Differentiating the moment-generating function at zero givesThe second derivative is the second raw moment, so the variance is $M_X^{\prime \prime }(0)-(M_X^{\prime }(0){)}^2$. The assumed exponential integrability justifies these derivatives.

The moment-generating function of a real random variable $X$ isfor real $t$ where the expectation is finite. One may regard it as an extended nonnegative function elsewhere. In particular $M_X(0)=1$; finiteness on a neighborhood of zero is the usual hypothesis for recovering moments and using its local expansion.

Every $X_n$ has the same exponential distribution, so $X_n\stackrel{d}{\to }X$ for $X\sim \mathrm{Exp}(1)$. Part (ii) gives $Z_n-X_n\stackrel{p}{\to }0$. Applying Slutsky's theorem to $Z_n=X_n+(Z_n-X_n)$ yieldsThe limiting probability distribution does not depend on $c$. This is precisely the attraction to the exponential invariant law of a random affine recursion.

The two recursions use a synchronous coupling: they receive exactly the same fresh inputs. Subtraction cancels the additive term:The variable $U_n$ is independent of the preceding difference and has expected value $1/2$. Thus expectation of a product of independent random variables givesBy Markov's inequality, for every $\epsilon >0$,This proves the requested convergence in probability; in fact the difference has L1 convergence. It is synchronous contraction of a random affine recursion, and the estimate holds for every real $c$.

The answer isIt holds initially by construction. If it holds at step $n-1$, then $X_{n-1}$ is a function of $E_0,\dots ,E_{n-1},U_1,\dots ,U_{n-1}$, so it is independent of the fresh inputs $E_n,U_n$. Part (a)(iii) therefore gives $U_n(X_{n-1}+E_n)\sim \mathrm{Exp}(1)$. This completes the mathematical induction and identifies the stationary distribution of the recursion.

Put $W=X+Y$. The independent pair $(U,W)$ has the same joint distribution as $(V,W)$ from part (ii), because both first coordinates are uniform and independent of the same sum law. HenceConsequentlyThe rate is one. This multiplication identity is the mechanism behind the exponential invariant law of a random affine recursion.

Use the inverse change of variables for probability densitiesIts absolute Jacobian determinant isThereforeSince ${\int }_0^{\infty }w{e}^{-w}dw=1$, its marginal densities are $f_V(v)=1_{(0,1)}(v)$ and $f_W(w)=w{e}^{-w}{1}_{(0,\infty )}(w)$. The joint density factorizes as $f_V{f}_W$, so $V$ and $W$ are independent random variables, with $V$ uniform on $[0,1]$ and $W\sim \mathrm{Gamma}(2,1)$ in the shape-rate convention. This is beta-gamma independence with both shape parameters equal to one.

By independence of random variables, the joint probability density is the product of the two probability density functions of the rate-one exponential distributions:Its values on the boundary are immaterial, because the boundary has zero area.

Work on the single event of probability one where $Y_n/n\to \mu =2p-1>0$. On every sample path in that event, $Y_n\to +\infty$ and the unit-step property ensures that every positive integer $j$ is reached, so $T_j<\infty$. Starting from zero requires at least $j$ steps to reach $j$, hence $T_j\ge j$ and $T_j\to \infty$.

The convergent numerical sequence $Y_n/n$ on this fixed sample path also converges along its possibly random subsequence $n=T_j$. Since $Y_{T_j}=j$,Taking the reciprocal is valid because $\mu >0$. We obtainThis is a first-passage law of large numbers for an upward skip-free random walk. No independence between the hitting times and the path is needed: almost sure convergence already supplies convergence along every diverging subsequence on the same event.

Since $Y_n\to +\infty$ almost surely, only finitely many early values can attain the global minimum; it is a finite integer. Also $W\le Y_0=0$. For every nonnegative integer $m$, the unit-step property and translation invariance of the increments giveSubtracting consecutive tails yields the probability mass functionThus $-W$ has the geometric distribution counting failures before the first success, with success parameterThis support convention is essential: the minimum of an upward-biased random walk can equal its starting position, so $-W=0$ has positive probability.

The first-step analysis gives $h_0=1$ and, for $k\ge 1$,The minimal nonnegative solution for hitting probabilities theorem states that the probability of ever hitting a target is the pointwise smallest nonnegative solution of this transition recurrence with value one at the target. It follows by comparing any such solution with the probabilities of hitting within $n$ steps, then taking their increasing limit.

The characteristic equation $p{r}^2-r+q=0$ has distinct roots $1$ and $\beta =q/p\in (0,1)$. Hence every solution on the nonnegative integers with boundary value one has the formIf it is nonnegative for all $k$, taking $k\to \infty$ forces $A\ge 0$, and therefore $g_k-{\beta }^k=A(1-{\beta }^k)\ge 0$. The candidate $g_k={\beta }^k$ is itself nonnegative and satisfies the recurrence and boundary value. It is consequently the minimal solution, giving the biased random walk hitting probabilityOn the negative half-line the already established values $h_k=1$ satisfy the same recurrence away from zero. No unproved boundary condition at infinity has been imposed.

For $k=0$, the target is reached at time zero, so $h_0=1$. For $k<0$, the preceding almost sure convergence gives $Y_n\to +\infty$ with probability one. A sample path with unit upward and downward steps must visit every intermediate integer when it crosses from a negative value to a positive value. It must therefore visit zero at a finite time. ThusThe unit-step property matters: positive long-run displacement alone would not force a walk with larger upward jumps to land exactly on zero.

Put $q=1-p$ and ${\xi }_i=Y_i-Y_{i-1}$. The Markov property and the state-independent transition probabilities show that, conditional on the whole previous history, ${\xi }_i$ is $+1$ with probability $p$ and $-1$ with probability $q$. Multiplying these conditional probabilities for any finite step sequence proves that the ${\xi }_i$ are independent and identically distributed random variables, withBecause $Y_n=k+{\sum }_{i=1}^n{\xi }_i$, the strong law of large numbers givesIn particular, $Y_n\to +\infty$ almost surely. This stronger pathwise consequence will determine both eventual crossings and the asymptotic first-passage times.

For fixed $v>0$, the support inequalities become $v/2<u<v$. The marginal density is thereforeIt is nonnegative and integrates to $2-1=1$. Another interpretation is that $V=X+2Y$ sums independent exponential random variables of rates $1$ and $1/2$, so it has a hypoexponential distribution. The mean is $1+2=3$, which agrees with ${\int }_0^{\infty }v(e^{-v/2}-e^{-v})dv=4-1=3$.

Integrating the joint probability density across $u<v<2u$ gives the marginal densityFor every $u>0$, the conditional probability density is consequentlyHenceThis also follows from two independent exponential variables conditioned on their sum: given $X+Y=u$, the variable $Y$ is uniform on $(0,u)$, and $V=u+Y$. Conditioning on $U=u$ here is defined through the conditional probability density, rather than division by $\mathbb{P}(U=u)$, which is zero. The endpoint $u=0$ lies outside the positive marginal support and needs no specified conditional law.

The inverse linear transformation isThe positive support $x>0,y>0$ is exactly $u>0$ and $u<v<2u$. The independence and rate-one exponential distributions give $f_{X,Y}(x,y)=e^{-x-y}$ there. Since $x+y=u$, the change of variables for probability densities yieldsAlthough the density's value depends only on $u$, its support does not factor as a product of fixed intervals. Thus this expression does not imply independence of $U$ and $V$.

Set $\alpha =\arctan c\in (0,\pi /2)$. In polar coordinates, the event is $\sin \Theta >c|\cos \Theta |$, the angular sectorBecause $\Theta$ has the continuous uniform distribution on an interval of length $2\pi$, the probability is the sector's angular length divided by $2\pi$:The symmetry of the standard normal distribution in $Y$ makes the lower sector $Y<-c|X|$ equally likely. Their disjoint union is $|Y/X|>c$, apart from $X=0$, an event of probability zero. Therefore the requested cumulative distribution function isThis is the unit-scale half-Cauchy distribution. At $z=0$ there is no atom, and as $z\to \infty$ the cumulative distribution function tends to one.

The left panel uses $c=1$, giving a sector of angle $\pi /2$ and probability $1/4$. The right panel displays the domain $u<v<2u$ for the next change of variables for probability densities; a vertical section through it determines the conditional range.

The joint probability density of the two independent standard normal random variables is $(2\pi {)}^{-1}{e}^{-(x^2+y^2)/2}$. The polar coordinates map has absolute Jacobian determinant $r$, so the change of variables for probability densities givesAt the origin the angle may be assigned arbitrarily: that point has probability zero. The joint probability density factorizes as a Rayleigh distribution density $r{e}^{-r^2/2}$ and a continuous uniform distribution density $1/(2\pi )$. Hence $R$ and $\Theta$ are independent.

For $u>0$, substitute $s=r^2/2$ to findTherefore $R^2\sim \mathrm{Exp}(1/2)$, equivalently ${\chi }_2^2$, and applying a measurable function to $R$ preserves its independence from $\Theta$. Explicitly, with $S=R^2$,Thus $\Theta \sim \mathrm{Unif}[0,2\pi )$ and $S\sim \mathrm{Exp}(1/2)$ independently. This polar decomposition of a planar standard normal vector separates the uniformly distributed direction from its random squared distance.

Write $A_k=N_{2k}-N_k$ and $B_k=N_k$. The stationary increments and independent increments of the Poisson process make $A_k$ and $B_k$ independent random variables, each with Poisson distribution of parameter $k$. ThusMultiplication of the two moment-generating functions givesThe odd terms cancel in Taylor's theorem, so $\log M_k(t)=t^2+O(k^{-1})$. The limit $e^{t^2}$ is the moment-generating function of a normal distribution with mean zero and variance two. The same continuity theorem for moment-generating functions therefore yieldsIn particular, the two equal-length increments must be separated before using independence: $N_{2k}$ and $N_k$ themselves are dependent.

For fixed real $t$, Taylor's theorem for the exponential function givesSubstituting into the previous moment-generating function,The continuity theorem for moment-generating functions states that, if the moment-generating functions of $Z_n$ and $Z$ are finite on a common open interval containing zero and $M_{Z_n}(t)\to M_Z(t)$ throughout that interval, then $Z_n\stackrel{\mathrm{d}}{\to }Z$. Here finiteness holds for all real $t$, and the limit is the moment-generating function of a standard normal variable. ThereforeThis proves the required instance of the Poisson central limit theorem directly from moment-generating functions; no appeal to a general central limit theorem is needed.