# I Experimental bound on the scalar spectral index

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1. Jun 21, 2017

### shinobi20

Based on the paper by Visinelli (https://arxiv.org/abs/1605.06449),

He stated in page 6 that the scalar spectral index as given by the Planck 2013 data (https://arxiv.org/abs/1303.5076) is,

$n_s = 0.9655 \pm 0.0062~~$ ($68\%$ C.L.)

but when I looked into the Planck 2013 paper, I did not see that constraint, so I tried to read the much later paper by the Planck team which is the Planck 2015 (arxiv.org/abs/1502.02114) and I saw it, also it is very strange for Visinelli to cite the 2013 results given that his paper had been published in 2016, so it makes more sense for him to cite the 2015 paper by the Planck team.

The problem is, he also stated in Figure 1. on page 13 that "the light blue band showed the $95\%$ confidence region" but he did not state any range for it unlike the $68\%$ above. I tried to search for it in the Planck 2013 and 2015 paper but I can't seem to find the range given by the Planck team for the $95\%$ confidence region. Can anyone help me with this?

Also, I noticed that in Visinelli's paper, equation (3.10) seems wrong, I think it should be $n_s -1$ instead of $1-n_s$ since if you try to compute for $n_s$ it would give an answer that is out of the range (far out). Any comment?

2. Jun 21, 2017

### kimbyd

The $1-n_s$ looks like a typo to me as well, as all of their other equations on the page put it in terms of $n_s-1$. But since $n_s-1 = -(1-n_s)$, the end result won't be that dramatic if it's wrong: you'd get $n_s = 1.0345$ instead of $n_s = 0.9655$.

I wouldn't worry too much about them using the older result. They were probably working on the paper for a while, and when they started the new results weren't out. Or it could be a simple oversight. Either way, a slight difference in the value and error bars shouldn't make much difference to the results.

3. Jun 21, 2017

### shinobi20

Thanks for that but my question is where to find the $95\%$ C.L. bound for $n_s$? How did they come up with that if the Planck paper did not mention it?

4. Jun 21, 2017

### kimbyd

Why do you need the 95% confidence bound?

In the Gaussian approximation, the 95% confidence bound will be approximately twice the 68% confidence bound. One solution may be to just estimate it yourself using the released data products, but that's a pretty big task to perform.

5. Jun 21, 2017

### George Jones

Staff Emeritus
I don't know what the author actually did, but this does give the result shown in Figure 1 of the paper.

In more detail, assume: 1) Gaussian; 2) $n_s = 0.9655 \pm 0.0062$ (68.27% CL). These assumptions then give that $0.0062$ is one standard deviation, and that the 95.45% CL is given by two standard deviations. Consequently, $n_s = 0.9655 \pm 0.0124$ (95.45% CL), i.e.,

$$0.9531 < n_s < 0.9779,$$

which are the bounds in the figure.