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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