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

Similarly,

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

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