Formulating the likelihood in an LRT involving geometric and exp distribution

Question

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.)

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.