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 / b

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.