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

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 $x$.