# Confidence Intervals of not Gaussian functions

Is anybody know a good tutorial about how we calculate Confidence Intervals of not Gaussian functions? I give some example of what I kind of function I think about:

1st example: Let be $$X_1, X_2 \dots X_n$$ i.i.d. $$Exp(\lambda)$$ random variables, where is $$\lambda$$ unknown. We observe $$min(X_i)$$. I would like to understand, how we calculate the confidence interval for $$\Theta = 1/\lambda$$.

2nd. example: Let be $$X_1, X_2 \dots X_n$$ i.i.d. $$Uni(0,\Theta)$$ random variables, where is $$\Theta$$ unknown. We observe $$max(X_i)$$. I would like to understand, how we calculate the confidence interval $$\Theta$$.

I am OK, to calculate the CDF of these max, min etc. function, but after this, I do not know how to progress.

My statistic knowledge kind of limited, so I am looking for some explanation where they do not just give a branch of math formula but also they explain in plain English.

• Just a potentially relevant question: How do you construct a confidence interval for the mean of 5 iid Gaussian observations with known variance and unknown mean? – Math-fun Feb 22 at 9:30
• I do not really see, how this connect, but I think in the Gaussian case. Estimating the unknow $\mu$, $n=5$ and know $\sigma^2$. will be a Gaussian $E[\bar{X}_n] = \mu$ and $var(\bar{X}_n) = \sigma^2/n$. So for the $\mu$ we get a $N(\mu, \sigma^2/n )$ from here is rutin to find the confidence intervals. – arkhein Feb 22 at 9:53
• @Math-fun aims at the following: For normal variables you used your assumptions to determine the distribution of your estimator and using this distribution you arrive at a confidence interval. Can you do the same for your examples? – Stefan Feb 22 at 10:49
• For the first example, you can rescale $\sum_{i=1}^n X_i$ into a chi-square distributed pivotal quantity. Finding a pivot is also key in the second example. – Jarle Tufto Feb 22 at 12:52
• Ok, I write down what I did so far, but I still struggle. Let stay with the 2nd. example. So the $max(x_n)$ I will mark as $M$. This is our estimator for the $\Theta$. The $CDF(M)= (x/\Theta)^n$, so the $PDF(M) = (n*x^{n-1}/\Theta^n)$, which lead to $E[M] = \int_0^{\Theta} PDF(M) * x dx = (n*\Theta) /(n+1)$, I could count the Var(M) also, with the $E[M^2]-(E[M])^2$, but it is not looking nice, so I have doubt I am on the right way. Also the estimator looking biasd. – arkhein Feb 22 at 19:52