Solutions

2.27

a.

$f(x)=\frac{1}{\sqrt{2\pi}}e^{-x^{2}/2}$ has a unique mode at $x=0$.

b.

For

$$f(x) = \begin{cases}1 & 0 \le x \le 1 0 & \text{otherwise} \end{cases}$$

any $x \in [0,1]$ is a mode.

c.

Suppose, without loss of generality, that the point of symmetry is the origin. Suppose, without loss of generality, that the density is unimodal with mode $a \ge 0$. For $x \le y \le 0$ we have $x \le y \le a$ and therefore by unimodality $f(x) \le f(y)$ and $f(y) \le f(0)$. For $0 \le x \le y$ we have $f(x)=f(-x)$ and $f(y)=f(-y)$ by symmetry, and $f(-y) \le f(-x) \le f(0)$ by unimodality as before. So the point of symmetry 0 is a mode.

d.

The exponential distribution

$$f(x) = \begin{cases}e^{-x} & x \ge 0 0 & x < 0 \end{cases}$$

is unimodal with mode $x=0$ since $f(x)$ is non-decreasing on $(-\infty,0]$ and non-increasing (in fact strictly decreasing) on $[0,\infty)$. If you switch the equality sign to the other side, it is not unimodal.

2.30

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.

2.33

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.

2.38

L^ATEX2Html is giving me grief about this one-it is available in the pdf version.