- #1

fab13

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- TL;DR Summary
- I would like to know if 2 formulations about the agregation of 2 measures are equivalent with expectation and standard deviation.

I have two independant experiments have measured ##\tau_{1},\sigma_{1}## and ##\tau_{2},\sigma_{2}## with ##\sigma_{i}## representing errors on measures.

From these two measures, assuming errors are gaussian, we want to get the estimation of Ï

and its error (i.e with a combination of two measures).

We choose the maximum likelihood method with the pdf of 2 measures:

$$

f(\tau, \sigma)=\frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{1}{2} \frac{(\tau-\hat{\tau})^{2}}{\sigma^{2}}\right)

$$

One has to maximize the likelihood function:

$$

\mathcal{L}=\prod_{i=1}^{2} \frac{1}{\sqrt{2 \pi} \sigma_{i}} \exp \left(-\frac{1}{2} \frac{\left(\tau_{i}-\hat{\tau}\right)^{2}}{\sigma_{i}^{2}}\right)

$$

taking the following condition:

$$

\frac{\partial(-\log \mathcal{L})}{\partial \hat{\tau}}=0

$$

We get :

$$

\Rightarrow \hat{\tau}=\frac{\tau_{1} / \sigma_{1}^{2}+\tau_{2} / \sigma_{2}^{2}}{1 / \sigma_{1}^{2}+1 / \sigma_{2}^{2}}\quad(1)

$$

##\sigma_{\hat{\tau}}## is deducted from second derivate of ##\log{\mathcal{L}}## :

$$

\frac{1}{\sigma_{\hat{\tau}}^{2}}=\frac{1}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}\quad(2)

$$

For these both measures, equivalent number ##\tilde{N}## is defined by:

$$

\frac{\sigma_{1}}{\tau_{1}}=\frac{1}{\sqrt{\tilde{N}_{1}}} \quad \frac{\sigma_{2}}{\tau_{2}}=\frac{1}{\sqrt{\tilde{N}_{2}}}\quad(3)

$$

After, we can write :

$$

\hat{\tau}=\frac{\tilde{N}_{1} \tau_{1}+\tilde{N}_{2} \tau_{2}}{\tilde{N}_{1}+\tilde{N}_{2}}

$$

Finally, we have :

$$

\hat{\tau}=\frac{\tau_{1} /\left(\sigma_{1} / \tau_{1}\right)^{2}+\tau_{2} /\left(\sigma_{2} / \tau_{2}\right)^{2}}{1 /\left(\sigma_{1} / \tau_{1}\right)^{2}+1 /\left(\sigma_{2} / \tau_{2}\right)^{2}}\quad(4)

$$

In conclusion, we can say that in one case :

- case (1) : weighted by the square of inverse error (eq##(1)##)

and in another case :

- case (2): weighted by the square of relative error (eq##(4)##)

In the same time, on [Wikipedia], it is said that :

Population-based statistics [ edit ]

The populations of sets, which may overlap, can be calculated simply as follows:

$$

N_{X \cup Y}=N_{X}+N_{Y}-N_{X \cap Y}

$$

The populations of sets, which do not overlap, can be calculated simply as follows:

$$

\begin{aligned}

X \cap Y=\varnothing \Rightarrow & N_{X \cap Y}=0 \\

\Rightarrow & N_{X \cup Y}=N_{X}+N_{Y}

\end{aligned}

$$

Standard deviations of non-overlapping ##(X \cap Y=\varnothing)## sub-populations can be aggregated as follows if the size (actual or relative to one another) and means of each are known:

$$

\begin{aligned}

\mu_{X \cup Y} &=\frac{N_{X} \mu_{X}+N_{Y} \mu_{Y}}{N_{X}+N_{Y}} \\

\sigma_{X \cup Y} &=\sqrt{\frac{N_{X} \sigma_{X}^{2}+N_{Y} \sigma_{Y}^{2}}{N_{X}+N_{Y}}+\frac{N_{X} N_{Y}}{\left(N_{X}+N_{Y}\right)^{2}}\left(\mu_{X}-\mu_{Y}\right)^{2}}\quad(5)

\end{aligned}

$$

For example, suppose it is known that the average American man has a mean height of 70 inches with a standard deviation of three inches and that the average American woman has a mean height of 65 inches with a standard deviation of two inches. Also assume that the number of men, ##N##, is equal to the number of women. Then the mean and standard deviation of heights of American adults could be calculated as

$$

\begin{array}{l}

\mu=\frac{N \cdot 70+N \cdot 65}{N+N}=\frac{70+65}{2}=67.5 \\

\sigma=\sqrt{\frac{3^{2}+2^{2}}{2}+\frac{(70-65)^{2}}{2^{2}}}=\sqrt{12.75} \approx 3.57

\end{array}

$$

Any help is welcome

From these two measures, assuming errors are gaussian, we want to get the estimation of Ï

and its error (i.e with a combination of two measures).

We choose the maximum likelihood method with the pdf of 2 measures:

$$

f(\tau, \sigma)=\frac{1}{\sqrt{2 \pi} \sigma} \exp \left(-\frac{1}{2} \frac{(\tau-\hat{\tau})^{2}}{\sigma^{2}}\right)

$$

One has to maximize the likelihood function:

$$

\mathcal{L}=\prod_{i=1}^{2} \frac{1}{\sqrt{2 \pi} \sigma_{i}} \exp \left(-\frac{1}{2} \frac{\left(\tau_{i}-\hat{\tau}\right)^{2}}{\sigma_{i}^{2}}\right)

$$

taking the following condition:

$$

\frac{\partial(-\log \mathcal{L})}{\partial \hat{\tau}}=0

$$

We get :

$$

\Rightarrow \hat{\tau}=\frac{\tau_{1} / \sigma_{1}^{2}+\tau_{2} / \sigma_{2}^{2}}{1 / \sigma_{1}^{2}+1 / \sigma_{2}^{2}}\quad(1)

$$

##\sigma_{\hat{\tau}}## is deducted from second derivate of ##\log{\mathcal{L}}## :

$$

\frac{1}{\sigma_{\hat{\tau}}^{2}}=\frac{1}{\sigma_{1}^{2}}+\frac{1}{\sigma_{2}^{2}}\quad(2)

$$

For these both measures, equivalent number ##\tilde{N}## is defined by:

$$

\frac{\sigma_{1}}{\tau_{1}}=\frac{1}{\sqrt{\tilde{N}_{1}}} \quad \frac{\sigma_{2}}{\tau_{2}}=\frac{1}{\sqrt{\tilde{N}_{2}}}\quad(3)

$$

**- Question 1) Why we call this quantity ##\tilde{N}## as an "equivalent number" in**

eq##(3)##

- Question 2) This expression eq##(3)## is defined as being the relative error of

measure expressed by the statistical error due to the number of

events. Where does this definition of relative error of measure come from

? I mean, how to justify it ?eq##(3)##

- Question 2) This expression eq##(3)## is defined as being the relative error of

measure expressed by the statistical error due to the number of

events. Where does this definition of relative error of measure come from

? I mean, how to justify it ?

After, we can write :

$$

\hat{\tau}=\frac{\tilde{N}_{1} \tau_{1}+\tilde{N}_{2} \tau_{2}}{\tilde{N}_{1}+\tilde{N}_{2}}

$$

Finally, we have :

$$

\hat{\tau}=\frac{\tau_{1} /\left(\sigma_{1} / \tau_{1}\right)^{2}+\tau_{2} /\left(\sigma_{2} / \tau_{2}\right)^{2}}{1 /\left(\sigma_{1} / \tau_{1}\right)^{2}+1 /\left(\sigma_{2} / \tau_{2}\right)^{2}}\quad(4)

$$

In conclusion, we can say that in one case :

- case (1) : weighted by the square of inverse error (eq##(1)##)

and in another case :

- case (2): weighted by the square of relative error (eq##(4)##)

**Question 3) Are these 2 cases, rather formulations, are equivalent ? Are they 2 interpretations of a same quantity ##\hat{\tau}## ? If not, what's the link between these both expressions eq##(1)## and eq##(4)## ?**In the same time, on [Wikipedia], it is said that :

Population-based statistics [ edit ]

The populations of sets, which may overlap, can be calculated simply as follows:

$$

N_{X \cup Y}=N_{X}+N_{Y}-N_{X \cap Y}

$$

The populations of sets, which do not overlap, can be calculated simply as follows:

$$

\begin{aligned}

X \cap Y=\varnothing \Rightarrow & N_{X \cap Y}=0 \\

\Rightarrow & N_{X \cup Y}=N_{X}+N_{Y}

\end{aligned}

$$

Standard deviations of non-overlapping ##(X \cap Y=\varnothing)## sub-populations can be aggregated as follows if the size (actual or relative to one another) and means of each are known:

$$

\begin{aligned}

\mu_{X \cup Y} &=\frac{N_{X} \mu_{X}+N_{Y} \mu_{Y}}{N_{X}+N_{Y}} \\

\sigma_{X \cup Y} &=\sqrt{\frac{N_{X} \sigma_{X}^{2}+N_{Y} \sigma_{Y}^{2}}{N_{X}+N_{Y}}+\frac{N_{X} N_{Y}}{\left(N_{X}+N_{Y}\right)^{2}}\left(\mu_{X}-\mu_{Y}\right)^{2}}\quad(5)

\end{aligned}

$$

For example, suppose it is known that the average American man has a mean height of 70 inches with a standard deviation of three inches and that the average American woman has a mean height of 65 inches with a standard deviation of two inches. Also assume that the number of men, ##N##, is equal to the number of women. Then the mean and standard deviation of heights of American adults could be calculated as

$$

\begin{array}{l}

\mu=\frac{N \cdot 70+N \cdot 65}{N+N}=\frac{70+65}{2}=67.5 \\

\sigma=\sqrt{\frac{3^{2}+2^{2}}{2}+\frac{(70-65)^{2}}{2^{2}}}=\sqrt{12.75} \approx 3.57

\end{array}

$$

**- Question 4) Considering the expectations ##\mu_x## and ##\mu_y## are not the same, like in my 2 measures at the beginning (corresponding to ##\tau_1## and ##\tau_2##), can we say that eq##(2)## and eq##(5)## are equivalent ? i.e in the case where I have 2 measures at the beginning of my most.**Any help is welcome