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# Solutions to August 2015 problems

1. Let $X$ and $Y$ be random variables such that $X,Y,X+Y,X-Y$ all have the
same distribution. If the common distribution has finite mean show that $X=0$
a.s. Prove that the assumption on finiteness of the mean cannot be dropped.

Since $2E\left\vert X\right\vert =E\left\vert (X+Y)+(X-Y)\right\vert=E\left\vert (X+Y)\right\vert +E\left\vert (X-Y)\right\vert$ it follows that $X+Y=Z(X-Y)$ for some non-negative random variable $Z$. Hence $X(Z-1)=Y(1+Z)$. Noting that $\left\vert Z-1\right\vert \leq1+Z$ we get $\left\vert Y\right\vert (1+Z)\leq \left\vert X\right\vert (1+Z)$ which implies $\left\vert Y\right\vert \leq \left\vert X\right\vert$. Since both sides have the same mean we get $\left\vert Y\right\vert = \left\vert X\right\vert$

This implies $\left\vert X\right\vert(1+Z) = \left\vert X\right\vert\left\vert 1-Z\right\vert$ so $1+Z=\left\vert 1-Z\right\vert$ when $X\neq 0$. In other words $X\neq 0$ implies $Z=0$ and $X(0-1)=Y(1+0)$ or $Y=-X$. But $X=0$ implies $Y(1+Z)=0$ so $Y=0$. Hence $Y=-X$ in both cases and $X+Y=0$ a.s. It follows that $X=0$ a.s.

For the counterexample let $U,V$ be i.i.d. with characteristic function $e^{-\left\vert t\right\vert}$. Let $X=\frac{U+V}{2}$ and $Y=\frac{U-V}{2}$.

2. Prove that a function $f$ from one metric space to another is uniformly continuous if and only if $d(A,B)=0$ implies $d(f(A),f(B))=0.$ [$d(A,B)$ is defined as $\inf \{d(x,y) :x\in A,y\in B\}$].

[ Solution by S Kannappan]

If $f$ is not uniformly continuous then there exists $\epsilon >0$ such that for every $\delta >0$ we can find points $x$ and $y$ with $d(x,y)<\delta$ but $d(f(x),f(y))\geq 4\epsilon$ Let $\delta _{1}=1$. Let $d(x_{1},y_{1})<\delta$ but $d(f(x_{1}),f(y_{1}))\geq 4\epsilon$ Inductively define $\delta_{n},x_{n},y_{n}(n\geq 1)$ satisfying the conditions $\delta_{n}\downarrow 0,d(x_{n},y_{n})<\delta_{n},d(f(x_{n}),f(y_{n}))\geq 4\epsilon$ as follows: having found $\delta_{j},x_{j},y_{j}(j\leq n)$ let $0<\delta_{n+1}<\min \{\frac{\delta_{n}}{2},d(N(x_{n}),F(x_{n})),d(N(y_{n}),F(y_{n}))\}$ where $N(z)=\{x:d(f(x),f(z))<\epsilon \}$ and $F(z)=\{x:d(f(x),f(z))\geq 2\epsilon \}$ for any $z$ Note that $z\in N(z).$ If $F(z)=\emptyset$ then $d(f(x),f(z))<2\epsilon$ for all $x$ and $4\epsilon \leq d(f(x_{1}),f(y_{1}))\leq d(f(x_{1}),f(z))+d(f(y_{1}),f(z))$ if $u\in F(z)$ and $v\in N(z)$ for some $z$ then $d(f(u),f(v))\geq d(f(u),f(z))-d(f(v),f(z))>2\epsilon-\epsilon=\epsilon$. Hence $\delta_{n+1}$ is well defined. We can find $x_{n+1},y_{n+1}$ such that $d(x_{n+1},y_{n+1})<\delta_{n+1}$ and $d(f(x_{n+1}),f(y_{n+1}))\geq 4\epsilon$.

This completes the construction of the sequences $\{\delta_{n}\},\{x_{n}\},\{y_{n}\}.$ We note that if $A=\{x_{n}:n\geq 1\}$ and $B=\{y_{n}:n\geq 1\}$ then $d(A,B)=0.$ We get the desired contradiction by showing that $d(f(A),f(B))\geq \epsilon$. For this we have to show that $d(f(x_{m}),f(y_{n}))\geq \epsilon$ for all $m$ and $n$ For $m=n$ we already know that $d(f(x_{n}),f(y_{m}))\geq 4\epsilon$ We first prove the inequality for $m Suppose, if possible, $d(f(x_{m}),f(y_{n}))<\epsilon$ (*) $.$ Then $d(x_{n},y_{n})<\delta _{n}\leq \delta_{m+1} Note that $y_{n}\in N(x_{m})$ because $d(f(x_{n}),f(y_{m}))\leq \epsilon$. If $x_{n}\in F(x_{m})$ we would have $d(x_{n},y_{n}) a contradiction. Thus $x_{n}\notin F(x_{m})$ which means $d(f(x_{n}),f(x_{m}))<2\epsilon$. But then $4\epsilon \leq d(f(x_{n}),f(y_{n}))\leq d(f(x_{n}),f(x_{m}))+d(f(x_{m}),f(y_{n})).$ Once again assume that $d(f(x_{m}),f(y_{n}))<\epsilon$. Then $d(x_{m},y_{m})<\delta_{m}\leq \delta_{n+1} Also the assumption that $d(f(x_{m}),f(y_{n}))<\epsilon$ implies that $x_{m}\in N(y_{n}).$ Hence the previous inequality implies that $y_{m}\notin F(y_{n}).$ This means $d(f(y_{m}),f(y_{n}))<2\epsilon.$ But then $4\epsilon \leq d(f(x_{m}),f(y_{m})) \leq d(f(x_{m}),f(y_{n}))+d(f(y_{n}),f(y_{m}))<\epsilon+2\epsilon$.This contradiction completes the proof.

3. [ Contributed by Manjunath Krishnapur]

Let $(\Omega ,\mathcal{F},P)$ be a probability space. Suppose $\{\mathcal{F}_{n}\}$ is an increasing sequence of sigma algebras on $\Omega$ contained in $\mathcal{F}$ and $\{\mathcal{G}_{n}\}$ is a decreasing sequence of sigma algebras on $\Omega$ contained in $\mathcal{F}$ such that  $\bigcap\limits_{n}\mathcal{G}_{n}$ is trivial. [A sigma algebra is trivial w.r.t. a probability measure $P$ if every set in it has probability $0$ or $1$]. Let $X$ be a random variable on $(\Omega ,\mathcal{F},P)$ which is measurable w.r.t. $\sigma \{\mathcal{F}_{n},\mathcal{G}_{n}\}$ ( the sigma algebra generated by $\mathcal{F}_{n}\cup \mathcal{G}_{n}$) for each $n$. Does it follow that $X$ is measurable w.r.t. the completion of the sigma algebra generated by all the $\mathcal{F}_{n}^{\prime }$?

No! Let $\{Y_{n}\}$ be i.i.d. non-constant random variables, $\mathcal{F}_{n}=\sigma\{Y_{2},Y_{3},...,Y_{n}\},\mathcal{G}_{n}=\sigma \{S_{n},S_{n+1},...,\}$ where $S_{n}=Y_{1}+Y_{2}+...+Y_{n}$. Let $X=Y_{1}$. Since $X=S_{n}-\{Y_{2}+Y_{3}+...+Y_{n}\}$ it follows that $X$ is measurable w.r.t. $\sigma \{\mathcal{F}_{n},\mathcal{G}_{n}\}$ for each $n$. By Kolomogorov’s $0-1$ Law $\bigcap\limits_{n}\mathcal{G}_{n}$ is trivial. However sigma algebra generated by all the $\mathcal{F}_{n}^{\prime }$ is $\sigma \{Y_{2},Y_{3},...\}$ and $X$ is independent of this and hence not measurable w.r.t. this sigma field (or its completion) since it is assumed to be non-constant.