Suppose that $X_1, X_2, \ldots, X_n$ (with $n > 0$) is a random sample from a non-central exponential distribution with probability density function:

$$f(x | \lambda, k) = \lambda * e^{-\lambda*(x-k)}$$

Both the scale parameter ($\lambda$) and the shift parameter ($k$) are unknown, with: $k < x < \infty$ and $\lambda > 0$ .

For the previously mentioned sample ($X_1, X_2, ... , X_n$) and $k$, the maximum likelihood estimate (MLE) is:

$$K = min_i X_i$$

Question : how would a $95\%$ confidence interval (for the shift parameter, $k$) look like ? I'd imagine it must be something like $[K-\epsilon, K]$, but how can we figure out the value for "$\epsilon$".

Thank you !


2 Answers 2


Note that it's quite easy to work out the distribution of $K(X)=\;\stackrel{_\min}{_i}X_i$, and so to identify $Q(X,k)\,=\,k-K(X)$ as a pivotal quantity.

From there, you can immediately pass to a confidence interval, by placing the limits on $Q$ so that you get 95% of the probability inside them and then manipulating the resulting interval to make $k$ the subject of the pair of inequalities.

You might put all the risk on that one side, as you did, or split the probability evenly, or whatever other approach you choose (as you always can with confidence intervals).

Your choice should (on average) produce the shortest interval, I think, and makes good sense in this case, but it's not the only choice.


$f_X(x|\lambda, k) = n\lambda^n e^{-n\lambda(x-k)} I_{[k, \infty)}(x)$

Then pivoting the continuous pdf:

Let $k_L(x)$ and $k_U(x)$ satisfy:

$\int_{k_U(x)}^x n\lambda^n e^{-n\lambda(x-k_U(x))} dx = \frac{\alpha}{2}$ and $\int_{x}^\infty n\lambda^n e^{-n\lambda(x-k_L(x))} dx = 1-\frac{\alpha}{2}$

Thus, $1-e^{-n\lambda(x-k_U(x))} = \frac{\alpha}{2\lambda^{n-1}}$ and $e^{-n\lambda(x-k_L(x)} = \frac{1-\frac{\alpha}{2}}{\lambda^{n-1}}$

Solve for $k_U(x)$ and $k_L(x)$


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