Find Critical region for exponential distribution

Question

The lifetime in hours of each bulb manufactured by a particular company follows an independent exponential distribution with mean $ \theta $ . To test the hypothesis $ \ H_0 : 2000 $ versus $ \ H_1 : 1000 $ an experimenter sets up an experiment with 50 bulbs with 5 bulbs in each of 10 different locations to examine their lifetimes. To get quick preliminary results, the experimenter decides to stop the experiment as soon as one bulb fails at each location. Let Y, denote the lifetime of the first bulb to fail at location i, i = 1, 2, ..., 10.

Obtain the most powerful test of size 0.05 to test $ \ H_0 $, versus $ \ H_1 $ based on the available lifetimes.

My attempt with the help of the hint provided

Since $ \ Y_i = min(\ X_1,\ X_2 ,\ X_3 ,\ X_4,\ X_5) $

$ P(min(X_1,...,X_p) < t) = 1 - P(min(X_1,...,X_p) > t)= 1 - P(X_1 > t, ..., X_p > t) $

$ 1 - (1 - P(X_1 < t) \times ... \times (1 - P(X_p < t)) = 1 - \left(1 - P(X_i < t)\right)^p $

After this $ \ P(X_i < t) = \int_0^t e^-\theta d\theta $

$ \ 1- \ e^-t $ Now if I put this value of cdf in above equation

$ 1 - \left(1 - 1+ \ e^-t \right)^p $

$ 1 - e^-{p\theta} $

Now if I differentiate this cdf to obtain the pdf of $\ Y_i $ then it is $ exp(-p\theta) $

Here $ p = 5 $ So $\ Y_i $ ~ $ exp(-5\theta) $

$ \sum_{i=0}^{10} \ Y_i $ ~ $gamma(10,5\theta ) $

Answer 1

This seems like a HW question, so I'll try not give the full answer. Break down the question into two parts.

1. What does it mean he reports the time for the first light bulb to fall at location? That means that for location $i$ he reports the $Y_i = min(X_1, ..., X_5)$, how is it distributed given that $X_j \sim exp(\theta)$?

We will find the cumulative distribution function:

$P(min(X_1,...,X_p) < t) = 1 - P(min(X_1,...,X_p) > t)= 1 - P(X_1 > t, ..., X_p > t)$

Since we were given in the question that the bulbs are independent,

$1 - (1 - P(X_1 < t) \times ... \times (1 - P(X_p < t)) = 1 - \left(1 - P(X_i < t)\right)^p $

Since $X_i \sim exp(\theta)$,

$1 - (1 - \exp(-p \theta))$

Therefore, the minimum of the light bulb life

$Y_i \sim exp(p \theta)$

In the question you asked, $p = 10$ as 10 locations are reported. From here continue finding the MP test using the likelihood ratio (LR) and find for which statistic the LR is monotonic and how it is distributed.

The sum of exponential r.v is distributed as Gamma. See https://en.wikipedia.org/wiki/Gamma_distribution#Related_distributions