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Thread: The life of a device and the probability of using it

  1. #1
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    The life of a device and the probability of using it

    I am having issues with this exercise:

    The life of a device is $\mathrm{Exp}(1/a)$-distributed. If one wishes to use it on $n$ different, independent, $\mathrm{Exp}(1/(na))$-distributed occasions, what is the probability that this is possible?

    So I denote the life by $X$ and the time of the $n$:th occasion by $Y_n$. I have been trying to compute $P(Y_n \leq X)$ using the formula

    $$ F_{Y_n|X=x}(x) = \frac{\int_0^x f_{x,y}(x,z) dz}{\int_{-\infty}^{\infty} f_{x,y}(x,z)dz} $$,

    and I am no where near the answer that is $\left(\frac{n}{n+1}\right)^n.$ I have a feeling that I am way off but no idea how to get to the right answer. Or even for just $Y_1$ the answer is supposed to be $\frac{n}{n+1}$ but I just don't see how.
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  2. #2
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    Re: The life of a device and the probability of using it

    Quote Originally Posted by Mysad View Post
    I am having issues with this exercise:

    The life of a device is $\mathrm{Exp}(1/a)$-distributed. If one wishes to use it on $n$ different, independent, $\mathrm{Exp}(1/(na))$-distributed occasions, what is the probability that this is possible?

    So I denote the life by $X$ and the time of the $n$:th occasion by $Y_n$. I have been trying to compute $P(Y_n \leq X)$ using the formula

    $$ F_{Y_n|X=x}(x) = \frac{\int_0^x f_{x,y}(x,z) dz}{\int_{-\infty}^{\infty} f_{x,y}(x,z)dz} $$,

    and I am no where near the answer that is $\left(\frac{n}{n+1}\right)^n.$ I have a feeling that I am way off but no idea how to get to the right answer. Or even for just $Y_1$ the answer is supposed to be $\frac{n}{n+1}$ but I just don't see how.
    First let's prove the case for Y_1.

    In general, if $X_1$ and $X_2$ are independent exponential random variables with rates $\lambda_1$ and $\lambda_2$, respectively; then $P(X_1<X_2)=\frac{\lambda_1}{\lambda_1 + \lambda_2}$.

    Proof:
    $P(X_1<X_2|X_1=x) = P(X_2>x|X_1=x)=P(X_2>x)=e^{-\lambda_2 x}$ (Note the use of independence.)

    $\begin{align*}P(X_1<X_2) &= \int^\infty_0 P(X_1<X_2|X_1=x)f_{X_1}(x)dx\\ &= \int^\infty_0 e^{-\lambda_2x}\lambda_1 e^{-\lambda_1x}dx \\ &= \lambda_1 \int^\infty_0 e^{-(\lambda_1 + \lambda_2)x} dx \\ &= \frac{\lambda_1}{\lambda_1+\lambda_2} \end{align*}$

    Now, in our problem, we have independent exponential RVs with rates $a$ and $na$. So,

    $P(X>Y_1)=\frac{na}{na+a} = \frac{n}{n+1}$.

    Think for a moment here -- if we know $P(X>Y_1) = \frac{n}{n+1}$, what does the exponential distribution uniquely allow us to say about $P(X>Y_2 | X>Y_1)$?

    The answer is that the memoryless property allows us to conclude that $P(X>Y_1)=P(X>Y_2|X>Y_1)$. You can repeat this up to $Y_n$.

    Therefore, P(X>Y_n) = \left(\frac{n}{n+1} \right)^n.

    Note that $\lim_{n \rightarrow \infty} \left(\frac{n}{n+1}\right)^n = e^{-1}$.
    Last edited by abender; Jun 8th 2017 at 05:01 PM.
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