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Apr 18

## 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$$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$$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$$Y$ Values of $X$$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.