Q&A Bank part 3 Question 3.11

[Simon C]

Hello all

The answer to this question outlines using a binomial distribution to estimate the proportion of drives working after 47 days. It then goes on to equate this estimate to the exponential distribution.

The solution quotes:

“Expressed in terms of the exponential parameter λ, p is:”

My question is how are these two equal? i.e. p=exp(-200λ).

In particular I am not clear why we start integrating at 200 rather than 0.

Thanks
Simon

 

[John Lee]

The answer to this question outlines using a binomial distribution to estimate the proportion of drives working after 47 days. It then goes on to equate this estimate to the exponential distribution.

The solution quotes:

“Expressed in terms of the exponential parameter λ, p is:”

My question is how are these two equal? i.e. p=exp(-200λ).

In particular I am not clear why we start integrating at 200 rather than 0.

This kind of question is actually quite common.

Essentially the likelihood is the probability of us observing our data set of 3 less than 200 days and 47 greater than 200 days which is:

L = constant × [P(X<200)]³ × [P(X>200)]^47

The constant is for the different combinations - which 3 results from the 50 were less than 200 days.

Now we could set p = P(X>200) and then estimate this, but frankly I'd just skip this out as it's an unnecessary complication.

To obtain probabilities for an exponential distribution we can either integrate or use the distribution function (given on page 11 of the Tables).

P(X>200) = integral from 200 upwards of f(x)

or

P(X>200) = 1 - P(X<200) = 1 - F(200)

An exam question similar to this is S02 Q12

 

[Simon C]

Thanks John.

I followed where you were saying:
L = constant × [P(X<200)]³ × [P(X>200)]^47

However I'm still not clear on how you have gone from this to evaluating just P(X>200). Where has [P(X<200)]³ gone?

Thanks
Simon

 

[John Lee]

Thanks John.

I followed where you were saying:
L = constant × [P(X<200)]³ × [P(X>200)]^47

However I'm still not clear on how you have gone from this to evaluating just P(X>200). Where has [P(X<200)]³ gone?

OK - let's ignore the solution for a second and carry on just considering the exponential distribution.

Using l for lambda, we have:

L = constant × [P(X<200)]³ × [P(X>200)]^47
= constant × [1 - exp(-200l)]³ × [exp(-200l)]^47
= constant × [1 - exp(-200l)]³ × [exp(-9400l)]

ln L = constant + 3ln(1 - exp(-200l)) - 9400l

So d/dl ln L is:

600exp(-200l) - 9400
1 - exp(-200l)

Setting this equal to zero gives:

600exp(-200l) = 9400
1 - exp(-200l)

600exp(-200l) = 9400[1 - exp(-200l)]

Fiddling gives:

10000exp(-200l) = 9400

exp(-200l) = 0.94

-200l = ln 0.94

l = 0.000309

Hence the mean is 1/l = 3232 as before.

What the solution did was obtain the MLE for p (p hat = 0.94), then compare the two likelihoods and say "ah - but p is exp(-200l) - ie l = -(ln p)/200 so by using the invariance property of estimators l hat = -(ln p hat)/200"

Either method will give the marks use the one you feel happier with.

Hope this sorts it better!