Just need some answers checking:

(iii) im not sure what needs doing but i've shown my attempt

A continuous random variable $X$ has cumulative distribution function

$$F(x) = \begin{cases} 0 & x<1,\\ x^2(3-2x) & 0\le x \le 1,\\ 1 & x>1. \end{cases} $$

$(i)$ Calculate $\Pr(\frac13\ <\ X\ <\ \frac23)$

$(ii)$ Find the probability density function of $X$

$(iii)$ The symmetry of the probability density function on the interval $[0,1]$ implies that $E(X) = \frac12$. Find $\operatorname{var}(X)$


(i) $\Pr(a < X < b) = \Pr(X \le b)-\Pr(x \le a) = F_x(b)-F_x(a)$

Given this,

$\Pr(\frac13\ <\ X\ <\ \frac23) = F(\frac23)-F(\frac13) = ((\frac23)^2(3-2(\frac23))) - ((\frac13)^2(3-2(\frac13))) = \cdots = \frac{13}{27}$

So $\Pr(\frac13\ <\ X\ <\ \frac23) = \frac{13}{27}$

(ii) Given $Pr(a\ \leq\ X\ \leq\ b)\ =\ \int_a^b f(x) \, dx$ for finding density function

$\int_a^b x^2(3-2x)\,dx$

$\int_a^b 3x^2-2x^3\,dx$

$= x^3-\frac12x^4$

(iii) Given $E(x) = \frac12,\ (E(x)^2)=\frac14$

To get $E(x^2)$ you do $PDF \cdot x^2:= (x^3 - \frac12x^4) \cdot x^2 = x^6 - \frac12x^8$

$\int_{a}^{b}x^6 - \frac12x^8 = [\frac{x^7}{7} - \frac{x^9}{18}]^1_0$

$Var(x) = [(\frac{x^7}{7} - \frac{x^9}{18})-\frac14]^1_0$

$= \frac{11}{126}-\frac14 = -\frac{41}{252}$

  • $\begingroup$ You should have $F_X(b)-F_X(a)$, not $F_x(b)-F_x(a)$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Mar 15 '15 at 20:50

The set-up in (i) is right. I have not checked the arithmetic.

For (ii), the density function is the derivative of the cdf. You calculated an integral.

In (iii), to calculate $E(X^2)$, you need $\int_0^1 x^2f(x)\,dx$, where $f(x)$ is the density function. The numerical answer cannot be right, since variance is always $\ge 0$.

  • $\begingroup$ So for (ii) i differentiate $x^2(3-2x)$ and for (iii) i use the answer from (ii) and do the same method as done before? $\endgroup$ – Matt Dec 11 '12 at 3:09
  • $\begingroup$ Density is $6x-6x^2$ on $[0,1]$, and $0$ elsewhere. Now multiply by $x^2$. We get $6x^3-6x^4$. Integrate. Finally, subtract $1/4$. If I am doing arithmetic right in my head, should get $1/20$. $\endgroup$ – André Nicolas Dec 11 '12 at 3:13
  • $\begingroup$ Yeah that makes sense, i got $\frac1{20}$ too. just doing the intergral instead of the derivative got me confused. Thanks $\endgroup$ – Matt Dec 11 '12 at 3:18
  • $\begingroup$ In your solution to (iii), beside not having right density, you substituted $x^2$ in what you thought was the density, instead of multiplying the density by $x^2$. $\endgroup$ – André Nicolas Dec 11 '12 at 3:20

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