# Formulating the likelihood in an LRT involving geometric and exp distribution

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.)

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.

• 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? – StubbornAtom Mar 22 at 7:08
• And how does the number of trials $x_i$ become the the number of failures? – StubbornAtom Mar 22 at 7:18
• I've edited the question, hopefully answering your questions. – jbowman Mar 22 at 19:10