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B How to find this CDF

  1. Aug 15, 2016 #1
    Hello all,

    I have the following random variable ##\Gamma_m=\frac{a_m}{\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n+1}## where the random variables ##\{a_n\}## are independent and identically distributed random variables. The CDF of random variable ##a_n## if given by

    [tex]F_{a_n}(x)=1-\frac{1}{1+x}[/tex]

    Now I need to find the CDF of ##\Gamma_m##. I started like this:

    [tex]F_m(\gamma)=Pr\left[\frac{a_m}{\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n+1}\leq \gamma\right]=1-Pr\left[\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n<\frac{a_m}{\gamma}-1\right]=1-\int_{a_m}Pr\left[\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n<\frac{a_m}{\gamma}-1\right]f_{a_m}(a_m)\,da_m[/tex]

    where ##f_{a_m}(a_m)=1/(1+a_m)^2## is the PDF of the random variable ##a_m##. Apparently, ##Pr\left[\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n<\frac{a_m}{\gamma}-1\right]## is the CDF of the random variable ##\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n## for a given ##a_m##. How can I continue from here without resorting to Laplace or Fourier Transform? The reason why is that I need to write the CDF in terms of functions, because later I need to find its PDF.

    Thanks
     
    Last edited: Aug 15, 2016
  2. jcsd
  3. Aug 15, 2016 #2

    andrewkirk

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    Do you want the unconditional CDF of ##\Gamma_m##, or the CDF conditional on the values of all the ##a_n## except ##a_m##? The latter is easy. The former requires a multiple integration.
     
  4. Aug 15, 2016 #3
    I need the latter first. How can I find it? Any tips?
     
  5. Aug 15, 2016 #4

    andrewkirk

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    You worked out the following:
    $$F_{\Gamma_m}(\gamma)=Pr\left[\frac{a_m}{1+\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n}\leq \gamma\right] $$
    For the conditional case this is:
    $$F_{\Gamma_m|a_1,...a_{n-1},a_{n+1},...,a_K}(\gamma)=Pr\left[\frac{a_m}{1+\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n}\leq \gamma\ \middle|\ a_1,...a_{n-1},a_{n+1},...,a_K\right] $$
    Instead of the next step you take in your post, rearrange this to get ##a_m##, which is the only random variable in the conditional case, on its own on the left of the '##\leq##' sign. So you have
    $$F_{\Gamma_m|a_1,...a_{n-1},a_{n+1},...,a_K}(\gamma)=Pr\left[a_m\leq \gamma{\left(1+\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n\right)}\ \middle|\ a_1,...a_{n-1},a_{n+1},...,a_K\right] $$
    which is equal to
    $$F_{a_n}\left( \gamma{\left(1+\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n\right)}\right)$$

    So now you can just use the formula you wrote for ##F_{a_m}##.
     
  6. Aug 16, 2016 #5
    That's why I arranged it in a different way, because in your approach to find the unconditional CDF, you need to average over all ##K-1## random variables. In my arrangement, I just need to average over one random variable if I can find the CDF of the summation conditioned on ##a_m##.
     
  7. Aug 16, 2016 #6

    andrewkirk

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    You asked in post 3 about the conditional CDF which, per the above, is just:

    $$1-\frac1{ \gamma{\left(1+\sum\limits_{\substack{n=1\\n\neq m}}^Ka_n\right)}}$$

    As I stated above, the unconditional CDF requires a completely different calculation. That will involve a ##K##-deep nested integration.
     
  8. Aug 16, 2016 #7
    My ultimate aim as I stated in the first post is to evaluate the unconditional CDF, and in post 3 I said I need the conditional CDF first, and by the conditional CDF, I meant the conditional CDF I formulated in the last part of my first post, namely, the CDF of the summation of the random variables conditioned on ##a_m##. I formulated the problem this way with the ultimate goal of the unconditional CDF to be found as easy as possible.
     
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