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Apr 18

Solution to Question 1 from UoL exam 2017, Zone B




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 X be the amount you bet, and let Y 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. X is described by the table

Table 1. Values and probabilities for X

X Prob
1 1/2
2 1/2

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

EX=1*1/2+2*1/2=3/2.

Table 2. Values and probabilities for Y

Y Prob
1 1/4
-1 1/4
2 1/4
-2 1/4

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

EY=(1-1+2-2)/4=0.

To find the covariance, we use the shortcut

(1) cov(X,Y)=EXY-(EX)(EY)=EXY.

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

Table 3. Values and probabilities for the pair

Values of Y
Values of X 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 p_{ij} of joint outcomes (X_i,Y_j). The usual mean formula EXY=\sum p_{ij}X_iY_j applies and gives

cov(X,Y)=EXY=(1-1+4-4)/4=0.

(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 1_A a function which is 1 on the set A and zero outside it. Since X can be either 1 or 2, we have

(2) 1=1_{\{X=1\}}+1_{\{X=2\}}

identically on the sample space. Hence,

EXY= (replacing 1 by the sum (2)) =EX1_{\{X=1\}}Y+EX1_{\{X=2\}}Y

(pulling out constant values of X)

=1E1_{\{X=1\}}Y+2E1_{\{X=2\}}Y=0

because on the sets \{X=1\},\ \{X=2\} the outcomes "lose" and "win" have equal probabilities.

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