Distribution of Flashlight Usage

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$\displaystyle (\lambda)$ 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)

= $\displaystyle \lambda e^{-\lambda t}\lambda e^{-\lambda t} + \lambda e^{-\lambda t}\lambda e^{-\lambda t} + \lambda e^{-\lambda t}\lambda e^{-\lambda t}$

= 3$\displaystyle \lambda^2 e^{-2\lambda t}$

Could someone tell me if this is correct?

Re: Distribution of Flashlight Usage

Hey JohnDoe2013.

Hint: Sum of exponential random variables produces a Gamma(n,lambda) distribution.

Re: Distribution of Flashlight Usage

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 $\displaystyle -\lambda z e^{-\lambda z}$

while the Gamma distribution was $\displaystyle \frac{\lambda^\alpha z^{\alpha-1}}{(\alpha-1) !}}$ with $\displaystyle \alpha = 2$

I don't understand what happened to the negative sign in $\displaystyle -\lambda$?

Re: Distribution of Flashlight Usage

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 $\displaystyle \alpha = 2$.

However, while I know that n is the number of batteries, and I know what $\displaystyle \lambda$ is, in your formula, there is no $\displaystyle \alpha$?

Could you clarify what $\displaystyle \alpha$ and n should be?

Thx!

Re: Distribution of Flashlight Usage

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.

Re: Distribution of Flashlight Usage

Suppose the batteries have lifetimes $\displaystyle t_1 < t_2 < t_3$. Then, you can use the flashlight for at least $\displaystyle t_2$ hours. Here is how: If you start with batteries with $\displaystyle t_1$ and $\displaystyle t_2$ lifetimes, then after $\displaystyle t_1$ hours, you can swap out that battery for the one with $\displaystyle t_3$ hours and continue using the flashlight for $\displaystyle t_2-t_1$ more hours. Alternately, if you start with any other combination of batteries, you can use the flashlight for $\displaystyle \min\{t_1+t_2,t_3\}$ hours.

Re: Distribution of Flashlight Usage

Hi Chiro,

Thanks for the reply. Your reply confirms my understanding of the Gamma distribution. $\displaystyle \alpha$ 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.

Re: Distribution of Flashlight Usage

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.