## Solution to Question 1 from UoL exam 2017, Zone B (FN3142)

This is a relatively simple problem but surprisingly many students cannot answer it. I provide two answers, the first of which gives a general idea about how to solve such problems.

Imagine the following gamble. First, flip a fair coin to determine the amount of your bet: if heads, you bet $1, if tails you bet $2. Second, flip again: if heads, you win the amount of your bet, if tails, you lose it. For example, if you flip heads and then tails, you lose $1; if you flip tails and then heads you win $2. Let be the amount you bet, and let be your net winnings (negative if you lost).

(a) [5 Marks] Show that the covariance between X and Y is zero.

(b) [5 Marks] Show that X and Y are not independent.

## Bulletproof method

(a) When dealing with discrete random variables, a probability table is the best tool. is described by the table

Table 1. Values and probabilities for

Prob | |

1 | 1/2 |

2 | 1/2 |

For , the two bets have equal probabilities of 1/2, so the mean is

.

Table 2. Values and probabilities for

Prob | |

1 | 1/4 |

-1 | 1/4 |

2 | 1/4 |

-2 | 1/4 |

For , the possible outcomes are 1,-1,2,-2, and each of them has probability 1/4, so the mean is

.

To find the covariance, we use the shortcut

(1) .

For a pair of variables we need a two-way table

Table 3. Values and probabilities for the pair

Values of | |||||

Values of | 1 | -1 | 2 | -2 | |

1 | 1/4 | 1/4 | 0 | 0 | |

2 | 0 | 0 | 1/4 | 1/4 |

In the main body of the table we have probabilities of joint outcomes . The usual mean formula applies and gives

.

(b) For independent variables, all joint probabilities equal products of individual (marginal) probabilities. From Table 3 we see that this is not true, so the variables are not independent.

## Elegant method

For this method, we still need everything that led to equation (1) but don't need Table 3. Denote a function which is 1 on the set and zero outside it. Since can be either 1 or 2, we have

(2)

identically on the sample space. Hence,

(replacing 1 by the sum (2))

(pulling out constant values of )

because on the sets the outcomes "lose" and "win" have equal probabilities.

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