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Hey guys !!!!

My mother language is not English by the way. Sorry for spelling and gramme.

I'm curious to see if you can help me with my problem.I have already tried for almost a week and did not get to a solution. I also know, that the Maximum likelihood estimation is part of statistics and probability calculation. But since it is about formula transformation, I used the analysis forum.

The Maximum likelihood estimation formula is the following:

$L(\theta = p) = \prod\limits_{i=1}^N \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}}+ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{k_i+b}{2}} \cdot (1-p)^{\frac{k_i-b}{2}} \right]$

$= \prod\limits_i \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \left(\sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}} + \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{k_i+b}{2}} \cdot (1-p)^{\frac{k_i-b}{2}} \right)\right]$

where $p \in [0,1]$, $0^0:=1$, $a,b \in \mathbb{N}\backslash\{0\}$ with $a+b = r$

I'm looking forward to your ideas.

"Hints":

- I exclude first the $p^{\frac{k_i}{2}} \cdot (1-p)^{\frac{k_i}{2}}$

That gave me at least another product without $m$, that I was able to pull out of the sum. However the other $p$'s I was not able to pull out.

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A second, very similar Maximum likelihood estimation I need is the following

$L(\theta = p) = \prod\limits_{i=1}^N \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}}+\sum\limits_{m = 1}^{r-1} \frac{1}{r} \cdot 2^{h_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{h_i-1} \cdot \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{h_i+b}{2}} \cdot (1-p)^{\frac{h_i-b}{2}} \right]$

where $p \in [0,1]$, $0^0:=1$, $a,b \in \mathbb{N}\backslash\{0\}$ with $a+b = r$

My mother language is not English by the way. Sorry for spelling and gramme.

I'm curious to see if you can help me with my problem.I have already tried for almost a week and did not get to a solution. I also know, that the Maximum likelihood estimation is part of statistics and probability calculation. But since it is about formula transformation, I used the analysis forum.

The Maximum likelihood estimation formula is the following:

$L(\theta = p) = \prod\limits_{i=1}^N \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}}+ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{k_i+b}{2}} \cdot (1-p)^{\frac{k_i-b}{2}} \right]$

$= \prod\limits_i \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \left(\sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}} + \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{k_i+b}{2}} \cdot (1-p)^{\frac{k_i-b}{2}} \right)\right]$

where $p \in [0,1]$, $0^0:=1$, $a,b \in \mathbb{N}\backslash\{0\}$ with $a+b = r$

I'm looking forward to your ideas.

"Hints":

- I exclude first the $p^{\frac{k_i}{2}} \cdot (1-p)^{\frac{k_i}{2}}$

That gave me at least another product without $m$, that I was able to pull out of the sum. However the other $p$'s I was not able to pull out.

**********************************************************************

**********************************************************************

**********************************************************************

A second, very similar Maximum likelihood estimation I need is the following

$L(\theta = p) = \prod\limits_{i=1}^N \sum\limits_{m = 1}^{r-1}\left[ \frac{1}{r} \cdot 2^{k_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{k_i-1} \cdot \sin{\left(\frac{am \pi}{r}\right)} \cdot p^{\frac{k_i-a}{2}} \cdot (1-p)^{\frac{k_i+a}{2}}+\sum\limits_{m = 1}^{r-1} \frac{1}{r} \cdot 2^{h_i} \cdot \sin{\left(\frac{m \pi}{r}\right)} \cdot \cos{\left(\frac{m \pi}{r}\right)}^{h_i-1} \cdot \sin{\left(\frac{bm \pi}{r}\right)} \cdot p^{\frac{h_i+b}{2}} \cdot (1-p)^{\frac{h_i-b}{2}} \right]$

where $p \in [0,1]$, $0^0:=1$, $a,b \in \mathbb{N}\backslash\{0\}$ with $a+b = r$

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