4 2017-Spring

4.1 Paper

4.2 Problem 1

Let the three points be \((x_1,y_1); (x_2,y_2); (x_3,y_3)\) Expected area of the rectangle with sides parallel to the coordinate axes is: \(E[\big(\max{(x_1,x_2,x_3)}-\min{(x_1,x_2,x_3)}\big)\big(\max{(y_1,y_2,y_3)}-\min{(y_1,y_2,y_3)}\big)]\) where \(x_i ~ U(0,1)\), \(y_i ~ U(0,1)\) and \(x_i, y_i\) are independent.

Thus, expected area can be simplified to \(E[\big(\max{(x_1,x_2,x_3)}-\min{(x_1,x_2,x_3)}\big)]^2\)

Let \(\max{(x_1,x_2,x_3)\) be represents as \(x_{(1)}\) and \(\min{(x_1,x_2,x_3)\) as \(x_{(3)}\) Then \(P_{X_{(1)}}(X_{(1)} < x) = P(X_{1}<x)P(X_{2}<x)P(X_{3}<x) = x^3 \implies E[X_{(1)}]=\frac{3}{4}\)


\(P_{X_{(3)}}(X_{(3)} < x) = 1-(1-x)^3 \implies P_{X_{(3)}}(X_{(3)} = x) = 3(1-x)^2 \implies E[X_{(3)}]=\frac{1}{4}\)

Thus, expected area = \((3/4-1/4)^2= 1/16\)

4.3 Problem 2

\(f(x,y) g(\sqrt{x^2+y^2})\) Consider the transformation:

\[ \begin{align*} x &= r \cos(\theta)\\ y &= r \sin(\theta) \end{align*} \]

Then \(f(r \cos(\theta), r \sin(\theta)) = g(r)\)

Reverse tranformation gives: \[ \begin{align*} Z &= X/Y = \tan(\theta)\\ r^2 &= y^2\sec^2(\theta) \end{align*} \]

Since \(f(r, \theta) = g(r)\) is independnet of \(\theta\), \(\theta ~ U\). Let \(\theta ~ U(0,2\pi)\)

Now, \(\theta = \tan^{-1}((Z)\)

\(f_Z(z) = f_\theta(\tan^{-1}(z)) |\frac{\partial \theta}{\partial z}|+ f_\theta(\pi+\tan^{-1}(z)) |\frac{\partial \theta}{\partial z}| = \frac{1}{2\pi}\)

\(\frac{\partial \theta}{\partial z} = 1/sec^2(\theta) = \frac{1}{z^2+1}\)

Thus, $f_Z(z) = =

4.4 Problem 3

4.5 Part a

\(E[X_{n+1}^r|X_n] = \int_0^{cx_n} x^r \frac{1}{cx_n} dx = \frac{(cx_n)^r}{r+1}\)

4.6 Part b

For \(r=1\):

\(E[X_{n+1}|X_n] = \frac{cx_n}{2}\)

$EX_{n+1}= E[E[X_{n+1}|X_n] ]= $ Thus $ = ^n 0 $ as \(n \rightarrow \infty\) as < 1$

For \(r=2\):

$EX_{n+1}^2 = $

$ = ^n $ as \(n \rightarrow \infty\)

4.7 Part c

4.7.1 ToDO

4.8 Problem 4

4.8.1 Part a

All \(n\) boys in single block \(\implies\) rreat them as one unit. Now for \(m\) girls there are \(m+1\) spots where we can put this one ‘unit’ in \(m+1\) ways while the girls can be arranged in \(m!\) ways and among the unit the boys can be arranged in \(n!\) ways.

Hence, the required probability: \(\frac{(m+1)n!m!}{(n+m)!}\)

4.8.2 Part b

For \(n>m\) the probability is zero,

If \(n\leq m\) we arrange the girls first leaving \(m+1\) spaces for \(n\) boys which can be filled in \({m+1 \choose n} \times n!\) and the girls can be arranged in \(m!\) ways

Required probability: \(\frac{{m+1 \choose n}m!n!}{(n+m)!}\)

4.8.3 Part c

Define: \(I_i=1\) if \((i-1,i,i+1)=(g,b,g)\) and \(0\) otherwise.

Then \(EW = E\sum_{i=1}^{n+m-2} I_i = (n+m-2) EI_1 = (n+m-2)P(I_1=1) = (n+m-2) \frac{n(m)(m-1)}{(n+m)!}\)