Q.6.15 - A First Course in Probability

Question

The random vector (X, Y) is said to be uniformly distributed over a region R in the plane if, for some constant c, its joint density is f(x, y) =  c if(x, y) ∈ R 0 otherwise

(a) Show that 1/c = area of region R. Suppose that (X, Y) is uniformly distributed over the square centered at (0, 0) and with sides of length 2 

(b) Show that X and Y are independent, with each being distributed uniformly over (−1, 1).

 (c) What is the probability that (X, Y) lies in the circle of radius 1 centered at the origin? That is, find P{X² + Y² < 1}. 

Step-by-Step Solution

Answer

(a) The area of region R is $A\left(R\right)=\frac{1}{c}$.

(b) X and Y are independent.

(c) $P\left\{\right(X,Y)\in C\}=\frac{\pi }{4}$

Step 1: Content Introduction

If the chance that the random vector lies in CR, we say that (X, Y) is uniformly distributed throughout a region R of finite area A(R).

Step 2: Explanation (Part a)

Let R² be the two - dimensional plane.

Then observe that for $E\subseteq R^2$

$P\left\{\right(X,Y)\in E\}={\iint }_{E}f(x,y)dxdy={\iint }_{E\cap R}c·area(E\cap R)$

Since $f(x, y)$ is a joint density function,

We have

$1=P\left\{(X,Y)\in R^2\right\}=c·area\left(R^2\cap R\right)=c·area\left(R\right)$

Therefore,

The area of $R is 1 / c$.

Step 3: Explanation (b)

Let R be the real line.

For sets$A,B\subseteq R$

We have

$P(X\in A,Y\in B)={\int }_B{\int }_{A}f(x,y)dxdy={\int }_{B\cap \mid -1,1]}{\int }_{A\cap \mid -1,1]}\frac{1}{4}dxdy$ 

=$\frac{1}{4}\text{ length }(A\cap [-1,1\left]\right)\text{ length }(B\cap [-1,1\left]\right)$

And

$P\{X\in A\}={\int }_{-2}^{*}{\int }_{A}f(x,y)dxdy={\int }_{-1}^1{\int }_{AN-1,1]}\frac{1}{4}dxdy$ 

$=\frac{1}{2}\text{ length }(A\cap [-1,1\left]\right)$

And

$P\{Y\in B\}={\int }_{-\infty }^{\approx }{\int }_{B}f(x,y)dxdy={\int }_{-1}^1{\int }_{BN-1,1]}\frac{1}{4}dxdy$ 

$=\frac{1}{2}\text{ length }(B\cap [-1,1\left]\right)$

Thus,

$P\{X\in A,Y\in B\}=\frac{1}{4}\text{ length }(A\cap [-1,1\left]\right)\text{ length }(B\cap [-1,1\left]\right)$ 

$=P\{X\in A\}P\{Y\in B\}$

Therefore,

X and Y are independent.

Also,

The above formulas for$P\{X \in A\} and P\{Y \in B\}$ show that each of $X and Y$ are uniformly distributed over $[-1,1]$.

Step 4: Explanation (c)

Let C be the interior of the circle of radius 1 centered at the origin.

Then we are required to compute $P\left\{\right(X,Y)\in C\}$.

Thus,

$P\left\{\right(X,Y)\in C\}=\frac{1}{4}·area(C\cap R)=\frac{1}{4}·area\left(C\right)=\frac{1}{4}·\pi =\frac{\pi }{4}$