Question

The density function of X is given by f(x) = a + bx2 if 0 ? x ? 1 0 otherwise. Suppose also that you are told that E(X) = 3/5. (a) Find a and b. (b) Determine the cdf, F(x), explicilty.

Problem 4. The density function of X is given by f(z) = 0 otherwise. Suppose also that you are told that E(X-3/5. (a) Find a and b. b) Determine the cdf, F(r

Answer 1

According to the given question:

is a continious variable with the following probability density function as:

$f(x)=a+bx^{2}$ if   $0\leq x\leq 1$

$=0$ otherwise.

and given expectation is

$E(x)=\frac{3}{5}$

i) According to the question, from the property of density function as:

$1)\int_{-\alpha }^{\alpha }f(x)dx=1$

$2)f(x)\geq 0$ for all

Therefore from the property of density function:

$\int_{-\alpha }^{\alpha }(a+bx^{2})dx=1$

or, $\int_{0 }^{1 }(a+bx^{2})dx=1$

or, $\left |ax+\frac{bx^{2+1}}{3} \right |_{0}^{1}=1$

or, $(a+\frac{b}{3})=1....(i)$

from the expectation we get,

$E(x)=\int_{-\alpha }^{\alpha }x.(a+bx^{2})dx=\frac{3}{5}$

or, $\int_{0 }^{1 }x(a+bx^{2})dx=\frac{3}{5}$

or, $\left |a\frac{x^{2}}{2}+\frac{bx^{3+1}}{4} \right |_{0}^{1}=\frac{3}{5}$

or, $(\frac{a}{2}+\frac{b}{4})=\frac{3}{5}....(ii)$

from equation (i) and (ii) we get, from (ii)*2 we get,

$(a+\frac{b}{2})=\frac{6}{5}...(iii)$

now (i) -(ii) we get;

$(\frac{b}{3}-\frac{b}{2})=1-\frac{6}{5}$

or, $\frac{-b}{6}=-\frac{1}{5}$

or, $b=\frac{6}{5}$

Therefore a is determine as:

$a+\frac{b}{3}=1$

or, $a+\frac{2}{5}=1$

or, $a=1-\frac{2}{5}$

or $a=\frac{3}{5}$

Therefore the probability density fucntion is :

$f(x)=\frac{3}{5}+\frac{6}{5}x^{2}$ if   $0\leq x\leq 1$

$=0$ otherwise.

b) Now the cumulative density function (CDF) is obtained by integrating the probability density function as shown below:

$F(x)=\int_{-\alpha }^{x }f(t)dt$

   $=\int_{0 }^{x}(\frac{3}{5}+\frac{6}{5}t^{2})dt$

   $=\left |\frac{3t}{5}+\frac{6t^{3}}{15} \right |_{0}^{x}$

Thereore,

$F(x)=(\frac{3x}{5}+\frac{2x^{3}}{5})$

Therefore the required CDF is as follows:

$F(x)=0$ when

$=(\frac{3x}{5}+\frac{2x^{3}}{5})$ when   $0\leq x\leq 1$

$=1$ when    $x\geq 1$