Past exam of the mathematics course of the University of Oxford / Undated / Prelims / Mathematics III resit / 6

A joint probability density function integrates to one. HeresoThis value also makes the probability density function nonnegative everywhere.

Obtain each marginal density by integrating the joint probability density function over the other variable:Each marginal density is zero off its displayed interval and has total integral one.

On the product support $[0,1{]}^2$, the joint probability density function factors asThe equality also holds outside the support with both sides zero. The probability density function factorization criterion proves that $U$ and $V$ are independent random variables.

For each nonnegative integer $z$, the events $\{X=z-y,Y=y\}$, $0\le y\le z$, are pairwise disjoint and their union is $\{X+Y=z\}$. By countable additivity and then independence,This is the discrete convolution of the two probability mass functions. If the convention for natural numbers excludes zero, set both probability mass functions equal to zero there; the formula remains valid. Also $p_Z(z)=0$ for negative $z$.

Let $R$ denote the red die. Conditional on $R=r$, each white roll succeeds independently with probability $p_r=(7-r)/6$. Consequently $T\mid R=r$ has the geometric distribution on $\{1,2,\dots \}$ and expected value $1/p_r=6/(7-r)$. The law of total expectation givesThis is $2.45$ rolls; the averaging must be over conditional expected values, not over success probabilities before taking a reciprocal.

By the law of total probability,Bayes' theorem therefore gives the conditional probability mass functionThe six probabilities sum to one. A small red outcome is more likely after an immediate success because more white faces can meet its threshold.

For a geometric distribution on positive integers, the probability generating function is $ps/[1-(1-p)s]$. Conditional averaging over the red die yieldsAs a power series, its exact domain of absolute convergence isIndeed the slowest decaying component has failure probability $5/6$, so $\mathbb{P}(T=t)\ge (1/36)(5/6{)}^{t-1}$. At $|s|=6/5$, the terms of the power series fail to tend to zero; outside that disk they grow. All six component series converge inside it. In particular, the probability generating function is defined on the usual real interval $0\le s\le 1$, with $G_T(1)=1$. The rational expression beyond the convergence disk is only an algebraic continuation, not the defining expected value.