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The weights of n items are $X_1, \ldots, X_n$, assumed independent uniform distribution ${\cal U}(0, 1)$. Mary and John each have a bin (or suitcase) which can carry total weight 1. Mary likes to pack in her bin only the heaviest item. John likes to pack the items in order $1, 2, \ldots, n$, packing each item if it can fit in the space remaining. Denote by $W_M$ the weight of Mary's suitcase and by $W_J$ the weight of John's suitcase.

Find the pdf-s of $W_M$ and $W_J$ and then compute the expectations of these random variables.

Hint. You need to know that, for any $c > 0$ and any positive integer $k$ we have $\int\limits_{x_1,...,x_k \geq 0, x_1+...+x_k \leq c} \frac{c^k}{k!}$.

I am not sure where to begin for this problem, so if anyone could walk me through it or give me a place to start that would be greatly appreciated.

David G. Stork
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momo
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1 Answers1

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Mary's bag carries only the heaviest item. This is often called the $n$th order statistic. Check out: https://en.wikipedia.org/wiki/Order_statistic

One way to get the PDF for it, is by first getting the CDF: $P(\max(X_1,\ldots,X_n)\leq x) = P(X_1\leq x, \ldots, X_n \leq x) = P(X_1\leq x)\cdots P(X_n\leq x) = x^n.$ Can you get the PDF from here?

For John's bag, this is the sum of the first $k$ weights such that $\sum_{i=1}^k X_i \leq 1 < \sum_{i=1}^{k+1} X_i$. For this problem, I'd take some time to think about how it is related to this problem: Summing (0,1) uniform random variables up to 1 and go from there to get the PDF.

Math Helper
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  • The PDF is the derivative of the CDF, so the PDF would be $n\cdot$$x^{n-1}$, correct? And when considering John's bag, based on the problem you linked, the CDF would end up being the same as Mary's? – momo Nov 11 '20 at 22:58
  • Also, some help calculating the expected value would be appreciated. Would it be the same as the problem you linked, i.e. the expected value is e? – momo Nov 11 '20 at 23:18
  • Any other tips or hints you have for this problem would be great. Thanks~ – momo Nov 12 '20 at 15:39
  • Yes, that is the correct PDF for Mary's bag. To get the expected value, compute the integral $\int_0^1 x n x^{n-1} dx$. Note that this depends on the number of items. – Math Helper Nov 12 '20 at 17:48
  • John's bag is a bit more complicated to compute. In the case that the number of items was unbounded, then the expected number of items in his bag would be $e-1$ and the expected weight would be $(e−1)/2$, but it'll require a bit more work to incorporate the fact that the number $n$ of bags is finite. – Math Helper Nov 12 '20 at 22:45