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

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

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}

  • $\begingroup$ 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. $\endgroup$ – Jacky Chong Oct 1 '16 at 7:52

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