Past exam of the year 1 of the computer science course of the University of Oxford / 2022 / Discrete mathematics, continuous mathematics and probability resit / 7

For each integer $x$, the events $\{X=x,Y=y\}$, indexed by the integers $y$, are disjoint and their union is $\{X=x\}$. Therefore countable additivity gives the marginal probability mass functionThe nonnegative terms make this sum well-defined, even when the support is infinite.

The random variables are independent random variables if the events determined by one are independent of the events determined by the other. For integer-valued variables, this is equivalent to the factorizationIndeed, independence gives the factorization by taking singleton events. Conversely, summing it over $x\in A$ and $y\in B$, for any subsets $A,B\subseteq \mathbb{Z}$, gives $\mathbb{P}(X\in A,Y\in B)=\mathbb{P}(X\in A)\mathbb{P}(Y\in B)$. Thus the singleton condition implies independence of all events in question.

Interpret existence of the expectations as finite expected values, equivalently $\mathbb{E}|X|<\infty$ and $\mathbb{E}|Y|<\infty$. First justify the summations rather than assuming that a signed double series can be rearranged. Independence and Tonelli theorem giveConsequently $XY$ is an integrable random variable, and Fubini theorem permits summing the signed terms in either order:If extended infinite expectations are allowed instead, the phrase “exist” needs additional care: expressions such as $0\cdot \infty$ do not define an expectation. The finite-expectation convention avoids that ambiguity.

Summing the two nonzero entries with $X=1$ givesThe remaining probability is $\mathbb{P}(X=0)=1-p$, so $X$ has the Bernoulli distribution with parameter $p$.

For the other marginal, the two entries at $Y=0$ add to $e^{-p}$, the entry at $Y=1$ is $p{e}^{-p}$, and for $k⩾2$ the only nonzero entry is already $e^{-p}{p}^k/k!$. Thusand $Y$ has the Poisson distribution with parameter $p$. This also checks normalization of the joint probability mass function by summing the marginal.

Equality can occur only at the two pairs $(0,0)$ and $(1,1)$. Their combined probability isTaking the complementary probability givesSince $1+2p>0$, the supplied exponential bound impliesThis is a bound on the mismatch probability for this particular coupling of the two marginal distributions; the variables are not presumed independent.

The marginal distributions have means $\mathbb{E}[X]=p$ and $\mathbb{E}[Y]=p$. In the joint distribution, $XY$ is nonzero only at $(1,1)$: when $X=1$, the only possible values of $Y$ are $0$ and $1$. HenceThese quantities are finite; for example, $0⩽XY⩽1$. The covariance changes sign when $p=e^{-p}$.

There is no such value of $p$ in the specified interval. Indeed, $\mathbb{P}(Y=1)=p{e}^{-p}>0$, and the joint probability mass function givesThis contradicts independence for every $0<p<0.8$. Equivalently, $(1,2)$ has joint probability zero but the product of its marginal probabilities is $p{e}^{-p}{p}^2/2>0$.

At the suggested value $r=e^{-r}$, the covariance is zero, but the same conditional probability contradiction still applies. This is an explicit example of how uncorrelated Bernoulli and Poisson variables need not be independent. Zero covariance controls one mixed moment, whereas independence requires the entire joint probability mass function to factor.