Question about errors, Hubble's constant

[Matt atkinson]

Homework Statement

I am just looking through some old notes I have from for cosmology, and there's something cropped up that i can't seem to figure out:

Say I have two (or more) values for $H_o$ each with errors such as:

$$H_{o_1}=70^{+a+b}_{-c-d}$$
and

$$H_{o_2}=69^{+e+f}_{-g-h}$$

How would I go about calculating the weighted averaged (a,c,e,g are statistical errors. The rest are systematic errors) and then uncerstainty on the weighted average when for instance $a\neq c$.

Homework Equations

All the formula i found are along the lines of:

$$\bar{x}=(\sum^{N}_{i=1}x_i/\sigma_i^2)/(\sum^{N}_{i=1}1/\sigma_i^2)$$

$$\sigma_{\bar{x}}=\sqrt{1/(\sum^{N}_{i=1}1/\sigma_i^2})$$

The Attempt at a Solution

I've attempted to workout the top uncertainty on it's own, and likewise with the bottom but that doesn't seem the right way to do it.

 

[mfb]

To do it properly, you first have to know about the correlations between the systematic uncertainties. Then you can get the likelihood functions of the individual measurements, combine them, and then extract central value and uncertainties from that again.
If you just have access to the given numbers and expect that the correlation is small, the quick and dirty weighted average should give some reasonable approximation. The uncertainty of the weighted average follows from the usual uncertainty propagation.

 

[Matt atkinson]

Ah okay thank you, I have just been given numbers and no correlation and been told to make an assumption. So I should say that if i assume the correlation between systematic uncertainties is small.

So to work that out, let's say I have:

$$76.9^{+3.9+10}_{-3.4-8}$$
$$66^{+11+9}_{-10-8}$$
How would I go about using that in the formula I have above for the weighted average. for instance what would I use for $\sigma_1$ when its values for the upper and lower uncertainties differ.

 

[mfb]

I would probably use the average of the upwards and downwards uncertainty for the weights. If those the uncertainties are too asymmetric, this simplified approach will fail anyway.

 

[Matt atkinson]

Okay thankyou! ill give it a try now, i did have attempt at doing each on their own it gave a strange result so ill try taking the average.