]]>

]]>

random variables does it follow that .

assuming that all the products and expectations exist.

2) Let be measurable and be continuous for all in some set of positive measure. Show that is continuous. Show that measurability cannot be dropped. If be continuous for all in some set with a limit point can we conclude that is continuous? Does there exist a set of measure such that if is continuous for all in then If is necessarily continuous? If is at most countable show that there exist such that is continous for all in but is not continuous. If is continuous and is an entire function for all in some set with a limit point can we conclude that is entire?

3) Let and be projections onto closed subspaces and of a

Hilbert space Find a necessary and sufficient condition on and

for to be a projection.

]]>

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.

]]>

2. Prove that a function from one metric space to another is uniformly continuous if and only if implies where .

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. Let be a random variable on which is measurable w.r.t. for each . Does it follow that is measurable w.r.t. the completion of the sigma algebra generated by all the ?

Note:- A sigma algebra is trivial w.r.t. a probability measure if every set in it has probability or and is the sigma algebra generated by

]]>

a) What is Is the representation unique for

b) What is (n terms) Do we have uniqueness here?

c) If is any normed linear space of dimension greater than prove that

We claim that and that the representation is unique. Clearly Let . Then where For

any we have the representation If with and real then and It follows that Thus and Thus the representation is unique. Uniqueness of representation for follows from this.

The answer to (b) is

(n terms)

and the representation with for all is unique if and only if Clearly any is a sum of complex numbers with modulus at most and hence it can be written as a sum of elements of Now fix and suppose with for all Then and if then with for all . But for all so for all and Thus the representation is unique when We claim that it is not unique if There are infinitely many complex numbers such that and In fact one such number is where is the polar form of By continuity the inequality holds for any sufficiently close to . Any representation of as a sum of elements of gives a similar representation for Hence the representation is not unique.

c) In the case of complex scalars we can restrict to a one-dimensional subspace and in the case of real scalars we can look at a two dimensional subspace containing a given vector with .

Note that uniqueness fails: let and in . Then .

2. If are independent identically distributed positive random variables does it follow that assuming that all the products and expectations exist.

No! Let take values and with probabilty each. Then so . We claim that . Since it follows that is a non-negative martingale. Hence it converges a.s. We shall show that the limit is a.s. thereby completing the proof. Let . Then . By SLLN applied to we see that . Hence a.s. because .

3. Consider the inequalities and . Are these true when and are open sets in ? Are these true when and are compact sets in ?

Suppose and are compact. Translate by . Then the inequality does not change and the proof is therefore reduced to the case say. In this case ( because and (because and are disjoint) so . The reverse inequality fails when the Cantor set. It is not true that for all open sets: there is an open set such that and . Since we would have and choose compact sets such that and . Then .

]]>

]]>

]]>

]]>

]]>