# B How to find this CDF

1. Aug 15, 2016

### S_David

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

$$F_{a_n}(x)=1-\frac{1}{1+x}$$

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

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

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. Aug 15, 2016

### andrewkirk

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.

3. Aug 15, 2016

### S_David

I need the latter first. How can I find it? Any tips?

4. Aug 15, 2016

### andrewkirk

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 $F_{a_m}$.

5. Aug 16, 2016

### S_David

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 $a_m$.

6. Aug 16, 2016

### andrewkirk

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.

7. Aug 16, 2016

### S_David

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.