With all four external fermions of species 1, both final attachments are possible. The middle row of the figure gives the two tree-level Feynman diagrams:

For a fixed ordering of external fermion states, the fermionic sign from exchanging the final identical fermions givesHere $u_j=u_{s_j}(p_j)$ are Dirac spinors. In particular, the scattering amplitude changes sign when the complete final labels $3$ and $4$ are exchanged, as required for identical fermions.

Label the initial species-1 and species-2 fermions by $(p_1,s_1)$ and $(p_2,s_2)$, and the final species-1 and species-2 fermions by $(p_3,s_3)$ and $(p_4,s_4)$. There is exactly one tree-level Feynman diagram: the $t$-channel scalar propagator joins the continuous species-1 line $1\to 3$ to the continuous species-2 line $2\to 4$. Its exchanged four-momentum $q_t$ determines $t$, in the notation of Mandelstam variables:The top row of the figure displays this Feynman diagram. A crossed attachment $1\to 4$ would change species at an interaction vertex and is forbidden. There is no initial antifermion for an annihilation channel.

Let $\mathcal{G}=T^{-1}(\mathcal{F})=\{T^{-1}(A):A\in \mathcal{F}\}$. Inverse images commute with complements and countable unions, so this is a sub-sigma-algebra of $\mathcal{F}$. The random variable $X\circ T$ is $\mathcal{G}$-measurable.

The assumption $\mathbb{P}\circ T^{-1}=\mathbb{P}$ says that $T$ is a measure-preserving transformation. Consequently $X\circ T$ has the same probability distribution as $X$ and is integrable. It also implies $\mathbb{E}[W\circ T]=\mathbb{E}[W]$ for every integrable random variable $W$, first for indicator functions, then for simple functions, and then by positive and negative approximation.

To identify a conditional expectation, take any $A\in \mathcal{F}$. The assumed identity with the bounded random variable $Z=1_A$ givesEvery member of $\mathcal{G}$ has the displayed form $T^{-1}(A)$, so this verifies all defining conditional expectation identities. ThereforeApply part (iii), since this conditional expectation has the same probability distribution as $X$. We concludeThis is invariance from integral duality for a measure-preserving transformation; neither invertibility of $T$ nor a second moment of $X$ is required.

We first establish the lower truncation identity as well. The pair $(-X,-Y)$ has the same hypotheses as $(X,Y)$, because conditional expectation is linear and negation preserves equality of probability distributions. Apply part (ii) to this pair and multiply by $-1$ to getFor every real $t$,Using this identity, part (ii), the lower truncation identity, and linearity of conditional expectation givesIn particular the argument does not assume that a function obtained by scalar clipping is a convex function or a concave function.

Write $C_a(t)=(t\wedge a)\vee (-a)$. The bounded random variables $C_a(X)$ and $C_a(Y)$ have the same probability distribution, and the displayed identity identifies the latter as the conditional expectation of the former. They are square-integrable, so part (i) gives $C_a(X)=C_a(Y)$ almost surely. Take $a=1,2,\dots$ and discard the countable union of exceptional null sets. At every remaining outcome, let $a\to \infty$ to obtainThus conditional expectation preserving a distribution forces equality even when only integrability is assumed.

The function ${\varphi }_a(t)=t\wedge a$ is a concave function, and $X\wedge a$ is integrable since $|X\wedge a|\le |X|+a$. The concave form of the conditional Jensen inequality givesThe random variables $X$ and $Y$ have the same probability distribution, so $\mathbb{E}[X\wedge a]=\mathbb{E}[Y\wedge a]$. Using the law of total expectation, the nonnegative difference between the right and left sides of the inequality has zero expectation. It must vanish almost surely. ThereforeThe equality comes from the distributional assumption as well as the conditional Jensen inequality.

Assume $X$ is square-integrable. Since $X$ and $Y$ have the same probability distribution, $Y$ is also square-integrable and $\mathbb{E}[Y^2]=\mathbb{E}[X^2]$. The product $XY$ is integrable by the Cauchy-Schwarz inequality.

We use the defining test-function identity for conditional expectation: for bounded $\mathcal{G}$-measurable $H$, $\mathbb{E}[XH]=\mathbb{E}[YH]$. Take $H=Y{1}_{\{|Y|\le m\}}$ and let $m\to \infty$. The dominated convergence theorem, with bounds $|XY|$ and $Y^2$, givesThis is the extension of the conditional expectation pull-out identity to this square-integrable multiplier. ConsequentlyA nonnegative random variable with zero expectation is zero almost surely. HenceGeometrically, this is the equality case in the L2 space orthogonal projection interpretation of conditional expectation.

The bounded-time optional sampling theorem says that for an integrable discrete-time martingale $(U_n)$ and a bounded stopping time $S$, $\mathbb{E}[U_S]=\mathbb{E}[U_0]$. It requires no assumption about an eventual unbounded stopping time.

Apply this theorem with $S=n\wedge \tau$, or simply take the expectation of the stopped martingale in part (ii). For $N>0$ we obtainSince $P_{n\wedge \tau }\ge 0$,The monotone convergence theorem now yields $\mathbb{E}[\tau ]\le 3abc/N<\infty$. In particular $\tau <\infty$ almost surely, which has been established without assuming that optional stopping is valid at $\tau$.

For each sample path with finite $\tau$, $P_{n\wedge \tau }$ eventually equals $P_{\tau }=0$. Also $0\le P_{n\wedge \tau }\le N^3$. The dominated convergence theorem gives $\mathbb{E}[P_{n\wedge \tau }]\to 0$. Taking the limit of a sequence in the earlier identity therefore provesFor example, initial holdings $(1,1,1)$ give expectation one, as the first transfer must eliminate its giver. This computes the first elimination time, rather than the later time when one player owns all tokens.

After $\tau$, the token product remains zero, whereas the deterministic term $nN/3$ continues to increase. Combining this with the pre-elimination calculation in part (ii),The stochastic process is adapted and integrable at each fixed time, and the displayed conditional expectation is nonnegative. Thus $(M_n)$ is a submartingale.

For $N>0$ it is neither a martingale nor a supermartingale. If $a,b,c>0$, there is positive probability that the first player transfers a token to the second in each of the next $a$ rounds. Until the last such round all three remain active, so this event has probability $6^{-a}>0$ and entails $\tau \le a$. At time $n=a$ the displayed conditional expectation is therefore strictly positive on an event of positive probability, ruling out both the zero drift required of a martingale and the nonpositive drift required of a supermartingale. If an initial holding is zero, $\tau =0$ and the same conclusion follows immediately. The degenerate case $N=0$ gives a constant martingale.

Put $P_n=X_n{Y}_n{Z}_n$. Before $\tau$, all three players are active. Each unordered pair is selected with probability $1/3$, and each direction of transfer then has conditional probability $1/2$. Given the current holdings $(x,y,z)$, a transfer within the first pair has mean productFor transfers within the other two pairs the analogous values are $xyz-y$ and $xyz-x$. Averaging over pairs givesAt and after elimination the product is zero, since an eliminated player never re-enters. In particular the stopped process has incrementThe indicator function is ${\mathcal{F}}_n$-measurable, and the expression in parentheses has zero conditional expectation on its support. The stopped process is adapted and integrable: at fixed $n$, its absolute value is bounded by $N^3+nN/3$. HenceThis is the product martingale for three-player token exchange.

A discrete-time stopping time for a filtration $({\mathcal{F}}_n)$ is a random variable $\tau$ taking values in $\{0,1,\dots \}\cup \{\infty \}$ such that $\{\tau \le n\}\in {\mathcal{F}}_n$ for every $n$. It means that whether stopping has occurred by time $n$ can be determined from the information available then.

For the first elimination time,Each event in this finite union belongs to ${\mathcal{F}}_j\subseteq {\mathcal{F}}_n$, by the definition of the natural filtration. Therefore $\tau$ is a stopping time. This verification does not require first knowing whether $\tau$ is finite almost surely.

Choose $Z_{\infty }$ as the pointwise limit of a sequence of $(Z_n)$ where it exists, and as zero otherwise. The convergence event and every $Z_n$ are $\sigma (X)$-measurable, so this is an exactly $\sigma (X)$-measurable version, bounded by $L$. Part (a)(ii), the Doob-Dynkin lemma, supplies a Borel measurable function $g$ on $[0,1]$ such that $Z_{\infty }=g(X)$. Applying scalar clipping to $g$ with bounds $[-L,L]$ preserves that equality and ensures $|g(u)|\le L$ everywhere.

Part (c)(ii) and the conditional expectation formula from part (b)(iv) now giveSince the indices $n$ form a countable family, these identities may be taken on one common full-probability event.

For fixed $n,k$, both the slope and the displayed average are constant on the atom $A_{n,k}$, whose probability is $2^{-n}>0$. An equality holding almost surely of these constants forces the deterministic equalitySum over $0\le k<j$. The telescoping sum givesThus the desired identity holds at every dyadic endpoint, including $1$, without evaluating the undefined slope at $X=1$.

Finally, given $x\in [0,1]$, take dyadic endpoints $x_n\to x$. The Lipschitz condition implies $|f(x_n)-f(x)|\le L|x_n-x|$, and the bound on the Borel measurable function $g$ givesPassing to the limit of a sequence provesThis is a martingale proof of the integral representation of a Lipschitz function. It constructs a bounded Lebesgue integrable function representing the increments of $f$, and in particular proves absolute continuity of $f$. No prior assumption that $f$ has a derivative is needed.

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.

Each $Z_n$ is a bounded Borel measurable function of $X_n$, including the specified value at $X_n=1$. Therefore $(Z_n)$ is an adapted process for $({\mathcal{F}}_n)$ and consists of integrable random variables.

Fix an atom of a sigma-algebra in ${\mathcal{F}}_n$ with positive probability, corresponding to the dyadic interval $[u,u+\delta )$, with $\delta =2^{-n}$. By the uniform distribution, each half has conditional probability $1/2$. The conditional expectation of $Z_{n+1}$ on that atom is thereforeThe middle values cancel. On the zero-probability atom the conditional expectation can be chosen as zero, agreeing with our version of $Z_n$. Henceso $(Z_n)$ is a martingale. This is the dyadic slope martingale associated with the Lipschitz function $f$.

For $1\le j\le n$, repeated use of the floor function identity givesThus the filtration satisfies ${\mathcal{F}}_n=\sigma (X_n)$. Each positive-probability atom of a sigma-algebra here has the formwith probability $2^{-n}$; $\{X=1\}$ is a possible additional null set atom. Averaging over each dyadic interval gives a versionFor a globally defined version of the right side, assign any value, for example zero, on $\{X=1\}$.

Indeed, the displayed random variable is ${\mathcal{F}}_n$-measurable and bounded by $\sup |h|$. On each $A_{n,k}$ its integral isby the uniform distribution. Every member of ${\mathcal{F}}_n$ is a union of the displayed sets and the possible zero-probability atom. Summing these identities therefore verifies the defining conditional expectation property on all of ${\mathcal{F}}_n$.

A family $({\xi }_i{)}_{i\in I}$ of integrable random variables is uniformly integrable whenThe Lipschitz condition gives, on $\{X<1\}$,Our chosen zero value on $\{X=1\}$ satisfies the same bound. Hence the random variables $(Z_n)$ are integrable, and for $K\ge L$ every tail expectation in the definition vanishes. Therefore the family is uniformly integrable.

Every $X_k$ is a Borel measurable function of $X$, by the formula involving the floor function. ThusConversely, part (i) gives the everywhere identity $X={\lim }_{k\to \infty }{X}_k$. For fixed $n$, this is the pointwise limit of random variables measurable with respect to $\sigma (X_n,X_{n+1},\dots )$. Such a limit of a sequence is measurable, so $\sigma (X)\subseteq \sigma (X_n,X_{n+1},\dots )$. ConsequentlyThese are increasing-precision observations even when the first finitely many are discarded; they do not behave like the independent random variables whose tail sigma-algebra appears in Question 1(b).

The floor function gives the deterministic boundfor every outcome, including $X=1$. Thus all three convergence assertions hold: there is pointwise, and hence almost sure convergence, as well as convergence in probability and convergence in L2. In detail, for $\epsilon >0$ the probability $\mathbb{P}(|X-X_n|>\epsilon )$ is zero as soon as $2^{-n}\le \epsilon$; alsoThe uniform distribution even gives the exact second moment of the error by integrating over the dyadic partition:

We prove the Doob-Dynkin lemma here by simple function approximation. Choose $C<\infty$ such that $|Z|\le C$ everywhere, and letEach $Z_m$ is a finite-valued $\sigma (X)$-measurable function and $Z_m\to Z$ pointwise. Write its level sets as $A_{m,j}$, with associated values $z_{m,j}$. By part (i), choose Borel sets $D_{m,j}$ with $A_{m,j}=X^{-1}(D_{m,j})$. Define the Borel measurable functionsAlthough the chosen $D_{m,j}$ need not partition all of $\mathbb{R}$, their inverse images do partition $\Omega$, so $g_m(X)=Z_m$ exactly.

Apply scalar clipping to each $g_m$ with bounds $[-C-1,C+1]$, preserving this equality, and define $g(x)={\mathrm{limsup}}_m{g}_m(x)$. A bounded limit superior of Borel measurable functions is a finite Borel measurable function. At every $\omega \in \Omega$ the limit of a sequence exists and givesThus the factorization is exact for a bounded, exactly $\sigma (X)$-measurable function. If hypotheses are instead given only up to null sets, the corresponding conclusion is a factorization holding almost surely after choosing a representative.

The sigma-algebra generated by a random variable $X$ is the smallest sigma-algebra on $\Omega$ making $X$ a measurable function into $(\mathbb{R},\mathcal{B}(\mathbb{R}))$. Equivalently, it is the intersection of all sigma-algebras containing the inverse images $X^{-1}(C)$ of Borel sets $C$.

Let $\mathcal{D}=\{X^{-1}(C):C\in \mathcal{B}(\mathbb{R})\}$. This collection is already a sigma-algebra: $X^{-1}(\mathbb{R})=\Omega$,The right-hand sets are in $\mathcal{D}$, since the Borel sigma-algebra is closed under the indicated operations. Any sigma-algebra making $X$ measurable must contain every member of $\mathcal{D}$. Conversely $\mathcal{D}$ makes $X$ measurable by definition. HenceThis equality holds exactly as a collection of sets, without completing the sigma-algebra.

Suppose the events $(B_n)$ are mutually independent events. Then $B_k$ is independent of ${\mathcal{F}}_{k-1}$, first by factorization against finite intersections of previous events and then by part (a). ThereforeIf ${\sum }_k\mathbb{P}(B_k)=\infty$, the sum in part (ii) is infinite almost surely, and consequentlyThis is the Second Borel-Cantelli lemma. In fact part (ii) also yields probability zero in the convergent case for these independent events.