Suppose that $X_1,X_2,…,X_n$ is a random sample from the exponential distribution with rate $\lambda\gt0$.

Contruct a 95% confidence interval for $\lambda$.

Step One: Choose a statistic.

Choose one whose distribute you know and is one that depends on the unknown parameter.

We have the sample mean:

\[\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i\]

We know that:

  • $\sum_{i=1}^{n}X_i\sim\Gamma(n,\lambda)$
  • $X\sim\Gamma(\alpha,\beta)\quad c\gt0$
    $\implies cX \sim\Gamma(\alpha,\beta/c)$

  • So, if we take the sample mean of $n$ iid random variable, we get a Gamma distribution below:

    \[\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i\sim\Gamma(n,n\lambda)\]

Step Two: Find a function of the statistic and the parameter you are trying to estimate whose distribution is known and parameter free.

\[\bar{X}\sim\Gamma(n,n\lambda)\]

To get rid of the unknown parameter $\lambda$

\[\lambda\bar{X}\sim\Gamma(n,n)\]

How do we find critical values for the $\Gamma(n,n)$ distribution? We can get them by using R:

qgamma(0.975,n,n)
qgamma(0.025,n,n)

Or we can do by transforming into chi-squared

Suppose that $X_1,X_2,…,X_n$ is a random sample from the exponential distribution with rate $\lambda\gt0$.

Contruct a 95% confidence interval for $\lambda$.

Step One: Choose a statistic.

Choose one whose distribute you know and is one that depends on the unknown parameter.

We have the sample mean:

\[\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_i\]

Step Two: Find a function of the statistic and the parameter you are trying to estimate whose distribution is known and parameter free.

\[\begin{align} \bar{X}\sim\Gamma(n,n\lambda)&\implies\lambda\bar{X}\sim\Gamma(n,n)\\ &\implies 2n\lambda\bar{X}\sim\Gamma\bigg(n,\frac{1}{2}\bigg)\\ &=\Gamma\bigg(\frac{2n}{2},\frac{1}{2}\bigg)\\ &=\chi^2(2n) \end{align}\]

Step Three: Find appropriate critical values. We put 0.025 in both tails:

png

Step Four: Put your statistic form Step Two between the critical values and solve for the unknown parameter “in the middle”.

\[2n\lambda\bar{X}\sim\chi^2(2n)\\ \Downarrow\\ \chi_{0.975,2n}^2\lt2n\lambda\bar{X}\lt\chi_{0.025,2n}^2\\ \Downarrow\\ \frac{\chi_{0.975,2n}^2}{2n\bar{X}}\lt\lambda\lt\frac{\chi_{0.025,2n}^2}{2n\bar{X}}\]