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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<n. Suppose, if possible, d(f(x_{m}),f(y_{n}))<\epsilon (*) . Then d(x_{n},y_{n})<\delta _{n}\leq \delta_{m+1}<d(N(x_{m}),F(x_{m})) 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})<d(N(x_{m}),F(x_{m}))\leq d(y_{n},x_{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}<d(N(y_{n}),F(y_{n})). 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.


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