Can any one tell, how i can get formula for median of geometric distribution with this probability mass function formula $P_{r}=\frac{ (1-\alpha) \alpha^{CW}} { 1- \alpha^{CW} }. \alpha^{-r}$
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$\begingroup$ What are $C$ and $W$? The PMF of $X \sim$ Geometric($p$) is of the form : $P(X=k)=(1-p)^kp$ when the support is $\{0,1,2,3,\ldots \}$ and $P(X=k)=(1-p)^{k-1}p$ when the support is $\{1,2,3,\ldots \}$. So here it appears $(1-p) = \alpha$ and $p = (1-\alpha)$. $\endgroup$– oliverjonesCommented Jan 14, 2021 at 2:51
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$\begingroup$ CW is a paramter, i.e. number of object to choose from. It is truncated geometric probability distribution. $\endgroup$– user3696623Commented Jan 14, 2021 at 2:55
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$\begingroup$ I need median for my question that I asked here math.stackexchange.com/questions/3983425/… $\endgroup$– user3696623Commented Jan 14, 2021 at 2:56
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1 Answer
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Utilize the following:
Find $m$ such that
(1) $\displaystyle P(X \le m) \ge \frac{1}{2}$
(2) $\displaystyle P(X \ge m) \ge \frac{1}{2}$
The median will be any such $m$. This can be calculated using the PMF.
Edit
According to Truncated Distributions
The median of a truncated distribution will be
$\displaystyle F^{-1}\bigg( \frac{F(a) + F(b)}{2} \bigg )$
answered Jan 14, 2021 at 3:01
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$\begingroup$ Thanks for providing feedback. please tell me how can I get the median in terms of $\alpha$ . According to answer given to me, I need value of $\alpha$ that makes curve similar to exponential distribution. Also what $\lambda$ value should I put when equating medians of both truncated geometric and exponential distributions. Can you help me on my question here. math.stackexchange.com/questions/3983425/… $\endgroup$ Commented Jan 14, 2021 at 5:04
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$\begingroup$ $F$ is cumulative function, what a & b are ? $\endgroup$ Commented Jan 14, 2021 at 8:07
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$\begingroup$ Ok got it that 'a' and 'b' are constant (range), Do I need to put some value for 'a' and 'b' ?. Also $\alpha$ is not here so how can I get value of $\alpha$ that gives same curve as exponential distribution as suggested in math.stackexchange.com/questions/3983425/… $\endgroup$ Commented Jan 14, 2021 at 15:39
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$\begingroup$ My exponential distribution has this formula $P(X=x)=\frac{r^x-r^{x-1}}{r^N-1},\,r>1$ and $P(X\le x)=\frac{r^x-1}{r^N-1} $ and trunacted geometric probability distribution has this formula $P_{r}=\frac{ (1-\alpha) \alpha^{N}} { 1- \alpha^{N} }. \alpha^{-r}$. Where 'N' is number of element to choose from and 'r'=probbaility of picking a number r from N numbers. Now to solve for $\alpha$ i need median of both distributions equated. What would be the final equation to solve for $\alpha$ and get its value in range 0 to 1? $\endgroup$ Commented Jan 14, 2021 at 15:57
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$\begingroup$ You can use my answer to calculate these medians. For exponential use the first part of my answer and for truncated geometric use the part I added that applies to truncated distributions. $\endgroup$ Commented Jan 14, 2021 at 16:24