# probability – Determining the confidence interval

We observe a simple random sample $$(X_1,ldots , X_n$$), where the distribution of observed property $$X$$ is given by its density: $$f(x)=frac{1}{xsigma sqrt{2pi}}e^{-frac{(ln x-m)^2}{2sigma^2}}, x>0$$.

Determine an 80% confidence interval for $$sigma$$, if we know:

Mean value of logarithms of the values from the dataset that we get deviates from its expected value no more than $$frac{sigma}{2}$$, with probability 0.98758;

If $$(x_1, ldots ,x_n)$$ is our dataset (a realisation of our sample), $$sum_{i=1}^n ln x_i=3.139797$$ and $$sum_{i=1}^n (ln x_i)^2=35.63178$$.

My attempt:

It’s not hard to see that, if $$Y=ln X$$, then $$Y$$ has $$mathscr{N}(m,sigma^2)$$ distribution. Now we work with $$Y$$. Let
$$overline{Y}_n=frac{1}{n}sum_{i=1}^n Y_i$$
$$overline{S}_n^2=frac{1}{n}sum_{i=1}^n (Y_i-overline{Y}_n)^2$$.
We know that $$frac{noverline{S}_n^2}{sigma^2}$$ has $$chi_{n-1}^2$$ distribution.

If we look for confidence interval in the form:
$$p{U_nleq sigma leq V_n}=0.8$$, we can actually demand $$p{frac{noverline{S}_n^2}{V_n^2}leq frac{noverline{S}_n^2}{sigma^2} leq frac{noverline{S}_n^2}{U_n^2} }=0.8$$.

We can demand:
$$p{frac{noverline{S}_n^2}{V_n^2}>frac{noverline{S}_n^2}{sigma^2}}=p{frac{noverline{S}_n^2}{U_n^2}.
This gives us (from what we know about $$chi_{n-1}^2$$ distribution):
$$V_n^2=frac{noverline{S}_n^2}{chi_{n-1,0.9}^2}$$
$$U_n^2=frac{noverline{S}_n^2}{chi_{n-1,0.1}^2}$$.

I assume that the info given in this problem means that one confidence interval for $$frac{1}{n}sum_{i=1}^n ln x_i-m$$, with confidence level $$0.98758$$ is $$(-frac{sigma}{2}, frac{sigma}{2})$$. But I don’t know how to move on from here.

We also know that for:
$$overline{S}_*^2=frac{1}{n}sum_{i=1}^n (Y_i-m)^2$$
$$frac{noverline{S}_*^2}{sigma^2}$$ has $$chi_n^2$$ distribution. I thought maybe we should use this because of this CI that has been given to us, but I don’t know how to do that.

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