Because $A\subseteq \mathcal{U}$, the set difference $A\setminus \mathcal{U}$ is empty and $\mathcal{U}\setminus A=\overline{A}$. Therefore the symmetric difference satisfiesThe disjoint sets $A$ and $\overline{A}$ have union $\mathcal{U}$, so

Use the definition $A\oplus B=(A\setminus B)\cup (B\setminus A)$ of symmetric difference. The algebraic laws for set difference and the empty set giveand

Complements are relative to the same universal set $\mathcal{U}$. Suppose $A\subseteq B$ and $x\in \overline{B}$. Then $x\in \mathcal{U}$ and $x\notin B$. If $x$ belonged to $A$, the assumed set inclusion would imply $x\in B$, a contradiction. Thus $x\notin A$ and $x\in \overline{A}$, provingThere is only one set inclusion in this conclusion; a reverse inclusion is neither needed nor generally true. The request for double inclusion applies to the equalities in (i) and (ii).

For the first set inclusion, take $x\in (A\cap B)\setminus C$. Membership in the intersection gives $x\in A$ and $x\in B$, while the set difference gives $x\notin C$. Consequently $x\in (A\setminus C)\cap (B\setminus C)$.

For the reverse set inclusion, an element of $(A\setminus C)\cap (B\setminus C)$ belongs to both $A$ and $B$ and does not belong to $C$. It therefore belongs to $(A\cap B)\setminus C$. By double inclusion proof,

We prove both set inclusions, as required by a double inclusion proof. If $x\in A\setminus (B\cup C)$, then $x\in A$ and $x\notin B\cup C$. Thus $x\notin B$ and $x\notin C$, so $x\in A\setminus B$ and $x\in A\setminus C$. This proves the forward set inclusion.

Conversely, if $x\in (A\setminus B)\cap (A\setminus C)$, then $x\in A$, $x\notin B$, and $x\notin C$. Hence $x\notin B\cup C$, giving $x\in A\setminus (B\cup C)$. Therefore

For any real $y$, put $k=\lfloor y\rfloor$. Since $Y$ is integer-valued,Continuity of the cumulative distribution function of $X$ therefore givesThe strict inequality $X<k+1$ matters before using the absence of point masses. The resulting probability distribution is discrete, with probability mass functionThus $Y$ has no ordinary probability density function with respect to length on the real line. Its cumulative distribution function is a step function; its derivative is zero almost everywhere and does not recover its probabilities. The floor function has converted a density into masses on the integers.

Since a square is nonnegative, $F_Y(y)=0$ for $y<0$. For $y\ge 0$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$. As $X$ has no point masses,Differentiating for $y>0$ gives the contributions of the two inverse branches:The value assigned to the probability density function at $y=0$ is arbitrary. It can be unbounded near $0$, but there is no atom there because $\mathbb{P}(X=0)=0$. Both positive and negative preimages must be included; keeping just one would generally lose probability.

Reversing sign reverses the inequality. Since $X$ has a probability density function, it has no point masses, soDifferentiating the cumulative distribution function givesalmost everywhere: the minus sign outside $F_X$ cancels the minus sign from the chain rule. If point masses were allowed, the cumulative distribution function would instead require the left limit $1-F_X((-y{)}^{-})$.

Translate the inequality inside the cumulative distribution function:Differentiating with the chain rule gives the probability density functionalmost everywhere. Translation shifts the probability distribution without changing its shape or its total probability.

Let $S(x)=\mathbb{P}(X>x)=1-F_X(x)$ for $x\ge 0$. Since $f_X=0$ on the negative half-line, $X\ge 0$ almost surely. We have $S^{\prime }=-f_X$ almost everywhere. On a finite interval, integration by parts givesThe lower endpoint contributes zero. The stated hypothesis gives $RS(R)\to 0$ as $R\to \infty$. Taking the increasing limits of the two nonnegative integrals proves the tail-integral identityThe equality allows the value $+\infty$; the boundary hypothesis alone does not force a finite expectation. In fact the formula for nonnegative $X$ also follows directly from Tonelli theorem, without that additional boundary hypothesis, by integrating $x={\int }_0^{\infty }{1}_{\{t<x\}}dt$ against the probability density function.

For an absolutely continuous probability distribution, its expectation isThis is finite when ${\int }_{-\infty }^{\infty }|x|f_X(x)dx<\infty$. More generally it is an extended real value if only one of the positive and negative parts diverges; it is undefined if both diverge. For nonnegative $X$ it always exists in $[0,\infty ]$.

The cumulative distribution function isFor an absolutely continuous probability distribution, the probability density function is a nonnegative integrable function $f_X$ withThus interval probabilities are integrals of $f_X$, and $F_X^{\prime }=f_X$ almost everywhere, in particular at continuity points of $f_X$.

The word “continuous” in the source is understood in this density sense. A merely continuous cumulative distribution function need not possess a probability density function; it can instead describe a singular continuous distribution.

For $n\ge 2$, $X_2+\cdots +X_n$ is independent of $X_1$ and has a Poisson distribution with parameter $(n-1)\lambda$. Part (ii) therefore givesIts probability at zero isFor $n=1$ the event being conditioned on is $X_1=1$, so the answer is $0$, also matching the displayed expression. The cancellation of $\lambda$ reflects the equal rates of all $n$ counts, not an approximation or large-$n$ limit.

For $0\le k\le n$, the event $\{X=k,X+Y=n\}$ is $\{X=k,Y=n-k\}$. Use independence in its numerator and the Poisson distribution of the sum in its denominator:Outside $0\le k\le n$ the conditional probability is zero. ConsequentlyThis binomial distribution includes $n=0$, when it is concentrated at zero. Conditioning on the total converts the independent counts into an allocation of that total between the two rates.

For independent random variables, the probability generating function of an independent sum is the product of the individual probability generating functions. ThusExpanding this last expression as $e^{-(\lambda +\mu )}{\sum }_{r\ge 0}(\lambda +\mu {)}^r{s}^r/r!$ identifies each probability coefficient. Therefore

By definition, the probability generating function isSubstituting the Poisson distribution and evaluating the exponential series givesThe series converges for every finite complex $s$; in particular it gives the probability generating function on $0\le s\le 1$.

The exponential series gives ${\sum }_{n=0}^{\infty }{\lambda }^n/n!=e^{\lambda }$. Thus the normalization constant for the Poisson distribution isEvery value $e^{-\lambda }{\lambda }^n/n!$ is nonnegative, and their sum is $e^{-\lambda }{e}^{\lambda }=1$. Hence this is a valid probability mass function.

Let $D$ be the number of days up to and including the first breakdown, and $M$ the event of a morning first breakdown. For $d=1,2,\dots$, conditional probability givesThusThis geometric distribution counts the day on which the first breakdown occurs. Its parameter is the probability of at least one breakdown during a day, rather than the morning breakdown probability $\alpha$ alone.

Let $J$ count successful journeys before the first breakdown. The probability of surviving the first $r$ journeys is the tail probability $\mathbb{P}(J\ge r)$. For $d\ge 0$, independence giveswhere again $q=(1-\alpha )(1-\beta )$. For a nonnegative integer-valued random variable, the tail-sum formula follows by writing $J={\sum }_{r\ge 1}{1}_{\{J\ge r\}}$ and interchanging nonnegative sums. ThereforeAs a separate first-day check using the law of total expectation, if $E=\mathbb{E}(J)$, then $E=\alpha \cdot 0+(1-\alpha )\beta \cdot 1+q(2+E)$. Solving gives the same finite value. This counts only successful journeys, excluding the failed journey.

Put $q=(1-\alpha )(1-\beta )$, the probability of surviving a full day, and $p=1-q=\alpha +\beta -\alpha \beta$. These lie in $(0,1)$. Let $M$ be the event that the first breakdown is in the morning. By independence, surviving $d-1$ complete days and then failing in the morning of day $d$ has probability $q^{d-1}\alpha$.

These events are disjoint, so summing the geometric series givesThere is no missing probability of never breaking down: survival for $d$ days has probability $q^d\to 0$.