I am given a distribution $$F(x)=\frac{1}{2}+\frac{x}{2(1+|x|)}$$ and I need to find the pdf. I did the following-

$$f(x)=F'(x)=\frac{2(1+|x|).1 -2.sgn(x).x}{2(1+|x|)^2}=\frac{2+2|x|-2|x|}{2(1+|x|)^2}=\frac{1}{(1+|x|)^2}$$

Is this correct?


1 Answer 1


As you apply the quotient rule on $\frac{u}{v}$, you have identify $v$ to be $2(1+|x|)$ but you did not square the $2$.

Notice that $$\int_{-\infty}^\infty \frac{1}{(1+|x|)^2}\, dx=2.$$

Upon squaring the $2$ as well in the denominator, you should obtain the pdf to be $\frac1{2(1+|x|)^2}$.

  • $\begingroup$ Thank you, is that all? $\endgroup$
    – thedumbkid
    Mar 1, 2021 at 7:03
  • $\begingroup$ seems to be so. $\endgroup$ Mar 1, 2021 at 7:20

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