Past exam of the options I for mathematics and philosophy course of the University of Oxford / 2013 / N / 1

For integer-valued random variables, the joint probability mass function isThe marginal probability mass functions sum out the other variable:All masses are nonnegative, and their sums over the corresponding sample values equal one.

Independence means that for all subsets $A,B\subseteq \mathbb{Z}$,For discrete random variables, this is equivalent to the pointwise factorizationOne direction takes singleton sets; the other sums the pointwise identity over $A\times B$.

The five specified masses sum to one, so there are no other possible outcomes. Symmetry gives $\mathbb{E}X=0$, while $\mathbb{E}Y=1/4$. The two terms with $Y=1$ have opposite $X$ values and equal masses, soNevertheless $\mathbb{P}(X=0,Y=1)=0$, whereas $\mathbb{P}(X=0)\mathbb{P}(Y=1)=(1/2)(1/4)=1/8$. Therefore $X$ and $Y$ are not independent. They are uncorrelated random variables, illustrating that zero covariance alone does not imply independence.

Linearity of expectation and the vanishing of cross covariances for independent variables giveThe expectation formula does not require independence; independence is used for the variance sum.

Write ${\bar{X}}_n=n^{-1}{\sum }_{j=1}^n{X}_j$ and apply the stated second-moment Markov inequality to $Y={\bar{X}}_n-\mu$. The preceding calculation gives $\mathbb{E}{Y}^2={\sigma }^2/n$. Thus for every $\epsilon >0$,This is convergence in probability to the common mean, proving the finite-variance weak law of large numbers. It also covers ${\sigma }^2=0$, when the bound is identically zero.