1. Let and be random variables such that all have the

same distribution. If the common distribution has finite mean show that

a.s. Prove that the assumption on finiteness of the mean cannot be dropped.

Since it follows that for some non-negative random variable . Hence . Noting that we get which implies . Since both sides have the same mean we get

This implies so when . In other words implies and or . But implies so . Hence in both cases and a.s. It follows that a.s.

For the counterexample let be i.i.d. with characteristic function . Let and .

2. Prove that a function from one metric space to another is uniformly continuous if and only if implies [ is defined as ].

[ Solution by S Kannappan]

If is not uniformly continuous then there exists such that for every we can find points and with but Let . Let but Inductively define satisfying the conditions as follows: having found let where and for any Note that If then for all and if and for some then . Hence is well defined. We can find such that and .

This completes the construction of the sequences We note that if and then We get the desired contradiction by showing that . For this we have to show that for all and For we already know that We first prove the inequality for Suppose, if possible, (*) Then Note that because . If we would have a contradiction. Thus which means . But then Once again assume that . Then Also the assumption that implies that Hence the previous inequality implies that This means But then .This contradiction completes the proof.

3. [ Contributed by Manjunath Krishnapur]

Let be a probability space. Suppose is an increasing sequence of sigma algebras on contained in and is a decreasing sequence of sigma algebras on contained in such that is trivial. [A sigma algebra is trivial w.r.t. a probability measure if every set in it has probability or ]. Let be a random variable on which is measurable w.r.t. ( the sigma algebra generated by ) for each . Does it follow that is measurable w.r.t. the completion of the sigma algebra generated by all the ?

No! Let be i.i.d. non-constant random variables, where . Let . Since it follows that is measurable w.r.t. for each . By Kolomogorov’s Law is trivial. However sigma algebra generated by all the is and 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.