Answer to Question #213459 in Statistics and Probability for jonny

Question

1.The pdf of a random variable x is given by f(x)={k(1-x),0<x<1 both inclusive

{0, elsewhere

Find the value of the constant k,the cdf of x and the value of m such that G(x)=1/2

2.Find the cdf of a random variable y where the pdf is given by

(a)f(x)={1/3,0<X<1 both inclusive

{1/3,2<x<4 both inclusive

{0,elsewhere

(b)f(x)={x/2,0<x<1 both inclusive

{1/2,1<x<2 both inclusive

{(3-x)/2,2<x<3 both inclusive

3.If the cdf of a random variable y is given by F(X)=1-9/y² for y>or equal to3

and F(x)=0 for y<3 find the P(X<or equal to 5),p(x>8) and the pdf of x.

Answer 1

1. $1=\int f(t)dt=\int\limits_{0}^1k(1-t)dt=k(t-t^2/2)|_0^1=k/2$

Therefore, k=2.

The CDF (cumulative distribution function) is $F(x) =\int\limits_{-\infty}^x f(t)dt$.

$F(x)=0$, if $x<0$;

$F(x)=1$, if $x>1$;

$F(x)=\int\limits_{0}^x 2(1-t)dt=(2t-t^2)|_0^x=2x-x^2$, if $x\in[0,1]$.

The median of distribution is a such x, that $F(x)=1/2$.

$2x-x^2=1/2$

$(x-1)^2=1/2$

$x=x_1=1-1/\sqrt{2}$ , since the other root $x_2=1+1/\sqrt{2}>1$ and $F(x_2)=1$

2. (a) $f(x)=1/3$, if $X\in[0,1]\cup [2,4]$ and $f(x)=0$ elsewhere.

The CDF is $F(x) =\int\limits_{-\infty}^x f(t)dt$.

$F(x)=0$, if $x\leq 0$;

$F(x)=\int\limits_{0}^{x}\frac{1}{3} dt=x/3$, if $x\in[0,1]$;

$F(x)=1/3+\int\limits_{0}^{x}0 dt=1/3$, if $x\in[1,2]$;

$F(x)=\frac{1}{3}+\int\limits_{2}^{x}\frac{1}{3} dt=\frac{1}{3}+\frac{x-2}{3}=\frac{x+1}{3}$, if $x\in[2,4]$;

$F(x)=1$, if $x>4$.

(b) Given the pdf $f(x)=x/2$ , if $0\leq x\leq1$

$f(x)=1/2$ , if $1\leq x\leq2$

$f(x)=(3-x)/2$ if $2\leq x\leq 3$

is not a pdf of a random variable, since

$\int f(t)dt=\int\limits_{0}^{1}\frac{t}{2}dt+\int\limits_{1}^{2}\frac{1}{2}dt+\int\limits_{2}^{3}\frac{3-t}{2}dt=\frac{1}{4}+\frac{1}{2}+\frac{1}{4}$

The CDF (cumulative distribution function) is $F(x) =\int\limits_{-\infty}^x f(t)dt$.

$F(x)=0$, if $x\leq 0$;

$F(x)=\int\limits_{0}^{x}\frac{t}{2} dt=x^2/4$, if $x\in[0,1]$;

$F(x)=\frac{1}{4}+\int\limits_{1}^{x}\frac{1}{2} dt=\frac{1}{4}+\frac{x-1}{2}=\frac{2x-1}{4}$, if $x\in[1,2]$;

$F(x)=\frac{3}{4}+\int\limits_{2}^{x}\frac{3-t}{2} dt=\frac{3}{4}+\frac{1}{2}(x-2)-\frac{(x-2)^2}{4}=\frac{-x^2+6x-5}{4}$, if $x\in[2,3]$;

$F(x)=1$, if $x>3$.

3. If the cdf of a random variable Y is given by $F(y)=1-9/y^2$ for $y\geq 3$ and $F(x)=0$ for $y<3$, then $P(Y<5)=F(5)=1-9/5^2=0.64$.

The PDF equals to $f(y)=F'(y)$. Hence

$f(y)=18/y^3$, for $y\geq 3$

$f(y)=0$, for any other value of y.