# 4 2017-Spring

## 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 $$n \rightarrow \infty$$

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