Question 7.2: Joey manipulates a die to increase his chances of winning a ......
Joey manipulates a die to increase his chances of winning a board game against his friends. In each round, a die is rolled and larger numbers are generally an advantage. Consider the random variable X denoting the outcome of the rolled die and the respective probabilities P(X = 1 = 2 = 3 = 5) = 1/9, P(X = 4) = 2/9, and P(X = 6) = 3/9.
(a) Calculate and interpret the expectation and variance of X.
(b) Imagine that the board game contains an action which makes the players use 1/X rather than X. What is the expectation of Y = 1/X? Is E(Y) = E(1/X) = 1/E(X)?
(a) The probability mass function of X is
Using (7.16), we calculate the expectation as
E\left(X\right) =\sum\limits_{i=1}^{k}{x_{i}p_{i}} =x_{1}P\left(X =x_{1}\right) +x_{2}P\left(X =x_{2}\right)+ ·· ·+ x_{k}P\left(X =x_{k}\right). (7.16)
E(X) =1 · \frac{1}{9}+ 2 · \frac{1}{9}+ 3 ·\frac{1}{9}+ 4 ·\frac{2}{9}+ 5 ·\frac{1}{9}+ 6 ·\frac{3}{9}
= \frac{1 + 2 + 3 + 8 + 5 + 18}{9}= \frac{37}{9}≈ 4.1.
To obtain the variance, we need
E(X²) =1 · \frac{1}{9}+ 4 · \frac{1}{9}+ 9 ·\frac{1}{9}+ 16 ·\frac{2}{9}+ 25 ·\frac{1}{9}+ 36 ·\frac{3}{9}
= \frac{1 + 4 + 9 + 32 + 25 + 108}{9}= \frac{179}{9}.
Therefore, using Var(X) = E(X²) − [E(X)]², we get
Var(X) = \frac{179}{9} – \left(\frac{37}{9}\right)^{2}= \frac{1611 − 1369}{81}= \frac{242}{81}≈ 2.98.
The manipulated die yields on average higher values than a fair die because its expectation is 4.1 > 3.5. The variability of is, however, similar because 2.98 ≈ 2.92.
(b) The probability mass function of Y = \frac{1}{X} is:
The expectation can hence be calculated as
E(Y ) = E \left(\frac{1}{X}\right)= 1 · \frac{1}{9}+ \frac{1}{2} ·\frac{1}{9}+ \frac{1}{3} ·\frac{1}{9}+\frac{1}{4} ·\frac{2}{9}+\frac{1}{5} ·\frac{1}{9}+\frac{1}{6} ·\frac{3}{9}
= \frac{1}{9}+\frac{1}{18}+\frac{1}{27}+\frac{1}{18}+\frac{1}{45}+\frac{1}{18}=\frac{91}{270}.
Comparing the results from (a) and (b) shows clearly that E \left(\frac{1}{X}\right)\neq \frac{1}{E\left(X\right) }. Recall that E(bX) = bE(X). It is interesting to see that for some transformations T (X) it holds that E(T (X)) = T (E(X)), but for some it does not. This reminds us to be careful when thinking of the implications of transformations.
| x_{i} | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | \frac{1}{9} | \frac{1}{9} | \frac{1}{9} | \frac{2}{9} | \frac{1}{9} | \frac{3}{9} |
| y_{i} =\frac{1}{x_{i}} | 1 | \frac{1}{2} | \frac{1}{3} | \frac{1}{4} | \frac{1}{5} | \frac{1}{6} |
| P(\frac{1}{X}= y) | \frac{1}{9} | \frac{1}{9} | \frac{1}{9} | \frac{2}{9} | \frac{1}{9} | \frac{3}{9} |