Q.5.3 - A First Course in Probability

Question

 Consider the function

$f\left(x\right) = \left\{\begin{array}{ll}C(2x - x^3) & 0 < x < \frac{5}{2}\\ 0 & otherwise\end{array}\right.$

Could $f$be a probability density function? If so, determine

$C$. Repeat if $f\left(x\right)$were given by

$f\left(x\right) = \left\{\begin{array}{ll} C(2x - x3)& 0 < x < \frac{5}{2}\\ 0& otherwise\end{array}\right.$

Step-by-Step Solution

Answer

When $c=0,\pm 2 then f\left(x\right)<0$

Step 1 Given Information

$f\left(x\right) = \left\{\begin{array}{ll}C(2x - x^3)& 0 < x < \frac{5}{2}\\ 0& otherwise\end{array}\right.$

Step 2 Explanation

$2x-x^3 =-x(x-\sqrt{2}) (x+\sqrt{2})$

$c=0$

${\int }_{-\infty }^{\infty }f\left(x\right)dx=0 , {\int }_{-\infty }^{\infty }f\left(x\right)dx=1$

$c>0 , f\left(x\right) <0, x\in (\sqrt{2},\frac{5}{2})$

Step 3 Explanation

$c=2$

$f\left(x\right)= 2(2x-x^3) =4x-2x^3$

$f\left(x\right)<0,x\in (\sqrt{2},\frac{5}{2})$

$f(1.5) = 4\times 15-2\times {(1.5)}^3=-0.75$

$c<0, f\left(x\right)<0 (0,\sqrt{2})$

Step 4 Explanation

$c=-2$

$f\left(x\right)= -2(2x-x^3)$

        $=-4x+2x^3$

$f\left(1\right)=-4+(2\times 1)=-2$

$f\left(x\right) \ge 0$

Step 5 Explanation

$f\left(x\right)=\left\{\begin{array}{ll}C(2x - x^3)& 0 < x < \frac{5}{2}\\ 0& otherwise\end{array}\right.$

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

$c=0, {\int }_{-\infty }^{\infty }f\left(x\right)dx=0$

$c>0,f\left(x\right)<0 ,x\in (2,\frac{5}{2})$

Step 6 Explanation

$x=2.2$

$f(2.2) =4(2.2)-2(2.2) =-0.88$

$c<0 , x\in (0,2)$

$c=-2, f\left(x\right) =-4x+2x^2$

$f\left(1\right)=-4+2=-2<0$