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Chapter 6: Q.6.52 (page 274)

Let X and Y denote the coordinates of a point uniformly chosen in the circle of radius 1centered at the origin. That is, their joint density is f(x,y)=1蟺x2+y2鈮�1.

Find the joint density function of the polar coordinates R=(X2+Y2)1/2 and 胃=tan-1Y/X.

Short Answer

Expert verified

Joint density function of the polar coordinates,

fR,胃(r,胃)=r蟺

Step by step solution

01

Radius :

A straight line drawn from the center of a circle or sphere to the circumference.

02

Explanation : 

The random vector's density function,

f(x,y)=1蟺

For x2+y2鈮�1

Polar coordinates: R=(X2+Y2)1/2and 胃=tan-1Y/X.

The density function of a random vector, according to the assertion,

fX.Y(x,y)=1蟺for x2+y2鈮�1.

Now, apply the transformation

g(x,y)=(r,胃)=x2+y2,tan-1yx

We can express the density function of a random vector using the theorem about variable transformations (R,胃)=g(X,Y)as

fR,胃(r,胃)=fX,Y(x,y).det(鈭�g(x,y))-1

Then calculate,

鈭�g(x,y)=xx2+y2yx2+y2-yx2+y2xx2+y2

That yields

det(鈭�g(x,y))=x2+y2(x2+y2)32=1x2+y2=1r

Hence,

fR,胃(r,胃)=r.fX.Y(x,y)=r蟺for r鈭�(0,1)

Androle="math" localid="1647253987981" 胃鈭�(0,2蟺)

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Most popular questions from this chapter

Choose a number X at random from the set of numbers 1,2,3,4,5. Now choose a number at random from the subset no larger than X, that is, from 1...,X. Call this second number Y.

(a) Find the joint mass function of X and Y.

(b) Find the conditional mass function of X given that Y = i. Do it for i = 1,2,3,4,5.

(c) Are X and Y independent? Why?

The joint probability mass function of X and Y is given by

p(1,1)=18p(1,2)=14p(2,1)=18p(2,2)=12

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Let X1,...,Xnbe a set of independent and identically distributed continuous random variables having distribution function F, and let X(i),i=1,...,ndenote their ordered values. If X, independent of theXi,i=1,...,n, also has distribution F, determine

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