Here is a question from old exam papers I am having trouble with:

Suppose the lifetime of a bulb is distributed exponentially with mean life $\theta$ (in hours). Let $X_i$ be the number of trials required to get a bulb surviving at least $t$ (known) hours for the first time in the $i$th lot, where lot sizes are large enough. Discuss a likelihood ratio test for testing $H_0:\theta\ge\theta_0$ against $H_1:\theta<\theta_0$.

I am not sure what exactly my observed sample is supposed to be. So I am stuck at setting up the problem correctly.

Say I have $k$ lots of bulbs, the $i$th lot having size $n_i\,, i=1,2\ldots,k$.

Then is the observed sample some $\mathbf X=(X_1,X_2,\ldots,X_k)$, where $X_i=(X_{ij})_{i=1,\ldots,k\,;\,j=1,\ldots,n_i}$ is again a vector for each $i$ ? I am using $X_{ij}$ to denote the number of trials needed to get the $j$th bulb surviving for at least $t$ hours for the first time in the $i$th lot.

If I assume $X_1,X_2,\ldots,X_k$ are all independent samples, then my likelihood function should be

$$L(\theta\mid \mathbf x)=\prod_{i=1}^k\prod_{j=1}^{n_i}p(\theta)(1-p(\theta))^{x_{ij}-1}\quad,\,\theta>0$$

, where $p(\theta)=\int_t^{\infty}\frac{1}{\theta}e^{-z/\theta}\,dz=e^{-t/\theta}\,,t>0$ is the success probability.

But this might be wrong since I am supposed to have a sample consisting of only $X_1,\ldots,X_k$ where each $X_i$ denotes number of trials (so not vectors by themselves). In the above setup, this does not seem to be the case.

If this formulation is wrong, I would like a hint on how to arrive at the correct one. And also, where exactly am I supposed to apply a large sample approximation with large $n_i$s? Is this referring to Wilks' theorem on the asymptotic distribution of the log-likelihood ratio statistic?

(I am not looking for a solution to the actual problem.)


1 Answer 1


I would formulate your likelihood based on the geometric distribution. What you are trying to estimate is $p(\theta, t)$, where $p$ is the probability of a bulb surviving at least $t$ hours given the parameter $\theta$ of the exponential distribution, as you have realized. Given this formulation, you are observing $x_i$, the number of "failures" before your first (and only) success for each of $k$ lots; that's your data. Since your last trial within each lot is the success (as the last bulb tried is the one that survives for at least $t$ hours), the number of failures is just the number of trials $-1$.

The likelihood function for the $i^{th}$ lot is:

$$\mathcal{L}(p;x_i) = (1-p)^{x_i}p$$

and the likelihood for $k$ lots is just the product of the $k$ individual likelihoods (since the parameter $p$ is the same and the lots are independent):

$$\mathcal{L}(p; x) = (1-p)^{\sum_{i=1}^k x_i}p^k$$

Note that this is actually the likelihood function for a negative binomial $(k,p)$ distribution, because you have $k$ independent lots.

Hints: You can easily convert the likelihood ratio test for $\theta$ to one for $p$, since the relationship between $p$ and $\theta$ is strictly monotonic, or you can construct a likelihood ratio test for $p$ and transform it into one for $\theta$. You can use a large sample approximation for the distribution of the MLE of $p$ if $k$ and $\sum x_i$ are large enough, say, both $> 5$ (as a possibly-discredited rule of thumb), to simplify this latter approach to the more common Normal distribution problem.

  • $\begingroup$ Thanks. Is $k$ also supposed to be the total lot size as in my question? So I should not consider individual lots but only the entire lot as a whole? $\endgroup$ Commented Mar 22, 2019 at 7:08
  • $\begingroup$ And how does the number of trials $x_i$ become the the number of failures? $\endgroup$ Commented Mar 22, 2019 at 7:18
  • 1
    $\begingroup$ I've edited the question, hopefully answering your questions. $\endgroup$
    – jbowman
    Commented Mar 22, 2019 at 19:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.