Hello,

I need to fit the exponential decay law to some data with weights. I tried to use LsqFit.jl, however the `standard_errors`

function returns too low values. I discovered it by comparing my results with my friend and I also made the fit in R.

Hello,

I need to fit the exponential decay law to some data with weights. I tried to use LsqFit.jl, however the `standard_errors`

function returns too low values. I discovered it by comparing my results with my friend and I also made the fit in R.

Well, if you dont’t show us your code nobody will be able to help you. ):

I am sorry, I did not think it would be useful as I just called the relevant functions, but anyway here it is:

```
using LsqFit
x = [0, 975, 1767, 2682, 5959]
y = [5490.0, 4546.0, 3679.0, 3134.0, 1451.0]
weights = [109.0, 102.0, 96.0, 89.0, 71.0]
@. model(t, p) = p[1]*(0.5)^(t/p[2])
fit = curve_fit(model, x, y, weights, [0.1, 80])
err = round(standard_errors(fit)[2], sigdigits=1)
T = fit.param[2]
println("$T +- $err")
```

Just to clarify, the `weights`

variable contains the stderr of measurement of y.

Sorry, please provide a piece of code that is completely defined. I tried your code and get the message:

```
julia> @. model(t, p) = p[1]*(0.5)^(t/p[2])
model (generic function with 1 method)
julia> fit = curve_fit(model, x, y, weights, [0.1, 60])
ERROR: UndefVarError: x not defined
Stacktrace:
[1] top-level scope
@ REPL[3]:1
```

What are x, y and the weights?

In particular:

"

Do your best to make your example **self-contained** (“minimal working example”, **MWE** ), so that it runs (or gets to the error that you want help with) as is. This means including package loading (e.g. `using ThatPackage`

) and any **data** that the code operates on. If your data is large or proprietary, generate **example data** if possible and include that. This StackOverflow answer explains how a DataFrame can be turned into a string representation that can be copy/pasted.

"

This is probably the issue. In `LsqFit.jl`

, the weights are assumed to be the inverse of the variances.

1 Like

I edited the code in my previous post.

I tried that also, but the error is still suspiciously low.

Using the inverse of the variances as weights, we obtain ` using LsqFit`

:

```
p1 = 5541 +/- 86
p2 = 3135 +/- 110
```

Which from the QC plot below does not seem unreasonable.

What values do the other software packages compute?

1 Like

Here is my take on it

```
using LsqFit
x = [0, 975, 1767, 2682, 5959]
y = [5490.0, 4546.0, 3679.0, 3134.0, 1451.0]
weights = [109.0, 102.0, 96.0, 89.0, 71.0]
@. model(t, p) = p[1]*(0.5)^(t/p[2])
fit = curve_fit(model, x, y, [0.1, 80])
err = round(standard_errors(fit)[2], sigdigits=1)
T = fit.param[2]
println("$T +- $err")
@. model(t, p) = p[1]*(0.5)^(t/p[2])
fit = curve_fit(model, x, y, weights, [0.1, 80])
err = round(standard_errors(fit)[2], sigdigits=1)
T = fit.param[2]
println("$T +- $err")
@. model(t, p) = p[1]*(0.5)^(t/p[2])
fit = curve_fit(model, x, y, 1 ./ (weights .* weights), [0.1, 80])
err = round(standard_errors(fit)[2], sigdigits=1)
T = fit.param[2]
println("$T +- $err")
```

yielding

```
3158.802386807851 +- 100.0
3170.773189838312 +- 0.1
3134.79231889205 +- 100.0
```

It is much more clear to write:

```
σy = [109., 102., 96., 89., 71.] # standard deviations of y
w = 1 ./ σy.^2 # weights for LsqFit
```

1 Like

I see the problem now, I used `1/σ`

instead of `1/σ^2`

. Thank you, for your help!