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Suppose that the length of time Y it takes a worker to complete a certain task has the probability density function given by $f(y)=\begin{cases} e^{-(y-\theta)} &, y>\theta\\ 0 & ,elsewhere \end{cases}$ where θ is a positive constant that represents the minimum time until task completion. Let $Y_1, Y_2, . . . , Y_n$ denote a random sample of completion times from this distribution. Find

a the density function for Y(1) = min(Y1, Y2, . . . , Yn). b E(Y(1)).

Could anyone get me started on this since the random sample is not independent?

Michiel Van Couwenberghe
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afsdf dfsaf
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  • Let $j = \min_i(Y_i)$. What's the likelihood that the (independent of $Y_j$) samples are all greater than $Y_j$? (This is the intersection of the conditions that each of the $n-1$ samples is $ \geq Y_j$.) – Eric Towers Feb 12 '14 at 07:29

1 Answers1

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Hints:

  • The sample times are presumably independent.
  • Find the probability $Y_i \le x$ for a particular $i$
  • Find the probability $Y_i \gt x$ for a particular $i$
  • Find the probability $Y_i \gt x$ for all $i$
  • Find the probability $\min_i(Y_i) \gt x$
  • Find the probability $\min_i(Y_i) \le x$
  • Find the density function for $\min_i(Y_i)$
  • Find the expectation of $\min_i(Y_i)$
Henry
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  • could you show me how to find the probability for $ Y_i \le x$ for a particular i ? – afsdf dfsaf Feb 12 '14 at 15:49
  • $\displaystyle P(Y_i \le x) = \int_{y=\theta}^x e^{-(y-\theta)} dy = 1-e^{-(y-\theta)}$ for $x \gt \theta$ – Henry Feb 12 '14 at 16:07
  • just want to put the right position for all your hints, is it going to be let $U = min (Y_1, Y_2, ..., Y_n)$. $P[U \le t] = 1 - P[U > t] = 1 - P[Y_1 > t, Y_2 > t ,..., Y_n > t] = 1 - [1-F_Y (t)]^n $ Then we take the derivatives of $P[U \le t] = 1 - [1-F_Y (t)]^n$ to find the density function for $min_i (Y_i)$? – afsdf dfsaf Feb 12 '14 at 21:49
  • Yes, but $1 - [1-F_Y (t)]^n$ is not difficult to express in terms of $e,y,\theta,n$. Try it. – Henry Feb 12 '14 at 22:13
  • $P[U > t] = 1 - [-e^{-(y-\theta)}]^n$ Is this right? – afsdf dfsaf Feb 12 '14 at 22:53
  • density for $Y_{(1)}$ is $n[-e^{-(y-\theta)}]^{n-1} e^{-(y-\theta)}, y > 0$ for $E(Y_{(1)}) = 1-ye^{-(y-\theta)}-e^{-(y-\theta)} + y $, right? – afsdf dfsaf Feb 12 '14 at 23:21
  • You have a sign error or two at the start and can simplify: $P[Y_{(1)} \gt y] = e^{-n(y-\theta)}$, while the density for $Y_{(1)}$ is $ne^{-n(y-\theta)}$. Your statement for the expectation is very wrong as it should not include $y$ – Henry Feb 13 '14 at 00:02
  • The way I calculate density for $Y_{(1)}$ is $f_U (t) = n [F_y(t)]^{n-1} f_y(t) = n [1-e^{-(y-\theta)}]^{n-1} e^{-(y-\theta)}$, for $y > \theta$...does it look right to you...if you could you show me how to get to your density answer? – afsdf dfsaf Feb 13 '14 at 04:43
  • You should have $$P[Y_i \le y] = 1-e^{-(y-\theta)},$$ $$P[Y_i \gt y] = e^{-(y-\theta)},$$ $$P[Y_{(1)} \gt y] = [e^{-(y-\theta)}]^n = e^{-n(y-\theta)},$$ $$F_{Y_{(1)}}(y)= P[Y_{(1)} \le y] = 1 - e^{-n(y-\theta)},$$ $$f_{Y_{(1)}}(y)= ne^{-n(y-\theta)}$$ – Henry Feb 13 '14 at 08:05
  • Got it...before getting to the detail of my calculation for $E(Y_{(1)})$ ... I just want to check with my setup matches with yours first....$$ \int_\theta^y xe^{-(x-\theta)} dx $$? If this is different from yours, could you show yours? – afsdf dfsaf Feb 13 '14 at 14:59
  • Two points: (1) you have the original density there when you should have the density for the minimum; and (2) $y$ is not the correct top limit of the integration – Henry Feb 13 '14 at 15:12
  • For (1), after looking at your comment, I am a bit confused as to how to write out the density for the minimum. Could you show it?..For (2), what is the top limit is going to be .... since my thinking is for the support $ y > \theta $...so it should integrate up to t. – afsdf dfsaf Feb 13 '14 at 17:14
  • The density is $ne^{-n(y-\theta)}$ and the integral should go to infinity – Henry Feb 13 '14 at 21:07
  • if I integral this, my answer is infinity through...if not, could you show me your answer? – afsdf dfsaf Feb 13 '14 at 23:18
  • No. $\displaystyle \int_\theta^\infty ne^{-n(x-\theta)} dx = 1$ and $\displaystyle \int_\theta^\infty xne^{-n(x-\theta)} dx$ is also finite. – Henry Feb 13 '14 at 23:22
  • so your answer for $E(Y_{(1)})$ is what? – afsdf dfsaf Feb 13 '14 at 23:38
  • The shifted expectation of an exponential random variable with rate $n$ – Henry Feb 13 '14 at 23:41
  • this info does not relate to this problem though... – afsdf dfsaf Feb 13 '14 at 23:45
  • I did not see why $\int_\theta^\infty ne^{-(x-\theta)} dx $is 1. Could you explain that? – afsdf dfsaf Feb 13 '14 at 23:50
  • You are missing an $n$ in the exponent as in $\displaystyle \int_\theta^\infty ne^{-n(x-\theta)} dx = \left. -e^{-n(x-\theta)} \right|_\theta^\infty = 0 -(-1)= 1$. – Henry Feb 14 '14 at 00:44
  • you have to multiply by x inside this integral right? – afsdf dfsaf Feb 14 '14 at 01:38
  • For the expectation, yes: $\displaystyle \int_\theta^\infty xne^{-n(x-\theta)} dx$. Integration by parts will do it – Henry Feb 14 '14 at 07:34
  • However, when I integrate this, I got $x - \frac{1}{n}$...which does not make much sense to me... if that is different from yours, could you show me how did you get your answers? – afsdf dfsaf Feb 14 '14 at 21:01
  • You should have $\left. -\left(x+\frac1n\right) e^{-n(x-\theta)} \right|_\theta^\infty $ and your result should have $n$ and $\theta$ but not $x$ – Henry Feb 15 '14 at 01:04
  • The answer is $\theta - \frac{1}{n}$? – afsdf dfsaf Feb 15 '14 at 19:38
  • Close: $\theta + \frac{1}{n}$ – Henry Feb 15 '14 at 21:33
  • you are right as I mess up with one of the sign...finally... – afsdf dfsaf Feb 15 '14 at 21:38