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Consider a deck consisting of seven cards, marked 1, 2, ..., 7. Three of these cards are selected at random. Define an rv W by W ¼ the sum of the resulting numbers, and compute the pmf of W. Then compute E(W) and Var(W). [Hint: Consider outcomes as unordered, so that (1, 3, 7) and (3, 1, 7) are not different outcomes. Then there are 35 outcomes, and they can be listed.] (This type of rv actually arises in connection with Wilcoxon’s rank-sum test, in which there is an x sample and a y sample and W is the sum of the ranks of the x’s in the combined sample.)

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LammettHash

There are [tex]\dbinom73=\dfrac{7!}{3!(7-3)!}=35[/tex] possible outcomes. At the least, you can draw (1, 2, 3) to get a sum of 6, and at most (5, 6, 7) for a sum of 18. Then the PMF is

[tex]P(W=w)=\begin{cases}\dfrac1{35}&\text{for }w\in\{6,7,17,18\}\\\\\dfrac2{35}&\text{for }w\in\{8,16\}\\\\\dfrac3{35}&\text{for }w\in\{9,15\}\\\\\dfrac4{35}&\text{for }w\in\{10,11,13,14\}\\\\\dfrac17&\text{for }w=12\end{cases}[/tex]

That is, consider all the possible sums and their integer partitions. For example,

6 = 1 + 2 + 3

7 = 1 + 2 + 4

8 = 1 + 2 + 5 = 1 + 3 + 4

9 = 1 + 2 + 6 = 1 + 3 + 5 = 2 + 3 + 4

and so on.

Then the expected value is

[tex]\mathrm E[W]=\displaystyle\sum_wwP(W=w)[/tex]

[tex]\mathrm E[W]=\displaystyle\frac{(6+7+17+18)+2(8+16)+3(9+15)+4(10+11+13+14)+5(12)}{35}[/tex]

[tex]\mathrm E[W]=12[/tex]

We can find the variance via

[tex]\mathrm{Var}[W]=\mathrm E[W^2]-\mathrm E[W]^2[/tex]

[tex]\mathrm E[W^2]=\displaystyle\sum_ww^2P(W=w)[/tex]

[tex]\mathrm E[W^2]=\displaystyle\frac{(6^2+7^2+17^2+18^2)+2(8^2+16^2)+3(9^2+15^2)+4(10^2+11^2+13^2+14^2)+5(12^2)}{35}[/tex]

[tex]\mathrm E[W^2]=152[/tex]

[tex]\implies\mathrm{Var}[W]=152-12^2=8[/tex]

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