Past exam of the mathematics course of the University of Cambridge / 2009 / ia / Paper 2 / 9F (Probability)

Use the usual assumption that the two die scores are independent random variables. A fair score has expected value $7/2$ and variance $35/12$. Linearity of expectation and the variance of a sum therefore giveCount the ordered pairs of scores to obtain the probability mass functionsOn these attainable probability supports, the maxima occur at $x=7$ and $y=0$, each with probability $1/6$; the minima occur at $x=2,12$ and $y=-5,5$, each with probability $1/36$. Outside the respective probability support, the mass is zero.

They are not independent: $Y=0$ implies equal die scores and hence $X$ is an even number, so $P(X=7,Y=0)=0$, whereas $P(X=7)P(Y=0)=1/36>0$. In fact $\mathrm{Cov}(X,Y)=\mathrm{Var}(S_1)-\mathrm{Var}(S_2)=0$: this is an example of uncorrelated random variables that are dependent.

Again assume independent die scores. Their sum has the discrete convolution of the two score distributions, soAll terms are nonnegative. The arithmetic-geometric mean inequality applied to the two extreme-score contributions givesIf all eleven probabilities were equal, normalization would make each $1/11$, but the inequality would require $1/11\ge 2/11$. This contradiction proves the sum cannot be uniform on $\{2,\dots ,12\}$, whatever the individual die probabilities.