Hey JohnDoe2013.
Hint: Sum of exponential random variables produces a Gamma(n,lambda) distribution.
Suppose you have a flashlight that uses 2 batteries and a package of 3 new batteries. Suppose that each battery has an independent lifetime (in hours) with exponential distribution. Find the distribution of the number of hours T that you can use the flashlight.
My thoughts are:
P(T = t) = P(2 of the batteries have lifetimes of t hours)
= P(X = t, Y = t) + P(X = t, Z = t) + P(Y = t, Z = t)
=
= 3
Could someone tell me if this is correct?
Hi Chiro,
Thank you for responding.
I knew about the Gamma Distribution. I just wasn't sure whether n should be 2 or 3.
If n = 2, then basically it should be equivalent to the convolution of 2 of the random variables.
But then I noticed that the convolution was
while the Gamma distribution was with
I don't understand what happened to the negative sign in ?
Hi Chiro,
Forget about the earlier note. I made a mistake in the computation of the convolution. After correcting the computation, i do get the Gamma distribution with .
However, while I know that n is the number of batteries, and I know what is, in your formula, there is no ?
Could you clarify what and n should be?
Thx!
I think you need to adjust for the fact that you have three batteries for a two-battery device (all new).
Basically you should calculate the probability that any two batteries should fail and use that as the life-time of the flashlight.
Suppose the batteries have lifetimes . Then, you can use the flashlight for at least hours. Here is how: If you start with batteries with and lifetimes, then after hours, you can swap out that battery for the one with hours and continue using the flashlight for more hours. Alternately, if you start with any other combination of batteries, you can use the flashlight for hours.
Hi Chiro,
Thanks for the reply. Your reply confirms my understanding of the Gamma distribution. is basically equivalent to the n in your formula. Based on what I've read, n-1 is supposed to be the number of backup units or spares. In this case n-1 = 1 since we have one spare battery. Hence n = 2. So, we can basically use the Gamma with n = 2 to calculate the lifetime of the battery.
Hi SlipEternal,
Thank you for the response. Your response was one of the approaches I've seen which is to figure out how long 3 batteries can last. Alternatively, we can figure out how long before 2 batteries fail.
which is where the Gamma distribution comes up.. The reasoning then would be that the life of the battery is equivalent to the time to failure of any 2 batteries, which explains why in the Gamma, n-1 = number of backup units =1, so that n=2.