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Write $s=\sqrt{1-{\rho }^2}>0$. The inverse change of variables is $y=\rho x+sz$, whose Jacobian determinant has absolute value $s$. Completing the square givesThe change of variables formula therefore transforms the joint probability density function toIt factors into two normalized standard normal distribution densities. Thus $X$ and $Z$ are independent $N(0,1)$ variables; factorization proves the requested independence, rather than simply zero covariance.

The standard Gaussian random vector $(X,Z)$ has a rotation-invariant density. In polar coordinates, $X=r\cos \theta$, $Z=r\sin \theta$, its density with the area factor is $(2\pi {)}^{-1}r{e}^{-r^2/2}$, so the angular coordinate is uniform on an interval of length $2\pi$.

Put $\alpha =\arcsin \rho \in (-\pi /2,\pi /2)$, so $s=\cos \alpha$ andIn the right half-plane, choose $-\pi /2<\theta <\pi /2$. The additional condition $Y>0$ then requires $-\alpha <\theta <\pi /2$. This wedge has angle $\pi /2+\alpha$; its boundary has probability zero. ConsequentlyAt $\rho =0$ this is $1/4$, as required by independence. The limits as $\rho \to 1$ and $\rho \to -1$ are $1/2$ and zero, respectively. The angular calculation is also the geometric basis of the Gaussian sign-correlation identity.

Use the positive-support convention for the geometric distribution and put $q=1-p$. Differentiating the convergent geometric series ${\sum }_{k\ge 0}{q}^k=(1-q{)}^{-1}$ givesConsequentlySubtracting the square of the expectation from the second moment yields

For the die, this is the coupon collector problem. When $r$ scores are still missing, let $W_r$ count rolls up to and including the next newly observed score. Every fresh roll has success probability $r/6$, so $W_r$ has a geometric distribution of parameter $r/6$. These six successive waiting times are independent: conditional on the entire past at the start of a stage, its waiting-time distribution depends only on the number $r$, not on the identities of the missing scores. Iterated conditioning therefore factors their joint distribution. The total number of rolls is $N={\sum }_{r=1}^6{W}_r$.

Using linearity of expectation givesUsing variance additivity for independent random variables, or the coupon collector waiting-time variance formula just derived by this decomposition, givesHence the requested standard deviation isThe hint rounds the sum of squares from $53.69$ to $53.7$; using that rounded value gives $\sqrt{39}\approx 6.245$, consistent at the precision of the hint. The exact computation preserves the slightly more precise value above.

The sampling without replacement makes all $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ unordered pairs equally likely. The count of red-red pairs is $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, while the count of mixed pairs is $3(n-3)$. Thus their probabilities satisfyand cancellation gives $n-3=3$. Hence $\boxed{n=6}$, with three black socks. The count of black-black pairs is also $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$, soThe mixed, red-red and black-black probabilities are respectively $3/5,1/5,1/5$, which add to one and verify the given ratio.

Let $X$ be one score. It has the discrete uniform distribution on $1,\dots ,6$, so its expected value and second moment areThereforeFor independent throws, linearity of expectation and additivity of variance give $\mathbb{E}{Y}_n=7n/2$ and $\mathrm{Var}(Y_n)=35n/12$. The sample average consequently has expectation $7/2$ and variance $35/(12n)$.

The Chebyshev inequality follows by applying the Markov inequality to the nonnegative square of a centered random variable. It givesThis is at most $1/10$ if $n\ge 350/27$. Thus $n=13$ suffices, as does every larger integer. This is the smallest integer guaranteed by this particular Chebyshev inequality bound; it need not be the smallest integer satisfying the actual probability requirement.

Choose a uniformly random permutation of $[n]$ and take its initial segments. This produces a Uniformly random maximal chain in a Boolean lattice. A fixed $r$-set belongs to this chain with probability $r!(n-r)!/n!={\left(\genfrac{}{}{0ex}{}{n}{r}\right)}^{-1}$. An antichain meets the chain at most once, so taking the expected value of the number of meetings proves the LYM inequality:In particular the Sperner theorem bound follows by multiplying each summand by at most the largest binomial coefficient.

For the vector request, an ordering by ordinary subset need not work: distinct vectors may cancel. Instead, write $s_A={\sum }_{i\in A}{x}_i$ and construct a separated block decomposition for vector subset sums. A block is separated if distinct labeled members have subset sums at distance at least one. We prove that the labeled subsets of $[n]$ can be partitioned into separated blocks, withblocks of size $n-2r+1$, for $0\le r\le \lfloor n/2\rfloor$, with $\left(\genfrac{}{}{0ex}{}{n}{-1}\right)=0$.

For $n=0$ the one subset forms one block. Suppose a block of sums $z_1,\dots ,z_{\ell }$ has already been constructed for $n-1$ vectors, and let $x=x_n$. Choose $z_{\ast }$ minimizing $z_j\cdot x$ in the block. Replace the two copies of the block, with and without the new coordinate, by the following two blocks:The old labels are retained, so these partition the two copies even if equal sums occur in different blocks. Translation preserves separation in the first translated portion, and the second block is a subset of the old separated block. For the new cross pairs, Cauchy-Schwarz inequality givesThus both new blocks are separated. Their sizes are $\ell +1$ and $\ell -1$, omitting an empty block. The resulting size-profile recurrence is $b_{n,r}=b_{n-1,r}+b_{n-1,r-1}$, and Pascal's identity verifies the displayed binomial-difference formula, including the even middle boundary. The total number of blocks telescopes to $\left(\genfrac{}{}{0ex}{}{n}{\lfloor n/2\rfloor }\right)$.

A signed sum associated with $A=\{i:{ϵ}_i=1\}$ is $2s_A-{\sum }_i{x}_i$. If two signed sums lie in the open ball of radius one, their difference has norm less than two, so their subset sums have distance less than one. At most one can therefore lie in any separated block. Hence the Littlewood-Offord inequality isThe counting is by sign choices, not by distinct numerical values of the sums. The open-ball condition handles the equality-distance boundary correctly.

For the last request take real coefficients. Absorb their signs into the choices ${ϵ}_i$, so all coefficients are positive and at least one. A signed-sum real interval of length four corresponds to a subset-sum real interval of length two. A strict three-member inclusion chain $A⊊B⊊C$ would have $s_C-s_A\ge 2$, which cannot occur in that open real interval. Thus its set family is a k-Sperner family with $k=2$. Every maximal chain in a Boolean lattice meets it at most twice, yielding ${\sum }_r|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})\le 2$. Write $f_r=|{\mathcal{F}}_r|/(\genfrac{}{}{0ex}{}{n}{r})$, so $0\le f_r\le 1$ and ${\sum }_r{f}_r\le 2$. If $B_1\ge B_2$ are the two largest binomial coefficients and $f_{\ast }$ is the fraction at a largest level, then $|\mathcal{F}|\le B_1{f}_{\ast }+B_2(2-f_{\ast })\le B_1+B_2$. This proves the two-level Littlewood-Offord bound:It is the greatest possible number uniformly over choices of coefficients and center. For even $n=2m$, take every coefficient equal to one and center $a=-1$: the real interval $(-3,1)$ contains the sum levels $-2,0$, of multiplicities $\left(\genfrac{}{}{0ex}{}{n}{m-1}\right),\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. For odd $n=2m+1$, take center zero, giving levels $-1,1$, each of multiplicity $\left(\genfrac{}{}{0ex}{}{n}{m}\right)$. These attain the bound. For $n=0$ the only sign choice gives the separate maximum one. Attainment here is at the exhibited centers; it need not hold at every prescribed center.

Index the functions as $g_0,g_1,g_2,g_3$ and write $\Lambda ={\mathbb{E}}_{x,d}{\prod }_{j=0}^3{g}_j(x+jd)$. We first control $g_0$. Change variables to $y=x+3d$, average over $d$, and apply Cauchy-Schwarz in $y$, discarding the bounded factor $g_3(y)$. Expand the squared inner average with the second copy of $d$ written $d+h$, then put $x=y-3d$. This yieldsThe right side is real and nonnegative because it arose as an average of squared absolute values. Next use $y=x+2d$ and apply Cauchy-Schwarz in $(h,y)$, discarding the bounded $g_2$ derivative. Expanding with a displacement $k$ in $d$ givesFinally use $y=x+d$ and Cauchy-Schwarz in $(h,k,y)$, discarding the bounded $g_1$ double derivative. Expansion with displacement $r$ givesAll these changes of variables are bijections without any division in $G$.

Put $t_j=|G[j]|$, where $G[j]=\{z:jz=0\}$ is the torsion subgroup killed by an integer $j$. The variables $2k$ and $3h$ are uniform on $2G$ and $3G$, whose indices are $t_2,t_3$. Since the last squared integrand is nonnegative, its restricted average is at most $t_2{t}_3$ times the unrestricted average over the two increments. By part (i), that unrestricted average is $\Vert g_0{\Vert }_{U^3}^8$. Hence $|\Lambda |\le (t_2{t}_3{)}^{1/8}\Vert g_0{\Vert }_{U^3}$.

To control $g_1$, apply the same three eliminations in the order of slopes $3,0,2$. The surviving $g_1$ has increments $-2h,k,-r$. The last increment is uniform, and only the first is restricted to $2G$, so the resulting bound is $|\Lambda |\le t_2^{1/8}\Vert g_1{\Vert }_{U^3}$. Reversing the arithmetic progression by $(x,d)\mapsto (x+3d,-d)$ treats $g_3$ and $g_2$ respectively. This proves the four-term progression bound with torsion factors:For a fixed finite group, this positive constant tends to zero with $\alpha$. In ${\mathbb{Z}}_N$, $t_j=\gcd (j,N)$, so $\boxed{\beta =6^{1/8}\alpha \text{ suffices uniformly in }N}$. If multiplication by two and three are bijective, the sharper $\beta =\alpha$ holds.

A dependence on torsion is necessary if a bound uniform over all finite abelian groups is intended. In $G={\mathbb{F}}_2^m$, choose independent random signs $f(x)$. A cube with eight distinct vertices has sign product of expectation zero. Each of the twenty-eight possible collisions has probability $1/|G|$, so the expectation of $\Vert f{\Vert }_{U^3}^8$ is at most $28/|G|$. Some sign function therefore has norm at most $(28/|G|{)}^{1/8}$. But with factors $(f,1,f,1)$, every arithmetic progression is $(x,x+d,x,x+d)$ and its product is $f(x{)}^2=1$. Thus unbounded torsion obstructs uniform control of four-term progressions. The stated fixed-group bound, and the uniform cyclic-group bound needed in part (iii), are unaffected.

The expectations use normalized counting measure on a finite abelian group $G$. This is the finite-group convention implicit in the averaging notation; an arbitrary infinite group has no normalized counting probability measure on all its points. For $h\in G$, define ${\Delta }_{h}f(x)=f(x)\overline{f(x+h)}$, and let $\mathcal{C}$ denote complex conjugation. The Gowers U3 norm isAveraging first over $h_3$ gives the Derivative identity for the Gowers U3 normIndeed, if $u(x)={\Delta }_{h_1}{\Delta }_{h_2}f(x)$, the average over $(x,h_3)$ is ${\mathbb{E}}_{x,h_3}u(x)\overline{u(x+h_3)}=|{\mathbb{E}}_{x}u(x){|}^2$. The eighth root is consequently well-defined. Homogeneity follows because the cube has four unconjugated and four conjugated factors, so $\Vert af{\Vert }_{U^3}=|a|\Vert f{\Vert }_{U^3}$. If the norm is zero, every squared term in the finite average vanishes. Taking $h_1=h_2=0$ gives ${\mathbb{E}}_x|f(x){|}^4=0$, so $f=0$ pointwise. This proves positive definiteness.

For the triangle inequality, we prove the needed Gowers-Cauchy-Schwarz inequality. For eight functions $F_{\omega }$ on a product of finite sets $X_1\times X_2\times X_3$, form the box norm multilinear averageFix a coordinate $j$. After averaging its two variables separately, this is $\mathbb{E}A\overline{B}$, where $A$ uses all vertices with ${\omega }_j=0$ and $B$ those with ${\omega }_j=1$, with the remaining parity conjugations. The Cauchy-Schwarz inequality giveswhere $S_j^b$ replaces the function at each vertex by the original function with its $j$th index set to $b$. Each right-hand average is a squared absolute value averaged over the other coordinates, hence nonnegative. Apply this operation successively for $j=1,2,3$, to every resulting factor. The eight final families each use one original $F_{\omega }$ at all vertices. Taking roots gives the box Cauchy-Schwarz inequalityNow take $X_j=G$ and $F_{\omega }(a,b,c)=f_{\omega }(a+b+c)$. With $x=a_0+b_0+c_0$ and $h_j=x_j^1-x_j^0$, these independent variables induce uniform $(x,h_1,h_2,h_3)$: each such tuple has exactly $|G{|}^2$ preimages. Therefore each box norm is the corresponding $U^3$ norm, and the mixed cube obeysExpand the eight factors in the cube expression for $f+g$. Its $2^8$ mixed terms are bounded by this inequality, and their bounds sum to $(\Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}{)}^8$. Thus $\boxed{\Vert f+g{\Vert }_{U^3}\le \Vert f{\Vert }_{U^3}+\Vert g{\Vert }_{U^3}}$, completing the norm proof.

Two consequences will be useful below. Multiplication by a character of a finite abelian group leaves the norm unchanged, because the alternating character product around a three-dimensional cube equals one. Alsowhere the last inequality uses the $h=0$ term and $\Vert u{\Vert }_{U^2}^4={\sum }_{\chi }|\hat{u}(\chi ){|}^4\ge |\mathbb{E}u{|}^4$. Hence $\Vert f{\Vert }_{U^3}\ge |G{|}^{-1/8}\Vert f{\Vert }_2$.

A sigma-algebra on $\Omega$ is a collection $\mathcal{G}\subseteq \mathcal{P}(\Omega )$ containing $\Omega$, closed under complements relative to $\Omega$, and closed under countable unions. These properties also give closure under countable intersections by De Morgan laws.

Let $\mathcal{G}={\bigcap }_{r\in R}{\mathcal{F}}_r$. Every constituent sigma-algebra contains $\Omega$, so $\Omega \in \mathcal{G}$. If $A\in \mathcal{G}$, its complement belongs to every ${\mathcal{F}}_r$, hence to $\mathcal{G}$. Likewise, if $A_j\in \mathcal{G}$ for all $j$, then ${\bigcup }_j{A}_j$ belongs to every ${\mathcal{F}}_r$. Thus $\mathcal{G}$ is a sigma-algebra. For an empty indexing set, the usual intersection convention gives $\mathcal{G}=\mathcal{P}(\Omega )$, which is also a sigma-algebra.

A union need not work. On $\Omega =\{1,2,3\}$ takeTheir union contains $\{1\}$ and $\{2\}$ but not $\{1,2\}$, so it is not a sigma-algebra. Put $\mathcal{A}={\bigcup }_r{\mathcal{F}}_r$ and let $\mathfrak{C}$ be the collection of all sigma-algebras on $\Omega$ containing $\mathcal{A}$. This collection is nonempty because it contains the power set $\mathcal{P}(\Omega )$. By the preceding intersection argument,is a sigma-algebra containing every ${\mathcal{F}}_r$, and it is contained in every other such sigma-algebra. This proves existence and minimality.

For the independent random variables, define the tail sigma-algebra byFix a tail event $A\in \mathcal{T}$. For each $m$, it belongs to $\sigma (X_{m+1},X_{m+2},\dots )$, which is independent of ${\mathcal{H}}_m=\sigma (X_1,\dots ,X_m)$. To justify independence of the generated sigma-algebras, first use independence of finite collections on cylinder events, then extend twice by the pi-lambda theorem. In particular, $\mathbb{P}(A\cap C)=\mathbb{P}(A)\mathbb{P}(C)$ for every $C\in {\mathcal{H}}_m$ and every $m$.

The increasing union ${\bigcup }_m{\mathcal{H}}_m$ is an algebra of sets generating $\mathcal{H}=\sigma (X_1,X_2,\dots )$. The collection of $C\in \mathcal{H}$ satisfying this factorization is a Dynkin system: it contains $\Omega$, is closed under complements, and under countable disjoint unions. The pi-lambda theorem therefore extends the factorization to all $C\in \mathcal{H}$. Since $A\in \mathcal{H}$, taking $C=A$ gives $\mathbb{P}(A)=\mathbb{P}(A{)}^2$. Consequently,This is the Kolmogorov zero-one law, proved here directly.

For each fixed $m$, write $S_n=S_{m-1}+{\sum }_{j=m}^n{X}_j$ for $n\ge m$, with $S_0=0$. The random variable $S_{m-1}$ is finite, so $S_{m-1}/\sqrt{n}\to 0$. Adding a real sequence tending to zero changes neither the extended-real limit inferior nor the extended-real limit superior. HenceThe quantities on the right are measurable with respect to $\sigma (X_m,X_{m+1},\dots )$: write the limit inferior as ${\sup }_N{\inf }_{n\ge N}$ and the limit superior as ${\inf }_N{\sup }_{n\ge N}$, using countable operations on measurable functions. Therefore both threshold events belong to every such sigma-algebra, and hence are tail events, for every real threshold, including when either limit is infinite.

The final recurrence claim is false under the printed hypotheses. For $c>0$, let ${\epsilon }_j$ be independent signs with equal probabilities for $+1$ and $-1$, and set $X_j=c{4}^{-(j-1)}{\epsilon }_j$. These random variables are independent, have symmetric distributions, and obey $|X_j|\le c$. Nevertheless, the geometric series gives, on every sample path,Thus the proposed repeated-return event has probability zero, even though none of the increments is degenerate. The Kolmogorov zero-one law does not determine which of zero and one occurs.

An additional identical-distribution hypothesis makes the intended conclusion true. Here is a full proof of that qualified version, using the oscillation of a bounded centered iid random walk. Symmetry and integrability give $\mathbb{E}{X}_j=0$, so $S_n$ is a martingale for its natural filtration. If the common probability distribution is concentrated at zero, every $S_n$ is zero and the conclusion is immediate. Otherwise choose $\delta >0$ with $\mathbb{P}(|X_1|>\delta )>0$. The Borel-Cantelli lemmas then give $|X_n|>\delta$ infinitely often almost surely; thus $S_n$ cannot converge to a finite limit.

For $a>0$, let $T_a=\inf \{n:S_n\ge a\}$. Before $T_a$ the sum is below $a$, and at $T_a$ it is at most $a+c$. The stopped martingale $a+c-S_{n\wedge T_a}$ is therefore nonnegative. A stopped integrable martingale remains a martingale at these bounded times, and a nonnegative martingale converges finitely almost surely by the martingale convergence theorem. On $\{T_a=\infty \}$ this would force $S_n$ to converge finitely, which has just been ruled out. Hence $T_a<\infty$ almost surely. Apply the same argument to $-S_n$ and take a countable intersection over positive integer $a$. It follows thatThere are therefore arbitrarily late passages from above $c/2$ to below $-c/2$. A path with bounded increments of magnitude at most $c$ cannot make such a passage without visiting $[-c/2,c/2]$: jumping directly between its two open complementary half-lines would require an increment larger than $c$. There are infinitely many such visits. Thus with the added identical-distribution hypothesis, the intended answer is

For events $A_n$, define their limit superior byIt is the event that infinitely many $A_n$ occur. The two Borel-Cantelli lemmas are as follows: if ${\sum }_n\mathbb{P}(A_n)<\infty$, then $\mathbb{P}(A_{\infty })=0$, without any independence assumption; if the $A_n$ are independent and ${\sum }_n\mathbb{P}(A_n)=\infty$, then $\mathbb{P}(A_{\infty })=1$.

For the first assertion, the union bound gives, for every $N$,The tail of a convergent nonnegative series tends to zero, proving the first Borel-Cantelli lemma.

For the second assertion, independence and $1-u\le e^{-u}$ for $0\le u\le 1$ giveBy continuity of probability, $\mathbb{P}({\bigcap }_{n\ge N}{A}_n^c)=0$. The event that only finitely many $A_n$ occur is the countable union of these zero-probability events over $N$. Its probability is zero. This proves the second Borel-Cantelli lemma, and hence both requested assertions:

For $a>0$ and $n\ge 2$, let $A_n(a)=\{X_n^2>a\log n\}$. Symmetry of the standard normal distribution and the Mills ratio asymptotic just proved giveFor $a>2$ the series of these probabilities converges, since its terms are eventually bounded by a constant times the convergent p-series $n^{-a/2}$. For $0<a<2$ it diverges: choose $r$ strictly between $a/2$ and one, and use $(\log n{)}^{-1/2}\ge n^{-(r-a/2)}$ for sufficiently large $n$, reducing to the divergent p-series $n^{-r}$. At $a=2$ it still diverges, as the integral test givesFor $a>2$, the first Borel-Cantelli lemma makes $A_n(a)$ occur only finitely often. For $0<a\le 2$, independence of the random variables makes the events $A_n(a)$ independent, so the second Borel-Cantelli lemma makes them occur infinitely often almost surely. Apply these conclusions simultaneously at $a=2+1/k$ for positive integers $k$, and at $a=2$. On the resulting probability-one event, the limit superior is at most every $2+1/k$ and at least two. Therefore the Gaussian sample limsup at logarithmic scale is

The real L2 space on a probability space consists of equivalence classes of measurable functions $X:\Omega \to \mathbb{R}$ satisfying $\mathbb{E}[X^2]<\infty$, where two random variables represent the same element if they are equal almost surely. Its norm isPassing to equivalence classes makes this a genuine norm: $\Vert X{\Vert }_2=0$ exactly when $X=0$ almost surely. The Minkowski inequality gives the triangle inequality.

Let $(X_n)$ be a Cauchy sequence in this norm. Choose an increasing sequence of indices $n_k$ such thatThis is possible by choosing $n_k$ beyond a Cauchy threshold for tolerance $2^{-k}$. Set $D_k=X_{n_{k+1}}-X_{n_k}$. The Cauchy-Schwarz inequality on a probability space gives $\mathbb{E}|D_k|\le \Vert D_k{\Vert }_2$. By Tonelli theorem,Thus the series of absolute differences is finite almost surely. Define $X=X_{n_1}+{\sum }_{k\ge 1}{D}_k$ on this probability-one event, and give it the value zero on its complement. It is a measurable function, and $X_{n_k}\to X$ almost surely.

For $l>k$, the Minkowski inequality givesApply the Fatou lemma to the squared difference as $l\to \infty$. It follows thatIn particular $X-X_{n_1}\in L^2$, and therefore $X\in L^2$ by the Minkowski inequality. This proves convergence in L2 of the subsequence. To recover the entire sequence, for a prescribed $\epsilon >0$ choose a Cauchy threshold $N$ for $\epsilon /2$, then choose $k$ with $n_k\ge N$ and $\Vert X-X_{n_k}{\Vert }_2<\epsilon /2$. For every $n\ge N$,Hence $\boxed{X_n\to X\text{ in }{L}^2}$, proving completeness of Lp spaces for $p=2$ without assuming the desired completeness in the argument.

The martingale property gives $\mathbb{E}[X_n\mid {\mathcal{F}}_m]=X_m$. Both $X_m$ and $X_n$ are square-integrable, so their product is integrable by the Cauchy-Schwarz inequality. Using the ${\mathcal{F}}_m$-measurability of $X_m$ in conditional expectation givesThe pull-out step for the possibly unbounded $X_m$ can be justified by first truncating it, then passing to the limit in the L2 norm and using the Cauchy-Schwarz inequality. Expanding the square now yieldsThe left side is nonnegative. Thus $a_n=\mathbb{E}[X_n^2]$ is nondecreasing and bounded above by $M$, and so $a_n\to a\le M$. For $m\le n$,By symmetry of the norm difference this proves that $(X_n)$ is a Cauchy sequence in the L2 space. The completeness established in part (a) gives an $X_{\infty }\in L^2$ withThis is the L2-bounded martingale convergence theorem, derived from martingale-difference orthogonality and completeness. The same sequence also converges almost surely by part (b), since $\mathbb{E}|X_n|\le \sqrt{M}$; uniqueness of a limit in probability identifies that limit with this $X_{\infty }$.

The hitting time $T$ is a stopping time for the natural filtration, because $\{T\le n\}={\bigcup }_{k\le n}\{S_k=b\}$. The integer-valued increments are at most one. A path started at zero therefore cannot reach or exceed $b$ without first visiting $b$, soDefine $Y_n=b-S_{n\wedge T}$. It is nonnegative and integrable: $|S_{n\wedge T}|\le {\sum }_{j\le n}|X_j|$. Moreover,The indicator function is ${\mathcal{F}}_n$-measurable, and $X_{n+1}$ is independent of ${\mathcal{F}}_n$ with zero expected value. Taking conditional expectation therefore proves directly that $(Y_n)$ is a nonnegative martingale.

The general result used here is that a nonnegative discrete-time martingale has a finite almost-sure limit; this is the martingale convergence theorem, since $\mathbb{E}{Y}_n=\mathbb{E}{Y}_0=b$ bounds its L1 norm. On $\{T=\infty \}$ it follows that $S_n=b-Y_n$ converges finitely. A convergent sequence of integers is eventually constant, so $X_n=S_n-S_{n-1}$ would eventually be zero on this event.

However, the independent events $\{X_n=1\}$ all have the same strictly positive probability. The second Borel-Cantelli lemma says that they occur infinitely often almost surely. This contradicts eventual constancy on any positive-probability subset of $\{T=\infty \}$. ThusNotice that no lower bound on the negative increments was needed.

Values at the endpoint one do not affect any of the conclusions, since $\mathbb{P}(U=1)=0$. Set $b_n(1)=1$ to define each random variable everywhere, and choose a Borel measurable function agreeing with $f$ almost everywhere if necessary. These choices leave all integrals and almost-sure assertions unchanged.

Every dyadic interval at level $n+1$ lies in a dyadic interval at level $n$. Equivalently, $b_n(U)$ is a measurable function of $b_{n+1}(U)$, so ${\mathcal{F}}_n\subseteq {\mathcal{F}}_{n+1}$: these sigma-algebras form a filtration. Since the uniform distribution of $U$ gives $\mathbb{E}|f(U)|={\int }_0^1|f(u)|du<\infty$, the conditional expectations $X_n=\mathbb{E}[f(U)\mid {\mathcal{F}}_n]$ are integrable and adapted. The tower property of conditional expectation givesThus $(X_n)$ is a martingale, and part (a) proves that it is uniformly integrable.

For $I_{n,k}=[k{2}^{-n},(k+1)2^{-n})$, the event $\{U\in I_{n,k}\}$ is an atom of a sigma-algebra ${\mathcal{F}}_n$, with probability $2^{-n}$. The value of the conditional expectation there is thereforeConsequently $\boxed{X_n=f_n(U)\text{ almost surely}.}$ This also shows directly that ${\int }_0^1|f_n|\le {\int }_0^1|f|$.

By the uniformly integrable martingale convergence theorem, there is an integrable $Y$ such that $X_n\to Y$ almost surely and in the L1 norm. To identify this limit, put ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $0\le U-b_n(U)<2^{-n}$ away from the harmless endpoint, $U={\lim }_n{b}_n(U)$, and $U$ is ${\mathcal{F}}_{\infty }$-measurable. Conversely every $b_n(U)$ is a measurable function of $U$. Thus ${\mathcal{F}}_{\infty }=\sigma (U)$, and both $Y$ and $f(U)$ have ${\mathcal{F}}_{\infty }$-measurable versions.

For $A\in {\mathcal{F}}_m$, the defining property of conditional expectation gives $\mathbb{E}[X_n{1}_A]=\mathbb{E}[f(U)1_A]$ whenever $n\ge m$. Passing to the limit in the L1 norm yieldsThe increasing union ${\bigcup }_m{\mathcal{F}}_m$ is an algebra of sets generating ${\mathcal{F}}_{\infty }$. The class of events for which this integral identity holds is a Dynkin system, so the pi-lambda theorem extends the identity to every $A\in {\mathcal{F}}_{\infty }$. Since both random variables are measurable there, uniqueness of conditional expectation gives $Y=f(U)$ almost surely. We have proved that the dyadic conditional averages recover integrable functions:because a probability-one assertion for $U$ is a full-Lebesgue measure assertion for $x$. Likewise, the uniform distribution of $U$ turns convergence in L1 intoNo pointwise continuity of $f$ is required.