Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 4

A discrete random variable on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ is a real-valued measurable function $X:\Omega \to \mathbb{R}$ whose probability distribution is concentrated on some finite or countable set $S$:Measurability means $\{\omega :X(\omega )\le t\}\in \mathcal{F}$ for every real $t$. Its probability mass function is $p_X(x)=\mathbb{P}(X=x)$, and ${\sum }_{x\in S}{p}_X(x)=1$. Requiring a countable image everywhere, rather than outside a null set, is an equivalent convention after harmless modification on a null set.

Use the ordered outcomes of the two rolls as the sample space:Thus each elementary outcome has probability $1/36$. This is the usual probability space for two independent fair dice.

For an outcome $\omega =(i,j)$, the two random variables are the coordinate mapsTheir marginal probability mass functions are $p_X(i)=p_Y(j)=1/6$ on $\{1,\dots ,6\}$, and their joint probability mass function satisfiesOff that support, the same factorization holds with zeros. Hence $X$ and $Y$ are independent random variables, by the joint probability mass function criterion for independence.

There are three even faces, so $\mathbb{P}(A)=1/2$. The sum is even precisely when both faces have the same parity, giving $9+9=18$ outcomes and $\mathbb{P}(B)=1/2$. Both $A$ and $B$ occur precisely when both faces are even, giving nine outcomes. ThereforeThus $A$ and $B$ are independent events.

No: pairwise independence does not imply that the three are independent events. Take $C=\{Y\text{ is even}\}$. Then $\mathbb{P}(C)=1/2$, and $\mathbb{P}(A\cap C)=9/36=1/4$ by independence of the two rolls. Also $B\cap C$ is precisely the event that both faces are even, so $\mathbb{P}(B\cap C)=1/4$. Thus all three pairs of events are independent.

But $A\cap B\cap C$ also consists of the nine even-even outcomes, soThis violates the triple-intersection requirement for independent events.

The printed numerical example reverses the digits and contradicts both definitions. With the stated last-digit definition, $[Z]=17$ gives $X=7$, $Y=1$. Use those definitions throughout.

Put $q=e^{-\lambda }\in (0,1)$ and $N=\lfloor Z\rfloor$. The exponential distribution givesThe Euclidean division $N=10Y+X$ has unique remainder $X\in \{0,\dots ,9\}$ and quotient $Y\in \{0,1,\dots \}$. Thus the joint probability mass function isand zero otherwise. This is decimal digit decomposition of a geometric random variable. Following the erroneous example instead would require different definitions and would not answer the stated question.

Summing the joint probability mass function over $y$ with the geometric series gives the marginal probability mass functionDivide the joint probability mass function by this positive marginal probability mass function to obtain the conditional distribution:It is the geometric distribution counting failures before the first success, with success probability $1-q^{10}$. For the other nonnegative integers $x\ge 10$, $\mathbb{P}(X=x)=0$; the elementary conditional probability given $X=x$ is undefined and supplies no additional case.

Summing over the last digit yields the marginal probability mass functionTogether with the other marginal probability mass function, this givesThe factorization holds on the whole support and with zeros elsewhere, proving that $X$ and $Y$ are independent. The conditional distribution in ii likewise does not depend on the permissible value $x$.