After analyzing performance of a cooperative system, I get the following expression for the system outage probability:

$P = 1 - \frac{{{e^{ - 2\mu /{\beta _M}}}}}{{\Gamma \left( {{\alpha _3}} \right)\beta _3^{{\alpha _3}}}}\sum\limits_{{m_1} = 0}^{{\alpha _1} - 1} {\sum\limits_{{n_1} = 0}^{{m_1}} {\sum\limits_{{m_2} = 0}^{{\alpha _2} - 1} {\sum\limits_{{n_2} = 0}^{{m_2}} {\frac{{{\mu ^{{m_1} + {m_2} - {n_1} - {n_2}}}{{\left( {\vartheta /2} \right)}^{{n_1} + {n_2}}}\Gamma \left( {{\alpha _3} + {n_1} + {n_2}} \right)}}{{\left( {{m_1} - {n_1}} \right)!\left( {{n_1}} \right)!\left( {{m_2} - {n_2}} \right)!\left( {{n_2}} \right)!\beta _M^{{m_1} + {m_2}}{{\left( {\frac{\vartheta }{{{\beta _M}}} + \frac{1}{{{\beta _3}}}} \right)}^{{\alpha _3} + {n_1} + {n_2}}}}}} } } }$

What I want to do is to find an asymptotic approximation of the above expression when $\beta_M$ goes to infinity.

I have tried to put specific values of $\alpha_1$ and $\alpha_2$. I saw that term inside the sum with $m_1 = m_2 = n_1 = n_2 = 0$ dominates the others. Therefore, no matter what values of $\alpha_1$ and $\alpha_2$ are, I obtained: $P \approx {\rm O}\left( {\frac{1}{{{\beta _M}}}} \right)$.

I did plot the above expression with several sets of $\alpha_1$ and $\alpha_2$, I got: enter image description here

From the figure, it seems that $P \approx O\left( {\frac{1}{{\beta _M^{\min \left( {{\alpha _1},{\alpha _2}} \right)}}}} \right)$.

I tried over and over gain, but I still got $P \approx {\rm O}\left( {\frac{1}{{{\beta _M}}}} \right)$.

Do you have any ideas and suggestions?

Thank you very much.



I define $x = 1/\beta_M$. Then, giving $\alpha_1$ and $\alpha_2$ fixed values, I expand the sum and factorize the term $\left( {\frac{1}{{{\beta _3}}} + \vartheta x} \right)$ with the highest order equal to $(\alpha_3 +\alpha_1 + \alpha_2 - 2)$. I obtained:

$P = 1 - \frac{{{e^{ - 2\mu x}}}}{{\beta _3^{{\alpha _3}}{{\left( {\frac{1}{{{\beta _3}}} + 2x} \right)}^{{\alpha _3} + {\alpha _1} + {\alpha _2} - 2}}}}f\left( x \right)$,

where $f\left( x \right)$ is a polynomial of $x$ with the highest order equal to $\alpha_1*\alpha_2 - 1$.

Through numerical analysis, I observe that P can be safely approximated as follows

$P \approx 1 - \frac{{{e^{ - 2\mu x}}}}{{\beta _3^{{\alpha _3}}{{\left( {\frac{1}{{{\beta _3}}} + 2x} \right)}^{{\alpha _3} + {\alpha _1} + {\alpha _2} - 2}}}}g\left( x \right)$,

where $g\left( x \right)$ contains the terms in $f\left( x \right)$ which have order less than or equal to $\min \left\{ {{\alpha _1},{\alpha _2}} \right\} - 1$.

I hope this update may help someone to get some idea ^^.

  • $\begingroup$ it really seems that $m_1+m_2=0$ are the contributions which dominates the in the large $\beta_M$ limit. And $m_1=m_2=0$ is only one way to satisfy the equality $\endgroup$ – tired Sep 3 '15 at 13:23

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