Problem: Let Y denote the amount of milk (in gallons) remaining in a 1-gallon container on its expiration date. Suppose that the density function for Y is $f(y)=2y$ for $0 \le y \le 1$. Suppose that a random sample of 5 gallons of milk is taken and the amount of milk remaining on the expiration date is recorded.

a.) Find $\mu$

b.) Find the density function of the minimum, $Y_{(1)}$

I know that the density function for any order statistic, $k^{th}$ , is as follows

$$g_{(k)}(y)= \left(\frac{n!}{(k-1)!(n-k)!1!}\right)\left(F_Y(y)\right)^{k-1}\left(1-F_Y(y)\right)^{n-k}\left(f_Y(y)\right)$$

Part a.) We had a similar (less complicated) example in our notes where we found the density function for the median and then used that to find the Expected value. Is this the proper route to solve this problem as well? If so, I am confused as to what the median is in this example. Would it be the median amount of 1-gallon, $k=.5$, where $n=1$? Or the median amount of all 5 1-gallon containers combined, $k=2.5$, where $n=5$?

Part b.) How is the minimum $Y_{(1)}$ ? Is this using the sample size $n=5$? Is 1 referring to 1 gallon? Is 1 the minimum because the event $\left(Y_{(n)} \le y \right)$ will only occur iff the events $\left(Y_i \le y \right)$ for every $i = 1,2,3,...,n$?


(a) If by $\mu$ you mean $E(Y)$ then

\begin{align} E(y) &= \int_{y=0}^{1} yf_Y(y)\;dy \\ &= \int_0^1 2y^2\;dy \\ &= 2/3. \end{align}

You don't need the median at all for this.

(b) The formula you have for $g_{(k)}(y)$ is the way to go. Here, we have $n=5, k=1$ and $F_Y(y) = \int_0^y 2y\; dy = y^2$. If my calculation is correct, you should end up with pdf $g_{(1)}(y) = 10y(1-y^2)^4$.

You seem to be confused by the notation $Y_{(1)}$ for the first-order statistic random variable. We have $5$ samples and so, firstly, we have $5$ random variables: $Y_1, Y_2, Y_3, Y_4, Y_5,$ with $Y_1$ being the $Y$ value for the first sample, and so on.

Next, we have $5$ order statistic random variables: $Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)}, Y_{(5)}$, defined:

\begin{align} Y_{(1)} &= \text{Smallest value of $Y_1, Y_2, Y_3, Y_4, Y_5$} \\ Y_{(2)} &= \text{Second smallest value of $Y_1, Y_2, Y_3, Y_4, Y_5$} \\ Y_{(3)} &= \text{Third smallest value of $Y_1, Y_2, Y_3, Y_4, Y_5$} \\ Y_{(4)} &= \text{Fourth smallest value of $Y_1, Y_2, Y_3, Y_4, Y_5$} \\ Y_{(5)} &= \text{Largest value of $Y_1, Y_2, Y_3, Y_4, Y_5$}. \end{align}

  • $\begingroup$ How do you verify that is a valid density function? $\endgroup$ Dec 1 '15 at 6:23
  • $\begingroup$ @SalamiRolls We have $0\leq Y_{(1)}\leq 1$ so check if $1=\int_0^1 10y(1-y^2)^4\;dy$. $\endgroup$
    – Mick A
    Dec 1 '15 at 7:26

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