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Let $V={\sum }_{n\ge 1}\Vert X_n{\Vert }_{\infty }$. It is a nonnegative extended random variable, and $\mathbb{E}V<\infty$ implies $\mathbb{P}(V<\infty )=1$. Otherwise its expectation would be infinite. Intersect this event with the countably many probability-one events on which the individual sample paths are continuous.

For every path in that intersection, ${\sum }_n\Vert X_n(\cdot ,\omega ){\Vert }_{\infty }<\infty$. The Weierstrass M-test gives absolute and uniform convergence of ${\sum }_n{X}_n(t,\omega )$ on the entire stated time domain. A uniform limit of continuous functions is continuous. Therefore $Y$ has continuous paths almost surely. The finite expectation is used to obtain pathwise summability before applying this deterministic theorem.

Conditioning on the observed time. The native PDF defines ${\tau }_2$ as the first production time and then conditions on that production occurring at ${\tau }_2$. If the intended information is its numerical value ${\tau }_2=t$, the answer depends on $t$. We first compute this conditional probability using a conditional probability density; conditioning on an exact value of a continuous random variable means a density ratio, since $\mathbb{P}({\tau }_2=t)=0$.

Before ${\tau }_2$, no molecules of the second chemical species are present, so the joint-removal chemical reaction cannot occur. Molecules of the first chemical species arrive as a rate-$2$ Poisson process, and each independently starts a clock with a rate-$1/2$ exponential distribution for producing the first molecule of the second chemical species. Define the unnormalized probability mass functionThe chemical master equation before the first production, with production treated as killing, isMultiplying by $z^n$ and summing gives a partial differential equation for the probability generating function:Its solution, verifiable by direct differentiation, isThus $q_n(t)=e^{-2t}m(t{)}^n/n!$, the survival function is $S(t)=F(1,t)=e^{-2t+m(t)}$, and the conditional distribution of $A_1(t)$ given survival to $t$ is a Poisson distribution of mean $m(t)$.

Production in $(t,t+dt)$ from Markov chain state $n$ has joint probability $(n/2)q_n(t)dt+o(dt)$. Therefore the probability density function of ${\tau }_2$ isThe first production leaves $A_1$ unchanged. Taking the density ratio gives a size-biased distribution of the surviving count:In particular, under this reading the requested conditional probability isThe shift by one matters: an observation made exactly when production occurs preferentially selects larger counts, since their production reaction propensity function is larger.

Without observing the numerical time. If the conditioning merely repeats the definition of ${\tau }_2$, the requested probability is instead the unconditional count at that stopping time. Before production, $A_1$ increases by one at rate $2$, while production occurs at rate $n/2$ in Markov chain state $n$. The competing exponential clocks implyThe first factor requires a second arrival before production from count one; the second requires production before a third arrival. The first arrival itself is certain. Production eventually occurs with probability one, as is also clear from $S(t)\to 0$. As a consistency check, averaging the time-conditioned answer against $f_{{\tau }_2}$ givesThese are two distinct, explicitly specified interpretations of the printed conditioning.

For a discrete random variable taking nonnegative integer values, its probability generating function isIt is defined at least for $|s|\le 1$. The usual probability generating function is intended for nonnegative integer-valued random variables; a general discrete real-valued random variable need not have this form. The coefficient of $s^k$ recovers $\mathbb{P}(X=k)$.

The identity $XY=(S_1+S_2)(S_1-S_2)=S_1^2-S_2^2$ gives $\mathbb{E}[XY]=0$, since the two rolls have the same probability distribution. Also $\mathbb{E}[X]\mathbb{E}[Y]=7\cdot 0=0$. ThereforeThe two random variables are uncorrelated random variables, but that alone does not establish independence of random variables.

This numbered item specifies the model used below. A convenient sample space is the collection of $36$ ordered pairs $(s_1,s_2)\in \{1,\dots ,6{\}}^2$, each with probability $1/36$. The corresponding random variables are determined by the sum and difference. Inverting the two equations givesThus a candidate pair $(X,Y)=(i,j)$ is possible exactly when $(i+j)/2$ and $(i-j)/2$ are both integers between $1$ and $6$. In particular, $i$ and $j$ must have the same parity. This inverse description will determine the joint probability mass function.

Assume $a<b$. An increasing affine map sends the two endpoints to zero and one when $ca+d=0$ and $cb+d=1$. Solving givesIndeed, for $0\le t\le 1$, the cumulative distribution function of the transformed random variable iswhich is the cumulative distribution function of the continuous uniform distribution on $[0,1]$.

A decreasing affine map works too, by swapping the endpoints:These are the two possible choices, since the image interval must have endpoints zero and one. The case $a=b$ cannot be transformed into a nondegenerate continuous uniform distribution by deterministic constants.

For each random variable, the probability mass function givesLet $S_m={\sum }_{i=1}^m{X}_i$. By linearity of expectation and variance additivity for independent random variables,The independence is needed for the variance calculation, but not for the mean.

The random variable $Y=Z^2/2$ is nonnegative, so its cumulative distribution function is zero for $y<0$. If $y\ge 0$, thenThe last step uses the symmetry of the standard normal distribution, $\Phi (-z)=1-\Phi (z)$. Hence

The probability generating function of a nonnegative integer-valued random variable isusing $s^0=1$ also at $s=0$. It is always defined for $0\le s\le 1$, and indeed the power series converges absolutely for all real or complex $s$ with $|s|\le 1$, becauseIt may be defined beyond the unit disk where this power series converges absolutely; such a larger domain depends on the distribution. In particular $G_X(0)=\mathbb{P}(X=0)$ and $G_X(1)=1$.

For a nonnegative integer-valued random variable $X$, its probability generating function isAt $s=0$, use $0^0=1$ so $G_X(0)=\mathbb{P}(X=0)$. The series also converges absolutely for complex $|s|\le 1$; for $s$ near zero, its coefficients recover the probability mass function.

Use the measurable partition $B_n=\{N=n\}$ for $n\ge 0$, omitting any null cells. For $0\le s\le 1$, the random variable $s^X$ is bounded, so the law of total expectation applies without moment assumptions. Conditional on $N=n$, the sum contains exactly $n$ terms. Independence of $N$ from the whole sequence preserves their joint law under this conditioning; independence of the $Y_k$ then givesFor $n=0$ the empty sum is zero and the empty product is one, in agreement with the formula. Therefore the probability generating function of an independent random sum is

First $\mu >0$. If a nonnegative random variable had zero expected value, it would equal zero almost surely, which is impossible for a distribution having a probability density function on $[0,\infty )$: a singleton has probability zero. Thus the given division by $\mu$ is legitimate.

Because $x\ge 0$ and $f(x)\ge 0$, $g(x)=xf(x)/\mu \ge 0$. MoreoverExtend $g$ by zero for negative $x$. It is a probability density function, defining the size-biased distribution of $X$: larger observations receive weight proportional to their value.

For a nonnegative integer-valued random variable, its expectation is the probability-weighted sumAll terms are nonnegative, so this is always defined as an extended nonnegative number, possibly $+\infty$.

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.

For a nonnegative integer-valued random variable, its probability generating function is $G_X(s)=\mathbb{E}[s^X]={\sum }_{k\ge 0}\mathbb{P}(X=k)s^k$, at least for $|s|\le 1$. Insert the Poisson distribution masses and sum the exponential series:

**Chebyshev's inequality: if $X$ has finite mean $\mu$ and variance ${\sigma }^2$, then for $a>0$,Apply Markov's inequality to the nonnegative random variable $Y=(X-\mu {)}^2$ with threshold $a^2$. Since $\mathbb{E}[Y]={\sigma }^2$ and $\{|X-\mu |\ge a\}=\{Y\ge a^2\}$, the desired statement follows immediately.

For $Z=2(X+Y)+1$, independence givesThe initial factor $s$ accounts for the shift by one; substituting $s^2$ accounts for doubling the random variable. Its support consists of positive odd integers.

Conditioning on $Z=z$ has positive probability exactly when $z=2r+1$ for an integer $r\ge 0$. This is the same event as $X+Y=r$. The addition of independent Poisson random variables gives a Poisson distribution with parameter $\lambda +\mu$ for the sum. Therefore, for $0\le t\le r$,Thus $X\mid \{Z=2r+1\}\sim \mathrm{Binomial}(r,\lambda /(\lambda +\mu ))$. The case $r=0$ is a constant zero random variable. For even, nonpositive, or noninteger $z$, the conditioning event has zero probability, so an event-conditional probability distribution is not specified by the quotient formula. This is a two-component case of Poisson-multinomial conditioning.

Because $s>1$, the function $t\mapsto s^t$ is increasing and nonnegative. Thus $\{X\ge x\}=\{s^X\ge s^x\}$. Applying Markov's inequality to the nonnegative random variable $s^X$, whose expected value is finite by assumption, yieldsUnder the usual probability generating function convention $X$ takes nonnegative integer values. The argument itself works for any real random variable if $G_X(s)$ is understood as the finite expected value $\mathbb{E}[s^X]$.

For a nonnegative integer-valued random variable, its probability generating function isThe series converges absolutely at least for $|s|\le 1$, with $0^0=1$ in the $k=0$ term; it may converge on a larger domain.