Question

Check whether the following is a p.d.f. 

f(x) = ${\displaystyle \{\begin{array}{ll}x& \text{for} 0\le x\le 1\\ 2-x& \text{for} 1<x\le 2.\end{array}}$

Answer 1

Here, f(x) ≥ 0 ${\displaystyle \forall }$ x ∈[0, 2]
Also, f(x) is continuous.
Now consider,

${\displaystyle {\int }_0^{2}f\left(x\right)\cdot dx={\int }_0^{1}f\left(x\right)\cdot dx+{\int }_1^{2}f\left(x\right)\cdot dx}$

= ${\displaystyle {\int }_0^{1}x\cdot dx+{\int }_1^2(2-x)\cdot dx}$

= ${\displaystyle \frac{1}{2}{\left[x^2\right]}_0^1+2{\left[x\right]}_1^2-\frac{1}{2}{\left[x^2\right]}_1^2}$

= ${\displaystyle \frac{1}{2}[1-0]+2[2-1]-\frac{1}{2}[4-1]}$

= ${\displaystyle \frac{1}{2}+2-\frac{3}{2}}$
= 1
∴ f(x) is p.d.f. of r.v.X