1
$\begingroup$

If I have one random variable that represents hours worked per job X~exponential($\theta$). I have another random variable that represents how many jobs obtained per month Y~Poisson($\lambda$). Using bayes I have used inverse gamma on the exponential and gamma on the Poisson. I now have two gamma distributions. How would I now use those distributions that obtain the probability that the total number of hours worked in a year is Z? p(Z>2000) for example? I imagine I have to use joint probabilities such that $X \times Y > 2000$ and divide that by the total joint probabilities of the two distributions. Given that I have obtained these two distributions? Does anybody know how I would do this in R?

would I be using dgamma for each distribution? How would I then multiply them together to get the total range of possible total hours worked?

I have to multiply all the possible values of one distribution by all the possible values of the other distribution, so I imagine with my two random variables, my function would look like this.

$$f_{XY}(x,y) = xy$$

$$f_{XY}(x,y) = \text{dgamma()*dgamma()}$$ with appropriate parameters?

So I want something like

$$P(X \times Y > 2000) = \int\int f_{XY}(x,y)dxdy$$

$$P(X \times Y > 2000) = \int\int xy \: dxdy$$

Can anyone help?

$\endgroup$
  • $\begingroup$ You don't have enough information to obtain a unique answer or even a narrow range of them. You need to know or make strong assumptions about how the variables are associated. $\endgroup$ – whuber Oct 2 '18 at 15:09
  • $\begingroup$ I see @whuber. The only assumptions are that months are 30 days each really. if the mean number of jobs coming through is 15 and the mean length of a job is say 4 hours. i'm not sure what other assumptions I would need, but I have not been given any others for this question, except for independence and stationary increments. $\endgroup$ – Bucephalus Oct 2 '18 at 15:26
  • $\begingroup$ Independence is a very strong assumption (and likely not realistic in this case) and hasn't been stated in your question. $\endgroup$ – whuber Oct 2 '18 at 15:31
  • $\begingroup$ Yes, @whuber, a lot of undergrad university questions are not realistic, agreed. They are merely designed for pedagogical purposes. $\endgroup$ – Bucephalus Oct 2 '18 at 15:34
  • $\begingroup$ For this to make any sense, you'd be generating a different exponential variate for each job, not the same one (why would every job take exactly the same time?) -- so not $X\times Y$ but $\sum_{i=1}^Y X_i$. $\endgroup$ – Glen_b -Reinstate Monica Oct 2 '18 at 15:52
2
$\begingroup$

Total number of hours per year (Z) is a function dependent on the number of hours per job and the number of jobs per months (and we have 12 months).

Z = X(Y(12))

I would use iterate some this expression with your lambdas and rates:

rexp(rpois(12, lambda = 0.5), rate = 0.2)

To estimate the distribution of Z, and then calculate the probability of Z > 2000 from that distribution. Also, as sanity check, the total amount of hours cannot be above 365*24, so I would delete any estimation above 8760 hours of threshold. Example:

iterations <- 1000
Z <- vapply(seq_len(iterations), function(x){sum(rexp(rpois(12, lambda = 0.5), rate = 0.2))}, numeric(1L))
hist(Z)

enter image description here

And then you can calculate the probability with sum(Z >2000)/length(Z)

$\endgroup$
  • $\begingroup$ Thanks @Llopis. That's helpful. But I have to use the bayesian posterior I believe, to satisfy this question. $\endgroup$ – Bucephalus Oct 2 '18 at 15:23
  • $\begingroup$ I don't understand why you need the posterior probability. Could you explain why do you believe so? The problem as you define it is a composition of functions, not a joint distribution. $\endgroup$ – llrs Oct 2 '18 at 15:38
  • $\begingroup$ Because my lecturer wants it done that way. But it's a good question you pose because I was thinking this earlier today, can't it just be done similar to what you suggest? And then, what the hell do we need Bayesian statistics for? I but I'm too ignorant at this stage to answer these questions....except that this question is supposed to be interrogating our knowledge of simple bayesian techniques, hence the analytical priors. Thanks @Llopis. $\endgroup$ – Bucephalus Oct 2 '18 at 15:42
  • $\begingroup$ it's 1:45am here, I'm going to bed. As @whuber indicated, I probably have not given enough information to answer this satisfactorily. I will close the question tomorrow if you wish. $\endgroup$ – Bucephalus Oct 2 '18 at 15:44
  • 1
    $\begingroup$ I have got the assignment solution so yes I have. However the question wasn't exactly the same in the assignment, I simulated the question here when I asked it. So, once I go over it I will let you know. $\endgroup$ – Bucephalus Jan 2 at 15:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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