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

Construct a 95% confidence interval for $\lambda$ based on the minimum value in the sample.

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Let $Y_n=min(X_1,X_2,…,X_n)$.

The cdf for each $X_i$ is

\[F(x) = P(X_i\le x)=1-e^{-\lambda x}\]

The cdf for $Y_n$ is

\[\begin{align} F_{Y_n}(y)&=P(Y_n\le y)\\ &=P(min(X_1,X_2,...,X_n)\le y) \end{align}\]

If we choose $x$’s like the way below, there’s a lot of way for $x$’s to fall so that the minimum is less that or equal to $y$ it’s going to be hard to translate this about the minimum to a statement involving where the individual $x$’s fall with respect to this fixed number $y$.

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But it will be easier if we look at the complement of this event. The minimum in a sample is greater than a fixed number $y$, if and only if every number in the sample is greater than $y$.

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\[\begin{align} F_{Y_n}(y)&=P(Y_n\le y)\\ &=P(min(X_1,X_2,...,X_n)\le y)\\ &=1-P(min(X_1,X_2,...,X_n)\gt y)\\ &=1-P(X_1\gt y,X_2\gt y,...,X_n\gt y)\\ &\stackrel{indep}{=}1-P(X_1\gt y).P(X_1\gt y)...P(X_n\gt y)\\ &\stackrel{ident}{=}1-[P(X_1\gt y)]^n = 1-[1-F(y)]^n\\ &=1-[1-(1-e^{-\lambda y}]^n\\ &=1-[e^{-\lambda y}]^n \end{align}\]

If we take the derivative and we get the pdf of an exponential with paramenter $n\lambda$

\[f(Y_n(y) = \frac{d}{dx}F_{Y_n}(y)=n\lambda e^{-n\lambda y}\]

The minimum of $n$ iid exponential with rate $\lambda$ is exponential with rate $n\lambda$!

\[Y_n\sim exp(rate=n\lambda)\]

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

Construct a 95% confidence interval for $\lambda$ based on the minimum value in the sample.

Step One: Choose a statistic.

\[Y_n=min(X_1,X_2,...,X_n)\]

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

\[\begin{align} Y_n&=min(X_1,X_2,...,X_n)\sim exp(rate=n\lambda)\\ &=\Gamma(1,n\lambda)\\ &\implies n\lambda Y_n\sim\Gamma(1,1) = exp(rate=1) \end{align}\]

Step Three: Find appropriate critical values.

\[n\lambda Y_n\sim exp(rate=1)\]

Now we want to intergrate the exponential rate $1$ distribution up to some number that gives us area $0.95$ to the left

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To find the number which we’re calling question mask, we need to integrate the exponential rate $1$ pdf from 0 to the question mask and set that equal to $0.95$.

\[\int_0^? e^{-x}dx=0.95\\ \Downarrow\\ 1-e^{-?}=0.95\\ \Downarrow\\ ?=-\ln(0.05)\]

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

\[0\lt n\lambda Y_n\lt-\ln(0.05)\\ \Downarrow\\ \bigg(0,\frac{-\ln(0.05)}{nY_n}\bigg)\]

Where $Y_n=min(X_1,X_2,…,X_n)$.