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How do one go about solving this tricky question:

Given the pdf of $k_{o}$ as shown below $$f_{k_{o}}(k)=2\pi v ke^{-\pi v k^{2}}, \quad 0\leq k\leq \infty $$
and $q$ is given as $$ q = \zeta_{o} k_{o}^{n}, \quad 0\leq q\leq q_{u}$$ Prove that the pdf of $q$ is given as $$\eqalign{f_{q}(x) = &\, {2\pi v x^{{2 \over \eta} - 1}e^{-\pi v \left({x \over \zeta_{o}}\right)^{2 \over \eta}} \over \eta\zeta_{o}^{2 \over \eta}\int\limits_{0}^{q_{u}}{2\pi v \over \eta\zeta_{o}^{2 \over \eta}}y^{{2 \over \eta} - 1}e^{-\pi v \left({y\over\zeta_{o}}\right)^{2\over\eta}}\,\mathrm{d}y}\cr =& \, {2\pi v x^{{2 \over\eta}-1}e^{-\pi v \left({x\over\zeta_{o}}\right)^{2\over\eta}}\over\eta\zeta_{o}^{2\over\eta}\left(1 - e^{-\pi v \left({q_{u} \over \zeta_{o}}\right)^{2 \over \eta}}\right)}.\quad 0 \leq x\leq q_{u}.}$$

  • Note that $q \le q_0 \iff k_0 \le (q_0/\zeta_0)^{1/\eta}.$ Compute the conditional density of $k_0$ under this event. Then transform this density into one for $q$ in the standard way (for instance, see section 2 here). – stochasticboy321 Mar 10 '18 at 04:46
  • Thanks! for sharing that link! However, I am still struggling to get how it was applied here! – Abdulhameed Mar 10 '18 at 04:55
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    Okay, first forget about the intimidating constants, we'll deal with them later. In the notation of section 2, $X$ is $k_0,$ and $Y$ is $q$. (Consequently, $g(x) = \zeta_0 x^{\eta}$.) First work out the density for $q$ using this method, and pretending that its support is $[0,\infty)$ instead of $[0,q_u).$ Now, in the paper, the cutoff is imposed on top of this model - so they're imposing the condition $q \le q_u.$ – stochasticboy321 Mar 10 '18 at 05:07
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    (contd.) To get the new density of $q,$ we simply need to find the conditional density given this event, which is just the original density divided by the probability that $q \le q_u$. That gives you this big constant that you're seeing in the denominator (which, if you look up the Rayleigh distribution, you'll immediately see as its the CDF). – stochasticboy321 Mar 10 '18 at 05:08
  • I really don't think this needs an answer after that detailed a hint. Please try and solve it yourself - stuff like this is not very complicated if you've spent the time working a few such questions out in detail, and this is a great chance for you to get the practice in. I encourage you to post the solution here once you have it, in order to help the next person with a similar Q. – stochasticboy321 Mar 10 '18 at 23:31
  • Ok. Thanks I got it. – Abdulhameed Mar 11 '18 at 06:13

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