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Below is a problem I did. Part (a) is correct so you can skip over it. Part (b) is wrong. I want to know where I went wrong.

Problem:

Suppose that the length of time $Y$ that it takes a worker to complete a certain task has probability density function: $$ f(y) = \begin{cases} e^{-(y-\theta)} & \text{ for } y > \theta \\ 0 & \text{elsewhere} \end{cases} $$ where $\theta$ is a positive constant that represents the minimum time to task completion. Let $Y_1,Y_2,...Y_n$ denote a random sample of completion times from this distribution.
(a) Find the density function for $Y_{(1)} = min( Y_1,Y_2,... Y_n )$.
(b) Find $E( Y_{(1)} )$,

Answer: (a)

We know $f(y)$ and now we find $F(y)$. \begin{align*} F(Y) &= \int_{\theta}^Y e^{-(y-\theta)} \,\, dy \\ F(Y) &= \int_{\theta}^Y e^{\theta}e^{-y} \,\, dy \\ F(Y) &= -e^{\theta}e^{-y} \bigg|_{\theta}^{Y} \\ F(Y) &= -e^{\theta}e^{-Y} + e^{\theta} e^{-\theta} \\ F(Y) &= -e^{\theta}e^{-Y} + 1 \\ F(Y) &= 1 -e^{\theta}e^{-Y} \\ \end{align*} Let $g_1(y)$ be the density function the problem asks us to find. We have: $$ g_1(y) = n\left( 1 - F(y) \right)^{n-1} f(y) $$ This means that our answer will have a $n$ in it which is to be expected given how the problem is setup. \begin{align*} g_1(y) &= n\left( 1 - \left( 1 -e^{\theta}e^{-Y} \right) \right)^{n-1} e^{-(y-\theta)} \\ g_1(y) &= n\left( e^{\theta}e^{-Y} \right)^{n-1} e^{-(y-\theta)} \\ g_1(y) &= ne^{n\theta-yn} \end{align*} Hence the answer is: $$ f(y) = \begin{cases} ne^{-n(y - \theta)} & \text{ for } y > \theta \\ 0 & \text{elsewhere} \end{cases} $$ Now for part (b): \begin{align*} E(Y_{(1)}) &= \int_{\theta}^{\infty} y n e^{-n(y - \theta)} \, dy \\ E( Y_{(1)} ) &= ne^{n \theta} \int_{\theta}^{\infty} ye^{-ny} \,\, dy \\ \end{align*} Now we need to perform the following integration: $$ \int_{\theta}^{\infty} ye^{-ny} \,\, dy $$ We can do this using integration by parts with $u = y$ and $dv = e^{-ny} \,\, dy$. \begin{align*} du &= dy \\ v &= - \dfrac{ e^{-ny} }{n} \\ \int_{\theta}^{\infty} ye^{-ny} \,\, dy &= - \dfrac{ ye^{-ny} }{n} \bigg|_{\theta}^{\infty} + \int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy \\ % - \dfrac{ ye^{-ny} }{n} \bigg|_{\theta}^{\infty} &= 0 + \theta e^{ -n \theta } = \theta e^{ -n \theta } \\ % \int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy &= \dfrac{ e^{-ny} }{n^2} \bigg|_{\theta}^{\infty} \\ \int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy &= 0 - \dfrac{ e^{-n \theta } } {n^2} \\ \int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy &= - \dfrac{ e^{-n \theta } } {n^2} \\ \int_{\theta}^{\infty} ye^{-ny} \,\, dy &= \theta e^{ -n \theta } - \dfrac{ e^{-n \theta } } {n^2} \\ % E( Y_{(1)} ) &= ne^{n \theta} \left( \theta e^{ -n \theta } - \dfrac{ e^{-n \theta } } {n^2} \right) \\ E( Y_{(1)} ) &= n\theta - \dfrac{ 1 }{n } \end{align*} My answer is wrong. The book's answer is: $$ \theta + \dfrac{1}{n} $$ Where did I go wrong?

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    $\begingroup$ In the step where you plugged into $-\frac{ye^{-ny}}{n}\Big|_{\theta}^{\infty}$, you dropped the $\frac{1}{n}$ when plugging in the lower bound $\theta$, probably inadvertenly. And in the step where you solved $\int_{\theta}^{\infty} \frac{e^{-ny}}{n} \, dy$, the $\theta$ is the lower bound, so the minus sign cancels out and becomes positive. Be careful with the integral bounds! $\endgroup$ Commented Jun 20 at 3:44

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I am guessing this is why you asked the other question too xD.

Your calculation is mostly correct, but there's a sign error in the second integral.

If we go through the entire thing once, you have stuff correctly set up: $$\int_{\theta}^{\infty} ye^{-ny} \,\, dy = - \dfrac{ ye^{-ny} }{n} \bigg|_{\theta}^{\infty} + \int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy$$

The first term is correct: $$- \dfrac{ ye^{-ny} }{n} \bigg|_{\theta}^{\infty} = 0 + \frac{\theta e^{-n\theta}}{n} = \frac{\theta e^{-n\theta}}{n}$$

The error is in the second integral.

You wrote: $$\int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy = \dfrac{ e^{-ny} }{n^2} \bigg|_{\theta}^{\infty} = 0 - \dfrac{ e^{-n \theta } } {n^2} = - \dfrac{ e^{-n \theta } } {n^2}$$

but this should be: $$\int_{\theta}^{\infty} \dfrac{ e^{-ny} }{n} \,\, dy = \frac{1}{n} \int_{\theta}^{\infty} e^{-ny} dy = \frac{1}{n} \left[-\frac{e^{-ny}}{n}\right]_{\theta}^{\infty}$$

$$= \frac{1}{n} \left(0 - \left(-\frac{e^{-n\theta}}{n}\right)\right) = \frac{1}{n} \cdot \frac{e^{-n\theta}}{n} = +\frac{e^{-n\theta}}{n^2}$$

Basically when you evaluate $\left[-\frac{e^{-ny}}{n}\right]_{\theta}^{\infty}$, you get: $$0 - \left(-\frac{e^{-n\theta}}{n}\right) = +\frac{e^{-n\theta}}{n}$$

So the correct calculation should be: $$\int_{\theta}^{\infty} ye^{-ny} \,\, dy = \frac{\theta e^{-n\theta}}{n} + \frac{e^{-n\theta}}{n^2}$$

So finally I guess we can match the book's answer.

$$E(Y_{(1)}) = ne^{n\theta} \left(\frac{\theta e^{-n\theta}}{n} + \frac{e^{-n\theta}}{n^2}\right) = \theta + \frac{1}{n}$$

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  • $\begingroup$ You wrote: I am guessing this is why you asked the other question too xD. Your guess is correct. $\endgroup$ Commented Jun 20 at 3:52
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    $\begingroup$ Haha, cheers :P $\endgroup$ Commented Jun 20 at 3:53
  • $\begingroup$ The other question being math.stackexchange.com/questions/5076949/… $\endgroup$ Commented Jun 20 at 8:37
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Alternatively, you can make use of the gamma function:

If $t=n(y-\theta)$, then $y=\theta+\dfrac tn$ and $dy = \dfrac1n dt$; so, \begin{align} E[Y_{(1)}] &= \int_\theta^\infty nye^{-n(y-\theta)} \, dy \\ &= \int_0^\infty n \Big( \theta+\frac tn \Big) e^{-t} \, \frac1n dt \\ &= \theta \underbrace{\int_0^\infty e^{-t} \, dt}_{=\,\Gamma(1)\,=\,1} + \frac1n \underbrace{\int_0^\infty te^{-t} \, dt}_{=\,\Gamma(2)\,=\,1} = \theta + \frac1n \end{align}

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