Question

1.        The density function of X is given by f(x) = {a + bx^2, 0< x <1 0 ,otherwise. . If E(x)=3/5 find a,b.

Answer 1

$E(x) = \int\limits_{ - \infty }^{ + \infty } {xf(x)dx} = \int\limits_0^1 {x(a + b{x^2}} )dx \Rightarrow \int\limits_0^1 {\left( {ax + b{x^3}} \right)} dx = \frac{3}{5} \Rightarrow \frac{1}{2}a\left. {{x^2}} \right|_0^1 + \frac{1}{4}b\left. {{x^4}} \right|_0^1 = \frac{3}{5} \Rightarrow \frac{a}{2} + \frac{b}{4} = \frac{3}{5} \Rightarrow \frac{{2a + b}}{4} = \frac{3}{5} \Rightarrow 2a + b = \frac{{12}}{5}$

Next, we use the properties of the density function

$\int\limits_{ - \infty }^{ + \infty } {f(x)dx} = 1 \Rightarrow \int\limits_0^1 {(a + b{x^2}} )dx = 1 \Rightarrow a\left. x \right|_0^1 + \frac{1}{3}b\left. {{x^3}} \right|_0^1 = 1 \Rightarrow a + \frac{1}{3}b = 1$

We have the system

$\left\{ \begin{array}{l} a + \frac{1}{3}b = 1\\ 2a + b = \frac{{12}}{5} \end{array} \right. \Rightarrow \left\{ \begin{array}{l} 3a + b = 3\\ 10a + 5b = 12 \end{array} \right. \Rightarrow a = \frac{3}{5},\,\,b = \frac{6}{5}$

Answer: $a = \frac{3}{5},\,\,b = \frac{6}{5}$