1
$\begingroup$

Hi This is my first time posting a question on this website. Thank you advance for helping me out here.

My question is

Suppose the density of $X$ is $$f(x) = \frac{Kx^2}{(1 + x)^5}$$ when $x > 0$. Find the constant $K$ and the density of $Y = \frac{1}{(1 + X)}$.

++one more thing since this is pdf. This equals to 1.

I tried partial fraction expansion to do it but it was long and didn't lead to answer. For the density part I finished upto $\frac{-Kx^2}{(x+1)^3}$. However, I do not know K and could not finish it.

Thank you again

$\endgroup$
2
  • $\begingroup$ $f(x) = Kx^2=(1 + x)^5$? $\endgroup$
    – msm
    Oct 1 '16 at 7:40
  • $\begingroup$ sorry its Kx^2/(1+x)^5 $\endgroup$
    – King Kong
    Oct 1 '16 at 7:41
0
$\begingroup$

Hint: Use $u$-sub first to get \begin{align} \int \frac{Kx^2}{(1+x)^5}\ dx = \int \frac{K(u-1)^2}{u^5}\ du. \end{align}

$\endgroup$
6
  • $\begingroup$ thanks I will try and get back to you! $\endgroup$
    – King Kong
    Oct 1 '16 at 7:42
  • $\begingroup$ sorry what is happening to K? because I approach same thing when K is not present before substitution. $\endgroup$
    – King Kong
    Oct 1 '16 at 7:46
  • $\begingroup$ so with K added in, I still need to use partial fraction expansion right? $\endgroup$
    – King Kong
    Oct 1 '16 at 7:48
  • $\begingroup$ OH I think it i got it now. make them separate and integrate them one by one $\endgroup$
    – King Kong
    Oct 1 '16 at 7:52
  • $\begingroup$ That's correct. $\endgroup$ Oct 1 '16 at 7:52

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.