5
$\begingroup$

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?

$\endgroup$

1 Answer 1

5
$\begingroup$

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 $\frac1{2(1+|x|)^2}$.

$\endgroup$
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.