The uniqueness theorem for moment-generating functions states that if $M_X$ and $M_Y$ are finite and equal on an open real interval containing zero, thenFor the convergence theorem for moment-generating functions, suppose that the moment-generating functions $M_{X_n}$ are finite on a common open real interval $(-a,a)$, $a>0$, and $M_{X_n}(t)\to M_X(t)$ for every $|t|<a$, where $M_X$ is finite there. ThenOne may equivalently assume convergence to a finite function $M$ on that interval: the theorem then asserts existence of a unique limiting probability distribution with moment-generating function $M$ as well as convergence in distribution.

For a Borel set $A\subseteq \mathbb{R}$, the joint probability density givesFor each $x\ne 0$, substitute $z=xy$. The change of variables for probability densities contributes $dy=dz/|x|$, with the absolute value accounting for the reversed orientation when $x<0$. The line $x=0$ has two-dimensional Lebesgue measure zero, so omitting it changes no probability. By the Tonelli theorem,The bracket is nonnegative and its integral over $\mathbb{R}$ is one. This proves the probability density of a product of random variables formula:As usual, a probability density function is determined only almost everywhere; exceptional infinite values are harmless.

For the independent random variables with the specified uniform distribution, $f(x,y)=b^{-2}$ on $(0,b{)}^2$ and zero elsewhere. When $0<z<b^2$, the conditions $0<x<b$ and $0<z/x<b$ are equivalent to $z/b<x<b$. ThereforeThe value assigned at zero is arbitrary. Substitution $s=z/b^2$ checks normalization of a probability distribution: ${\int }_0^1-\log sds=1$. This is the product of independent uniform random variables law.

The strong law of large numbers states that, for independent and identically distributed random variables $X_1,X_2,\dots$ with $\mathbb{E}|X_1|<\infty$,Thus, outside an event of probability zero, the sample mean converges to the common expected value.

Let $d=\gcd (a,b)>1$ and put $a^{\prime }=a/d$, $b^{\prime }=b/d$. Then $a^{\prime },b^{\prime }$ are coprime integers, and $p=b^{\prime }/(a^{\prime }+b^{\prime })$. The six requested revisions are as follows.

(b)(i): communicating classes. Every increment is divisible by $d$, so the residue class modulo $d$ is preserved. Conversely, if $j-i$ is divisible by $d$, Bézout's identity applied to $a^{\prime },b^{\prime }$ gives a path just as in (b)(i). Thus the original Markov chain is reducible on $\mathbb{Z}$, with exactly the $d$ closed communicating classesThe specified walk starting at zero stays in $d\mathbb{Z}$.

(b)(ii): period of a state in a Markov chain. The return equation $au=bv$ reduces to $a^{\prime }u=b^{\prime }v$. Hence $u=b^{\prime }k$, $v=a^{\prime }k$. For every state, the period of a state in a Markov chain is thereforeThis is still at least two, so no state is aperiodic.

(b)(iii): stationary distribution. The conclusion is unchanged: there is no stationary distribution on $\mathbb{Z}$. The maximum-mass proof in (b)(iii) applies verbatim, without using coprimality. Equivalently, each closed communicating class, after translating and dividing its state labels by $d$, is the same centered random walk with step sizes $a^{\prime },-b^{\prime }$ and has no stationary distribution.

(b)(iv): occupation of zero. On $d\mathbb{Z}$, the rescaled walk $S_n/d$ has exactly the transition probabilities of the coprime walk with parameters $a^{\prime },b^{\prime }$. The return-probability argument in (b)(iv) therefore makes this class null recurrent. The renewal process proof of the occupation limit applies unchanged:Starting in another residue class would make every visit to zero impossible, also giving occupation proportion zero.

(c)(i): modified communicating classes. These are unchanged: $C=\{-c-b,\dots ,c+a\}$ is one finite closed communicating class, and all other states are transient singleton classes. The original proof used consecutive reset destinations and never used coprimality. Resetting to the $b$ consecutive integers $c-b+1,\dots ,c$ connects different residue classes, so it would be incorrect to split $C$ according to residue classes modulo $d$.

(c)(ii): limiting distribution. The column-sum calculation, the irreducible Markov chain property on a finite class, positive self-transition probabilities, and geometric entrance argument also required no coprimality. Thus for every initial state the same convergence in distribution holds:In particular, the denominator $2c+a+b+1$ is unchanged; it is not divided by $d$.

The history $\sigma (S_0,\dots ,S_n)$ is contained in $\sigma (Y_1,\dots ,Y_n)$. The increment $Y_{n+1}$ is independent of this history by the given independence of random variables. ThereforeThis depends only on the current state and not on time. It is precisely the time-homogeneous Markov property, so $S_n$ is a Markov chain with transition probabilitiesThe conditional identity is an identity that holds almost surely; histories of zero probability require no division by their probabilities.

The support condition $-y<x<y$ forces $y>0$. As a check, the joint probability density integrates toFor any real $x$, the marginal density of $X$ isDividing the joint probability density by this marginal density gives the conditional density:Thus, conditionally on $X=x$, $Y-|x|$ has the rate-two exponential distribution.

For $U=X^2$, use the two inverse branches $x=\sqrt{u}$ and $x=-\sqrt{u}$. Each branch has absolute Jacobian determinant $1/(2\sqrt{u})$. Both lie in the original support exactly when $u>0$ and $y>\sqrt{u}$. The change of variables for probability densities therefore yieldsNo atom of a measure is created at $u=0$, since $\mathbb{P}(X=0)=0$. The value of the probability density function at this boundary may be assigned arbitrarily. Checking the integral over $0<u<y^2$ gives $4y{e}^{-2y}$, which integrates to one.