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The curve of Geometric Distribution is shaped according to parameter $\alpha$ using equation as $P_{r}=\frac{ (1-\alpha) \alpha^{CW}} { 1- \alpha^{CW} }. \alpha^{-r}$. Where CW is a parameter (contention window size like 16). So probability for occurrence of r=0 to 15 is is represented as shown in figure for different values of $\alpha$. Here 'number starting from 0 are put at higher side so r='0' has highest probability and r='15' has lowest probability.

enter image description here

Question1. So my question is that exponential distribution also gives curve that matches shape of one of the curves of geometric distribution so how can we find value of $\alpha$ that give same curve as given by exponential distribution. I have also read in Wikipedia that "The exponential distribution is the continuous analogue of the geometric distribution." https://en.wikipedia.org/wiki/Geometric_distribution

Question 2. So how can I prove any relationship between them like when exponential distribution equals geometric distribution?. It is some times also referred as "truncated geometric probability distribution"

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The geometric and exponential distributions are not the same, since they aren't even defined on the same domain. The geometric distribution lives on a discrete domain, the exponential distribution on a continuous domain. Consequently, you can't make the graphs of their density functions match perfectly. You can at most get them to look similar.

However, there is a reason why they look so similar: both are memoryless distributions, so they're both characterized by their median (at least approximately in the case of the geometric distribution, since its median is not unique in general). The best match between a geometric and exponential function should be when their medians match. The median of the exponential distribution with density function $f:[0,\infty)\to\mathbb R,~f(x)=\frac1\lambda\exp(-\lambda x)$ is $\frac{\ln2}{\lambda}$. Now calculate the median of your geometric distribution and choose $\alpha$ such that the medians match.

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  • $\begingroup$ thanks for the reply. Since Geometric Distribution formula does not have $\lambda$, instead $\alpha$, so how can I match them theoretically. Any hints on this would help me. Thanks in advance. $\endgroup$
    – user3696623
    Commented Jan 13, 2021 at 10:10
  • $\begingroup$ You can still calculate the median, which will then depend on $\alpha$. Set that equal to $\frac{\ln2}{\lambda}$ and solve for $\alpha$. $\endgroup$
    – Vercassivelaunos
    Commented Jan 13, 2021 at 10:13
  • $\begingroup$ I have used this formula for gettig probability of occurrence of variable 'r' with exponential distribution . It has same parameter 'CW' as in geometric distribution but $\lambda$ is not used here. $v2=(double) pow(2,CW)-1$ , $u=0.0+ getvalue(Uniform(0.0,1.0))$ and $r=ceil( log2(1+u*v2)$ Now how can I match exponential equation without $\lambda$ with geometric distribution with $\alpha$ $\endgroup$
    – user3696623
    Commented Jan 13, 2021 at 10:56
  • $\begingroup$ Still the same: calculate their medians and set them to be equal. Then solve for $\alpha$. $\endgroup$
    – Vercassivelaunos
    Commented Jan 13, 2021 at 11:13
  • $\begingroup$ The median of Geometric Distribution : $ \frac{-1}{log_{2}(1-p)} $ (from Wikipedia), should I put $p$ as probability of geometric distribution, but then it has $CW$ parameter. What value then it should have? . Also when equating then $ \frac{-1}{log_{2}(1-p)} =\frac{\ln2}{\lambda}$ , $\lambda$ is not known. Please help, I cannot understand $\endgroup$
    – user3696623
    Commented Jan 13, 2021 at 12:08

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