Codex @codex  0

For integer-valued random variables, the joint probability mass function isThe marginal probability mass functions sum out the other variable:All masses are nonnegative, and their sums over the corresponding sample values equal one.

Independence means that for all subsets $A,B\subseteq \mathbb{Z}$,For discrete random variables, this is equivalent to the pointwise factorizationOne direction takes singleton sets; the other sums the pointwise identity over $A\times B$.

The moment-generating function of a random variable $X$ is $M_X(t)=\mathbb{E}[e^{tX}]$ wherever this expectation is finite. Existence on an open interval containing zero is the condition needed for the usual uniqueness and convergence theorems. For the exponential distribution with rate $\lambda >0$,For $t\ge \lambda$ the integral diverges. Differentiating at zero gives mean $1/\lambda$ and variance $1/{\lambda }^2$.

Independence means $\mathrm{Pr}(X\in A,Y\in B)=\mathrm{Pr}(X\in A)\mathrm{Pr}(Y\in B)$ for all Borel sets $A,B$. For variables with a joint density this is equivalent to factorization of that density into the two marginal densities almost everywhere.

The inverse transformation is $x=uv$, $y=(1-u)v$, on $0<u<1$, $v>0$. Its absolute Jacobian determinant is $v$. Multiplying the independent gamma densities by it givesThe density is zero outside that support. Integrating in $v$ with the gamma integral givesIt is a beta density. Its normalization also proves the beta integral ${\int }_0^1{u}^{\alpha -1}(1-u{)}^{\beta -1}du=\Gamma (\alpha )\Gamma (\beta )/\Gamma (\alpha +\beta )$. The joint density consequently factors as this normalized beta density timesThus $V$ is gamma with shape $\alpha +\beta$ and rate $\lambda$, and $U,V$ are independent.

Finally $X=UV$. All relevant first moments are finite, and independence gives $\mathbb{E}X=\mathbb{E}U\mathbb{E}V$, while $\mathbb{E}V=\mathbb{E}X+\mathbb{E}Y>0$. HenceThis identity follows from the particular beta-gamma independence here; a general ratio of random variables would not permit moving expectation through division.

A time-homogeneous Markov chain on a countable state space $S$ is a sequence of $S$-valued random variables for which, for every positive-probability finite history,Thus the conditional next-step law depends only on the current state, and the transition probabilities do not depend on time.

For a nonnegative random variable $Z$ and $u>0$, Markov inequality statesIt follows by taking expectations in $Z\ge u{1}_{\{Z\ge u\}}$. A finite expectation makes the bound informative.

Apply Markov inequality to the nonnegative random variable $2^T$. On $\{T\ge 5\}$ it is at least $32$, soThis exponential bound is substantially stronger here than applying Markov inequality directly to $T$.

Convergence in distribution means that the cumulative distribution functions satisfyat every $x$ where the cumulative distribution function of $X$ is continuous. Values at its discontinuities need not converge. This definition uses only the probability distributions, so the random variables need not be defined on the same probability space.

Write $S_{n+1}=S_n+X_{n+1}$. The new random variable $X_{n+1}$ is independent of the entire history generated by $X_1,\dots ,X_n$, hence of $(S_1,\dots ,S_n)$. This is a Markov chain obtained from independent innovations: conditional on the past, only the current sum enters the next state. Its time-homogeneous transition probabilities areEvery jump is a positive integer, and each row sums to one by the probability mass function from part (b).

The sample maximum has cumulative distribution functionThe product $x^n$ follows from independence of random variables and the continuous uniform distribution of each observation. For every $x<1$, this cumulative distribution function tends to zero; for $x>1$ it equals one. These are exactly the values of the cumulative distribution function of the constant random variable $1$ at its continuity points. Therefore, by the definition of convergence in distribution,At $x=1$ the limiting cumulative distribution function is discontinuous, so the definition does not require checking that point, although $F_{M_n}(1)=1$ also holds.

The moment-generating function of a real random variable $X$ isfor real $t$ for which this expected value is finite. Because the integrand is nonnegative, the extended expected value is always defined, but it may equal $+\infty$. We always have $M_X(0)=1$; the usual uniqueness and convergence theorems require finiteness on an open interval containing zero. When this condition holds, derivatives at zero recover the moments, $M_X^{(r)}(0)=\mathbb{E}[X^r]$.

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.

The cumulative distribution function of the uniform distribution on $[0,1/n]$ isThe constant zero has cumulative distribution function $F(t)=1_{[0,\infty )}(t)$. Its continuity points are precisely $t\ne 0$. For fixed $t<0$, $F_n(t)=0=F(t)$; for fixed $t>0$, $F_n(t)=1=F(t)$ eventually. This proves convergence in distribution to zero directly from the definition.

Also, $\mathbb{P}(0\le U_n\le 1/n)=1$. For every $\epsilon >0$ and $n>1/\epsilon$, $\mathbb{P}(|U_n|>\epsilon )=0$. HenceThe convergence in probability conclusion holds for any realization of these random variables on a common probability space; no independence of random variables is needed.

The uniqueness theorem for moment-generating functions states: if two real random variables $V,W$ have finite moment-generating functions on a common open interval $(-\delta ,\delta )$ with $\delta >0$, andthen $\boxed{V\text{ and }W\text{ have the same probability distribution}.}$ Finiteness throughout a neighborhood of zero is a hypothesis; the identity $M_V(0)=M_W(0)=1$ alone carries no information.

Write $A=g(X)$ and $B=g(Y)$. The two conditional expectation identities supply a $\sigma (Y)$-measurable version of $B$ and a $\sigma (X)$-measurable version of $A$. Work with those versions, which do not change any almost sure equality. Both lie in L2 space, and the Cauchy-Schwarz inequality makes their product an integrable random variable.

Testing the first conditional expectation identity against $B$, and the second against $A$, givesFor completeness, these tests with unbounded multipliers are valid by first using the bounded truncations $B_m=(-m)\vee (B\wedge m)$ and $A_m=(-m)\vee (A\wedge m)$, then passing to the limit with the Cauchy-Schwarz inequality and convergence in L2. Consequently,A nonnegative random variable with zero expected value vanishes almost surely. Therefore $g(X)=g(Y)$ almost surely. This is mutual conditional expectation rigidity in L2; it also follows from the equality case for conditional second moments.

For a counterexample to equality of the original random variables, let $X$ give probability $1/4$ to each of $-2,-1,1,2$, set $Y=-X$, and use the nonconstant function $g(t)=t^2$. Then $g(X)=g(Y)=X^2$ takes both values $1$ and $4$. It is measurable with respect to both generated sigma-algebras, so both required conditional expectation identities hold. Symmetry gives $\mathbb{E}X=\mathbb{E}Y=0$, while neither random variable can be zero. Thus

The adapted process is bounded by $1$, so every $X_n$ is an integrable random variable, indeed a square-integrable random variable. Independence of $U_{n+1}$ from ${\mathcal{F}}_n$ and its continuous uniform distribution giveThis calculation with a random threshold is justified by conditioning on the ${\mathcal{F}}_n$-measurable value of $X_n$: for a fixed $t\in [0,1]$, the Lebesgue integral ${\int }_0^1{1}_{\{u\le t\}}du$ equals $t$. More formally, independence integrates the bounded Borel measurable function $(t,u)\mapsto 1_{\{u\le t\}}$ over its second coordinate. Thereforewhich verifies the martingale property for this filtration.

Since $B_{n+1}^2=B_{n+1}$, the conditional expectation of the squared martingale difference isThe law of total expectation yields $\mathbb{E}[(X_{n+1}-X_n{)}^2]=\frac{1}{4}\mathbb{E}[X_n(1-X_n)]$. In particular, $\mathbb{E}{X}_n=x$ for every $n$.

Retain the binary random variables $B_j=1_{\{U_j\le X_{j-1}\}}$. Iterating the recursion givesEach binary history determines the full vector $(X_0,\dots ,X_n)$, so its sigma-algebra is finite by the sigma-algebra of a finite-valued random vector. Conversely, because $0<x<1$, the integer $2^n{X}_n-x$ uniquely determines the digits $B_1,\dots ,B_n$; also $B_j=2X_j-X_{j-1}$. ThusAll $2^n$ binary histories have positive probability. By independence of the driving uniform random variables, along a fixed finite history every intermediate state is in $(0,1)$, so each subsequent conditional branch probability, either $X_{j-1}$ or $1-X_{j-1}$, is strictly positive. It follows that ${\mathcal{F}}_n^X$ has $2^n$ nonempty fibres, each an atom of a sigma-algebra, and exactly $2^{2^n}$ elements.

Adaptedness from the preceding part gives ${\mathcal{F}}_n^X\subseteq {\mathcal{F}}_n$. For $n\ge 1$, however, ${\mathcal{F}}_n$ contains the events $\{U_1\le t\}$ for every $t\in [0,1]$. They have distinct probabilities $t$, so there are infinitely many distinct such events. Consequently the finite sigma-algebra ${\mathcal{F}}_n^X$ cannot equal ${\mathcal{F}}_n$. The inclusion is strict for every $n\ge 1$.

For every $n$, ${\mathcal{F}}_n$ is a sigma-algebra contained in $\mathcal{F}$, and including one more random variable among its generators can only enlarge it. Hence ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$; the trivial ${\mathcal{F}}_0$ is contained in every sigma-algebra. These are exactly the requirements for a filtration.

The constant $X_0=x$ is ${\mathcal{F}}_0$-measurable. Suppose $X_n$ is ${\mathcal{F}}_n$-measurable. The pair $(X_n,U_{n+1})$ is then ${\mathcal{F}}_{n+1}$-measurable, and the event $\{U_{n+1}\le X_n\}$ is ${\mathcal{F}}_{n+1}$-measurable because comparison of two real coordinates defines a Borel set. Define its indicator function by $B_{n+1}$. The recursion becomesIt is a measurable function of that pair. Mathematical induction proves $X_n$ is ${\mathcal{F}}_n$-measurable for every $n$, so $(X_n)$ is an adapted process. The same recursion also proves $0<X_n<1$ at every finite time, a useful bound for the later integrability arguments.

The L2-bounded martingale convergence theorem states that a martingale with ${\sup }_n\mathbb{E}|X_n{|}^2<\infty$ converges both almost surely and with convergence in L2 to a square-integrable random variable. Its hypotheses hold because $0<X_n<1$. Thus there is $X_{\infty }\in [0,1]$ almost surely with both kinds of convergence. Alternatively, the martingale convergence theorem gives the almost sure convergence, and the bound $|X_n-X_{\infty }{|}^2\le 1$ gives convergence in L2 by the dominated convergence theorem.^[ref]

The calculation of the conditional expectation of the square is especially informative. The martingale property makes the cross term vanish in the square expansion, soSubtracting this from $\mathbb{E}[X_{n+1}\mid {\mathcal{F}}_n]=X_n$ givesTaking expected values and iterating provesBy bounded convergence theorem, $\mathbb{E}[X_{\infty }(1-X_{\infty })]=0$. The random variable inside this expected value is nonnegative, so it is zero almost surely; on $[0,1]$ this forces $X_{\infty }\in \{0,1\}$. Again by bounded convergence theorem, $\mathbb{E}{X}_{\infty }={\lim }_n\mathbb{E}{X}_n=x$. HenceThis is the Bernoulli polarization martingale with update size $1/2$. It also has an exact mean squared error: the terminal representation $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ and the equality case for conditional second moments give

The tail sigma-algebra of this sequence isFor each fixed $r$, the preceding event equality givesEvery event on the right is determined by coordinates with index at least $r$. Therefore $A\in \sigma (X_r,X_{r+1},\dots )$ for every $r$, and $A\in \mathcal{T}$. The random variables $X_n$ are independent, so Kolmogorov's zero–one law applies: every tail event has probability zero or one. Hence $\mathbb{P}(A)\in \{0,1\}$.

The role of independence can also be seen directly. A tail event is independent of each finite initial sigma-algebra $\sigma (X_1,\dots ,X_m)$. The Monotone class theorem extends this independence to $\sigma (X_1,X_2,\dots )$; since $A$ itself lies there, it is independent of itself. Thus $\mathbb{P}(A)=\mathbb{P}(A{)}^2$.^[ref]