# Finding probability density function with unknown values

I am not sure if this is the place to ask, have tried to read up on probability density function (PDF) in order to answer this question but to no avail.

How do I go about starting on this?

How do I generate a function based on the given $\lambda$ and that small $t$ symbol.

What is $X \sim Ga(n,\lambda)$?

An explanation before the answer would be nice. Appreciate all the help needed.

Regards

• This same question sheet appeared here a few days ago but the question seems to have disappeared (it's a pity to have lost it as I put some effort into the previous question that was posted). Please add the self-study tag to this question. – Glen_b -Reinstate Monica Oct 31 '13 at 15:13
• Ga(n,$\lambda$) is the gamma distribution. As a hint: Think about how you would simulate the queue process? To start, imagine the different scenarios that the second customer to arrive would face: server still serving cust 1, server just finished cust 1, server has been idle t<$\tau$, server has been idle t>$\tau$. That should help you think through how to model this. – user31668 Oct 31 '13 at 16:47
• if customer 2 arrive before server has finished serving customer 1, then the time in between serving the two of them would just be τ since the server will immediately serve customer 2 after the non-service period? if customer 2 arrive after server has finished serving customer 1, then the time in between would be τ + some value? what do i use to represent this value. how do i connect the both of the case? where did the exponential λ come in in this case – Matthew Oct 31 '13 at 17:16
• Please read the self-study tag wiki info. ... Now consider two simpler cases. First, if $\tau=0$, do you know how to write the pdf for the intercustomer time? Second, if the nonservice time was a constant $\tau=1$ and the service time was a constant $\theta=1$ do you know how to write down the pdf for that? – Glen_b -Reinstate Monica Oct 31 '13 at 22:14
• @Glen_b Correct me if i am wrong, if $\tau$ = 0, then the pdf would simply be dependent on the random variable $\lambda$ which gives the pdf: $\int \dfrac{1}{e^{\lambda}}\ast$$e^{\dfrac{1}{e^{\lambda}} t} dt. With a non zero \tau, it would be \int \dfrac{1}{e^{\lambda}+\tau}\ast$$e^{(\lambda+$\tau$)*t}$ dt – Matthew Nov 1 '13 at 11:53

One general approach which you can adapt for part (a):

Consider a random variable $X$ and a constant, $c$, where $X\sim f_X(x), a\leq x\leq b$, say.

Let $Z=X+c$.

$P(Z\leq z) = P(X+c\leq z) = P(X\leq z-c) = F_X(z-c); a\leq z-c\leq b$.

You just need to get the pdf from the cdf and get the limits sorted out.

Note that you don't need to integrate anything to answer (a) unless you don't know the cdf for an exponential ($F_X(x)=1−e^{−λx};x>0,λ>0$).

If you know it, you just start by writing down the cdf; then $P(X≤z−c)$ simply requires substitution, not integration. So it's nothing more than 'substitute then differentiate'... but you must take care with the limits on the random variable, since $Z$ is clearly defined over a different set of values than $X$ is; what's needed there is clear from the last part of the line of algebra in my question.

In part (b), note that $nτ$ is just a constant. It's the same problem as before (a shifted distribution), but with a gamma instead of an exponential; the hint in my answer above is general enough to work for both.

The only new trick is finding the MGF, which should involve a straightforward manipulation of the integral for the MGF to get something times a Gamma MGF.

... and (b)(ii) simply involves recognizing the relationship between parts (a) and (b).

• let me try to interpret your approach. i need to find the probability of random variable X when the waiting time falls in between some values, a and b. This is for the exponential serving time which makes use of the exponential distribution function. C would be the deterministic variable. So Z would be the total waiting time? The cdf would require me to integrate P(X≤z−c) with limits z-τ and differentiating it w.r.t z to get the pdf? – Matthew Nov 2 '13 at 15:53
• Ah, i would get the cdf for an exponential distribution either way. I get the idea for part (a) now. I would also like to inquire and clarify on how is part b.(ii) linked to b.(i). Since X is a gamma distribution, does the PDF of Tn follows some sort of distribution of Y which is derived from the gamma distribution? – Matthew Nov 3 '13 at 3:44
• Ah. I think I can finish this up more or less. Thanks for the help. – Matthew Nov 3 '13 at 7:05
• Just to clarify on b(i) and (ii): For b(i), I've only managed to manipulate it to obtain a constant times the pdf of Gamma Distribution which is equals to that constant times 1. It is quite different from the approach you gave and wondered if it is an alternative way. For b(ii), I realized that i can just use the answer for b(i) since I have already substituted Y=nt+X into the gamma distribution and the distribution is the sum of n exponential events. Am I finally done with this exercise? :) – Matthew Nov 4 '13 at 12:28
• I can't tell from your description of b(i) what you did. – Glen_b -Reinstate Monica Nov 4 '13 at 13:00