I Question about data analysis and error

[Arman777]

Let us suppose we have one constant variable $b \pm \delta b = 20 \pm 1$ and one function that depends on $x$, such as, $a(x) \pm \delta a$

The problem is I want the difference between $a(x)$ and $b$ to be $0$. Let me denote this difference as $c \pm \delta c$. To obtain a difference which is $0$, we can have only one condition $c \leq \delta c$.

We are also given a range of $x$ values.

So the problem is something like this.

Take a range of $x$ values (such as ${0,2}$)

For each value of $x$ in this range;

1) Calculate $a(x) \pm \delta a$

2) Take the difference between $a(x)$ and $b$;
$c \pm \delta c = (a(x) \pm \delta a) - (b \pm \delta b)$

3) if $c \leq \delta c$ (if the difference can be $0$), add it to an array that stores the values of the $x$.

Now after this we have some values of $x$ that satisfy $c \leq \delta c$. How can you find $x$ and $\delta x$ from this array

 

[BvU]

Some caveats are in order:

1) You want to be sure $\Delta a$ and $\Delta b$ are uncorrelated
2) You mean $\ | c | \leq \Delta c\$ of course.

And the 'for each $x$' can lead to a wrong result. Numerical example:

a = 400 ± 40, b = 360 ± 20, so c = 40 ± 45

And let a = 200 x $\Rightarrow$ x = 2.0 ± 0.2

Your array of x is, for example, 1.78, 1.79, ... 2.22.
Then the average is 2.0 and the standard deviation 0.13

Reason: your $x$ has a flat distribution instead of a normal distribution$\$

 

[Arman777]

1) You want to be sure Δa and Δb are uncorrelated
2) You mean |c|≤Δc of course.

Yes

Reason: your x has a flat distribution instead of a normal distribution

Yes I have noticed that...that seems like a trouble

And the 'for each x' can lead to a wrong result. Numerical example:

a = 400 ± 40, b = 360 ± 20, so c = 40 ± 45

And let a = 200 x ⇒ x = 2.0 ± 0.2

Your array of x is, for example, 1.78, 1.79, ... 2.22.

I mean a different way actually. Think like this,

You have a some range of $x$ values from $0$ to $2$. Then you are doing a loop such that
for each value of $x$:
$c \pm \delta c = (a(x) \pm \delta a) - (b \pm \delta b)$
if $|c|≤Δc$:
add x to an array

 

[Arman777]

I have revised the problem

 

[BvU]

I have revised the problem

Makes my reply quasi-incomprehensible !

Even so: some more editing is in order

 

[Arman777]

Makes my reply quasi-incomprehensible !

Even so: some more editing is in order

Waiting

 

[BvU]

Waiting for ? Ah - $\TeX$ details adjusted.

Then you are doing a loop such that

That doesn't tell us what the probability distribution of $x$ is going to be.

 

[Vanadium 50]

I have revised the problem

Please don't do that. Because...

Makes my reply quasi-incomprehensible !

 

[Arman777]

Please don't do that. Because...

sorry about that

 

[Arman777]

Waiting for ? Ah - $\TeX$ details adjusted.
That doesn't tell us what the probability distribution of $x$ is going to be.

Well I have printed the values and they were normally distributed

 

[BvU]

So how can that be ? Pure luck, or something else, like the dependency of $a$ on $x$ ?
Or some monte carlo principle ?

 

[Arman777]

So how can that be ? Pure luck, or something else, like the dependency of $a$ on $x$ ?
Or some monte carlo principle ?

Its actually has the similar idea as your. The relationship between $x$ and $a$ is linear. So as we increase $x$ at some point it hits $-\delta c$ and another point it hits $+\delta c$. The problem how can I calculate $x$ and $\delta x$ from the given data list.

This is the histpgram of the $x$ values.

 

[BvU]

Looks pretty uniform to me

 

[Arman777]

Looks pretty uniform to me

So mean(x_values) and std(x_values) are the correct solution ?

 

[BvU]

According to the numerical example, mean would be ok, st dev would be too optimistic

 

[Arman777]

According to the numerical example, mean would be ok, st dev would be too optimistic

I looked the uniform distribution std and it says
$$\sigma = \sqrt{\frac{(b-a)^2}{12}}$$

$b = max(xarray)$, $a = min(xarray)$

How to Calculate the Variance and Standard Deviation in the Uniform Distribution - dummies

 

[BvU]

Yes, I know. In the example I started with 2$\pm$0.2. The flat distribution gave a sigma of 0.13, about a factor of $\sqrt 3$ lower. Your procedure cuts at $\pm\sigma$ -- whereas a gauss distribution has , what was it again, 68% within $\pm\sigma$ and the remainder outside, contributing happily to the standard deviation.