Question

Additional Problem 3. If X is a continuous random variable having cdf F, then its median is defined as that value of m for which F(m) = 0.5. Find the median for random variables with the following density functions (a) f(r)-e*, x > 0 (c) f(x) 6r(1-x), 1.

Answer 1

Additional Problem 3

(a)

$f(x)=e^{-x}\;\;,\;\;x\geq0$

$\therefore \;F(x)=\int_{0}^{x}e^{-x}dx=1-e^{-x}$

$F(m)=0.5\Rightarrow 1-e^{-m}=0.5\Rightarrow e^{-m}=0.5\Rightarrow m=-ln\;0.5=0.69315$

(b)

$f(x)=1\;\;,\;\;0\leq x\leq1$

$\therefore \;F(x)=\int_{0}^{x}dx=x$

$F(m)=0.5\Rightarrow m=0.5$

(c)

$f(x)=6x(1-x)\;\;,\;\;0\leq x\leq1$

$\therefore \;F(x)=\int_{0}^{x}6x(1-x)dx=6\int_{0}^{x}(x-x^2)dx=3x^2-2x^3$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!F(m)=0.5\Rightarrow 3m^2-2m^3=0.5\Rightarrow 4m^3-6m^2+1=0 \Rightarrow (2m-1)(2m^2-2m-1)=0$

$\Rightarrow m=0.5\,,\,0.5\pm i$  

$\therefore m=0.5\;\;\;\;\;[\because \;0\nleq0.5\pm i\nleq1]$

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Additional Problem 6

$f(x)=\frac{2}{9}(x+1)\;\;,\;\;-1\leq x \leq2$

(a)

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!E(X)=\int_{-1}^{2}x\,f(x)\,dx=\int_{-1}^{2}\frac{2}{9}x(x+1)dx=\frac{2}{9}\int_{-1}^{2}(x^2+x)dx=\frac{2}{9}\left ( \frac{14}{3}-\frac{1}{6} \right )=1$

(b)

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!E(X^2)=\int_{-1}^{2}x^2f(x)dx=\int_{-1}^{2}\frac{2}{9}x^2(x+1)dx=\frac{2}{9}\int_{-1}^{2}(x^3+x^2)dx=\frac{2}{9}\left ( \frac{20}{3}+\frac{1}{12} \right )=\frac{3}{2}$

$Var(X)=E(X^2)-[E(X)]^2=\frac{3}{2}-1^2=\frac{1}{2}$

(c)

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!F_{X}(x)=\int_{-1}^{x}f(x)dx=\int_{-1}^{x}\frac{2}{9}(x+1)dx=\frac{2}{9}\left ( \frac{x^2}{2}+x+\frac{1}{2} \right )=\frac{x^2+2x+1}{9}=\left ( \frac{x+1}{3} \right )^2$

(d)

$P(X>0)=1-P(X<0)=1-F_{X}(0)=1-\left ( \frac{0&plus;1}{3} \right )^2=\frac{8}{9}$

(e)

$Y=\frac{X-1}{X&plus;1}$

$\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!E(Y)=\int_{-1}^{2}y\,f(x)\,dx=\int_{-1}^{2}\frac{2}{9}\left ( \frac{x-1}{x&plus;1} \right )(x&plus;1)dx=\frac{2}{9}\int_{-1}^{2}(x-1)dx=\frac{2}{9}\left ( 0-\frac{3}{2} \right )=-\frac{1}{3}$

(f)

$F_{X}(m_{p})=p\Rightarrow \left ( \frac{m_{p}&plus;1}{3} \right )^2=p\Rightarrow m_{p}=3\sqrt{p}\,-1$