Past exam of the probability course of the University of Oxford / 2023 / 2 / b

The independence of random variables $B$ and $U$ is needed for the intended mixture distribution in (ii) and (iii). It is absent from the printed assumptions. We use independence for the main formulas, and then give the answers under the literal assumptions.

The uniform distribution has probability density function $1/2$ on $(-1,1)$. Therefore, for $t\ne 0$,The value at zero follows either directly from the definition of the moment-generating function or by continuity. This answer needs no assumption about $B$.

If $B$ and $U$ are independent random variables, conditioning on the Bernoulli random variable $B$ giveswith the continuous value $M_Z(0)=1$. Indeed $Z=0$ when $B=0$, while the conditional probability distribution of $Z$ given $B=1$ is the original uniform distribution.

Under the literal assumptions alone, the law of total expectation gives onlyThis is not determined by $M_U$ and $p$. For an explicit counterexample, put $p=1/2$ and $B=1_{\{U>0\}}$. Then $B$ has the required Bernoulli distribution, butwhich differs from the independent-case formula.

Under independence of random variables, the law of a random variable $Z$ is the mixed probability distribution $(1-p){\delta }_0+p\mathrm{Unif}(-1,1)$. For negative $z$, only the uniform component can contribute to $\mathbb{P}(Z\le z)$; at zero the probability mass $1-p$ is included. Thus its cumulative distribution function isThe sketch consists of a horizontal segment at height zero before $-1$, a line of slope $p/2$ ending at the open point $(0,p/2)$, a jump of size $1-p$ to the closed point $(0,1-p/2)$, another line of slope $p/2$ ending at $(1,1)$, and a horizontal segment thereafter. The cumulative distribution function is right-continuous at the jump.

Under the literal assumptions, write $H(z)=\mathbb{P}(U\le z\mid B=1)$. The general answer is $F_Z(z)=(1-p)1_{\{z\ge 0\}}+pH(z)$, so the continuous part of the sketch is undetermined. The jump at zero is still exactly $1-p$, because $\mathbb{P}(U=0)=0$. In the dependent counterexample from (ii), $F_Z(z)=0$ for $z<0$, $F_Z(z)=(1+z)/2$ for $0\le z<1$, and $F_Z(z)=1$ for $z\ge 1$.