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Math Help - Distribution of Flashlight Usage

  1. #1
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    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 (\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)
    = \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 \lambda^2 e^{-2\lambda t}

    Could someone tell me if this is correct?
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  2. #2
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    Re: Distribution of Flashlight Usage

    Hey JohnDoe2013.

    Hint: Sum of exponential random variables produces a Gamma(n,lambda) distribution.
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    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 -\lambda z e^{-\lambda z}
    while the Gamma distribution was \frac{\lambda^\alpha z^{\alpha-1}}{(\alpha-1) !}} with \alpha = 2
    I don't understand what happened to the negative sign in -\lambda?
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    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 \alpha = 2.

    However, while I know that n is the number of batteries, and I know what \lambda is, in your formula, there is no \alpha?
    Could you clarify what \alpha and n should be?

    Thx!
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  5. #5
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    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.
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    Re: Distribution of Flashlight Usage

    Suppose the batteries have lifetimes t_1 < t_2 < t_3. Then, you can use the flashlight for at least t_2 hours. Here is how: If you start with batteries with t_1 and t_2 lifetimes, then after t_1 hours, you can swap out that battery for the one with t_3 hours and continue using the flashlight for t_2-t_1 more hours. Alternately, if you start with any other combination of batteries, you can use the flashlight for \min\{t_1+t_2,t_3\} hours.
    Last edited by SlipEternal; October 31st 2013 at 02:47 PM.
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  7. #7
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    Re: Distribution of Flashlight Usage

    Hi Chiro,

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