Consider a four state Markov chain with state space , initial state , and transition probability matrix
(a) Compute .
(b) Let . Compute .
If are independent uniformly distributed random variables on [0,1], then let be the second smallest among these numbers. Find a nonrandom sequence such that converges in distribution, and compute the limiting distribution.
The two terms on the right hand side look like the limit definition of the exponential function. Can we choose appropriately so that it is?
Let . Then
This is the distribution of .
Suppose that the real-valued random variables are independent, that has a bounded density (for , with respect to Lebesgue measure), and that is integer valued.
(a) Prove that has a density.
(b) Calculate the density of in the case where Uniform[0,1] and Poisson(1).
where the last equality follows from Monotone Convergence Theorem.
Hence, we have shown explicitly that has a density and it is given by .
When Uniform[0,1] and Poisson(1), we have and with support on .
Then from part (a), the density will be
Let be a Poisson process with unit rate, and let
where is the indicator of the event .
(a) Find a formula for in terms of .
(b) Show that if with a fixed constant, then in probability.
We know that is distributed as Poisson with parameter . So
If then we must have only finitely often. The probability of this even (from part a) is .
This decays to 0 for . Then clearly, we see that the probability of is equal to 1.
I don't know how to show the result for ...
Let and for where the r.v.'s are i.i.d. with .
(a) Prove that there exist constants such that and are martingales.
(b) If , then prove that almost surely and find .
(c) Prove that is not a uniformly integrable martingale.
We want . We can compute both sides of this equation explicitly.
Thus if we want this equality to hold we must have .
Similarly, if we want then
We can easily check that gives a trivial solution to the equation. Using the substitution we can find another solution for . We should get .
We've just shown that is a martingale. Thus, . Then since each is i.i.d., we can apply the Strong Law of Large Numbers to say almost surely. In other words, almost surely and so certainly almost surely.
Now to calculate . We introduce new notation: let .
by a symmetry argument.
So we can write .
But I don't know how to calculate ....
Recall from part (a) that the nontrivial solution for must be some negative number. Then almost surely by part (a) as well.
However, . This by the definition, means the martingale is not right closable. A martingale is right-closable iff uniformly integrable. Thus, we're done.