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I need to find the CDF of

[tex]\mathcal{X}=\sum_{l=0}^L|h(l)|^2[/tex]

where

[tex]h(l)[/tex]

is complex Gaussian with zeros mean and variance

[tex]\sigma^2_l[/tex]

In particular, I need to proof that:

[tex]\text{Pr}\left[\mathcal{X}\leq b\right]\doteq b^{L+1}[/tex]

where dotted equal means in asymptotic sense as b approaches 0.

I found the expression which is:

[tex]1-\sum_{l=0}^L\beta_l\exp\left(-\frac{b}{\sigma_l^2}\right)[/tex]

where betas are coefficients come from partial expansion, but I don't know how to prove that it is in the asymptotic sense equals to:

[tex]b^{L+1}[/tex]

How can I do that?

Thanks

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# The CDF of sum of i.ni.d exponential RVs

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