Question

CHAPTER 1,2/CLO 1,2

Answer 1

a. In order for $f$ to be a proper density function, its integral over its domain must evaluate to 1:

$\displaystyle\int_{-\infty}^\infty f(x)\,\mathrm dx=k\int_0^1(1-x)\,\mathrm dx=\frac k2=1\implies k=2$

$f$ also must be non-negative over its support, which is the case here.

I assume the instruction regarding the distribution function doesn't apply to parts (b) and (c).

(b) Integrate the density over the interval [0.1, 0.2]:

$P(0.1$

(c) Integrate the density over the interval [0.5, 1]:

$P(X>0.5)=\displaystyle\int_{0.5}^12(1-x)\,\mathrm dx=0.25$

The distribution function is obtained by integrating the density:

$F(x)=\displaystyle\int_{-\infty}^xf(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

(d) Using the distribution function, we have

$P(X$

(e) Using $F$ again, we get

$P(0.4$

(f) The mean is

$E[X]=\displaystyle\int_{-\infty}^\infty x\,f(x)\,\mathrm dx=\int_0^12x(1-x)\,\mathrm dx=\frac13$

The variance is

$V[X]=E[(X-E[X])^2]=E[X^2]-E[X]^2$

where

$E[X^2]=\displaystyle\int_{-\infty}^\infty x^2\,f(x)\,\mathrm dx=\int_0^12x^2(1-x)\,\mathrm dx=\frac16$

so that the variance is

$V[X]=\dfrac16-\left(\dfrac13\right)^2=\dfrac1{18}$