1}{2}-x ) - 0) dy \\ &= ((\frac{1}{2}) (\frac{1}{2}) - \frac{(\frac{1}{2})^2}{2} ) - (0 - 0) \\ &= \frac{1}{4} - \frac{1}{8} \\ &= \frac{1}{8} \end{split} \end{multline*}\]

Therefore, the probability that B+C < 1/2 = 1/8 or 12.5%

(b) BC < 1/2

B = 1 and C = 1/2. Let B = x, C = y. Then P(BC < 1/2) = P(xy < 1/2).

y < 1/2x.

\[\begin{multline*} \begin{split} \int_{0}^{1/2}\int_{0}^{1} \frac{1}{2} dxdy + \int_{1/2}^{1}\int_{0}^{1/2x} f(x,y) dxdy \\ &= \int_{0}^{1/2}\frac{y}{2}dy + \int_{1/2}^{1}\int_{0}^{1/2x} 1 dxdy \\ &= \frac{1}{2} + \int_{1/2}^{1}(\frac{1}{2y} - 0) dy \\ &= \frac{1}{2} + \frac{1}{2}( \ln(1) - \ln\frac{1}{2})\\ &= \frac{1}{2} + \frac{1}{2}( \ln\frac{1}{2})\\ &\approx .85 \end{split} \end{multline*}\]

Therefore, the probability that BC < 1/2 \(\cong\) 0.85 or 85%

(c) |B - C| < 1/2

Let B = x, C = y. Then P(|B - C| < 1/2) = P(|x - y| < 1/2).

part c illustration

To find the area shaded in violet, we just need to get the area of the square and subtract the area of the 2 triangles from it.

\[\begin{multline*} \begin{split} &= 1 - \frac{(\frac{1}{2})^2}{2} * 2 \\ &= \frac{3}{4} \end{split} \end{multline*}\]

Therefore, the probability that |B-C| < 1/2 = .75 or 75%

(d) max{B,C} < 1/2

Let B = x, C = y.

P(B < 1/2, C < 1/2)

= P(B < 1/2)*P(C < 1/2)

\(= (\frac{1}{2})^2\)

\(= \frac{1}{4}\)

Therefore, the probability that max{B,C} < 1/2 = .25 or 25%

(e) min{B,C} < 1/2

Let B = x, C = y.

P(B >= 1/2, C >= 1/2)

= 1 - P(B < 1/2, C < 1/2)

\(= 1 - \frac{1}{4}\)

\(= \frac{3}{4}\)

Therefore, the probability that min{B,C} < 1/2 = .75 or 75%