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Chapter 5: Q. 5.28 (page 216)

Consider the beta distribution with parameters (a,b). Show that

(a) when a>1and b>1, the density is unimodal (that is, it has a unique mode) with mode equal to (a-1)/(a+b-2)

(b) when a鈮�1, b鈮�1, and a+b<2, the density is either unimodal with mode at 0or 1or U-shaped with modes at both0and1;

(c) when a=1=b, all points in [0,1]are modes.

Short Answer

Expert verified

a.0,1is a maximum segment of xis unimodal

b. It is zero or one mode.

c. It is zero and one mode.

Step by step solution

01

Explanation (part a)

a.

Density function,

f(x)=1B(a,b)xa-1(1-x)b-1

Maximum of function,

g(x)=(a-1)logx+(b-1)log(1-x)

First derivative is,

g'(x)=a-1x-b-11-x

Second derivative is,

g''(x)=-a-1x2+b-1(1-x)2

For

a>1and b>1derivative is equal to zero

So,

g'(x)=0鈬�a-1x

=b-11-x鈬�x=a-1a+b-2

limx鈫�0g(x)=limx鈫�1g(x)=-鈭�

02

Explanation (part b)

The stationary point calculated in (0,1)is the lone contender for maximum in interval(0,1)in this situation (a).

However, we do have that.

g''a-1a+b-2=-(a-1)3(a+b-2)2+(b-1)3(a+b-2)2鈮�0

03

Explanation (part c)

c.

If a=b=1, X~Unif(0,1)is the result.

However, it is self-evident that every point between 0 and 1 is a mode.

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