This is a homework question for me, so I'm not looking for the full answer, however I am looking for some insight.

You're given:


Now, say you have a sufficiently smooth function f(t,W(t)). I want to find .

Here is what I have (using Ito's Lemma):






The problem I'm getting is that the answer should have a
term added to it. I just don't see how to do it. Is there a typo, and/or am I doing something wrong?

(in before Jason -- hopefully not presumptuous, I enjoy this thread a lot) I believe your error is here. Since W and X are not independent processes, you should have some sort of a cross-variation term.

Use integration by parts, and apply Ito's formula separately to the various terms you get.

I'm graduate student working on stochastic analysis, especifically asymptotic behavior of stochastic flows (i.e CLTs, etc)
Besides academia, is there any job prospect for me as a probabilist? If possible be specific ?

I know some people go into finance. What do they do specifically?

Do I need programing skills ( which I don't have)?

Sometimes, I feel I'm doomed.

My fve cents. Lets hope someone more competent takes the time to give you an better answer.


>I know some people go into finance. What do they do specifically?

Realized variance has been a popular topic (at least the first half of this decade) for mathematicians and engineers that work with finance - do a google search and youll find some good publications.

Here is a classic publication on the topic, the math is a bit outadet atm, but its a good introduction:

http://www.ssc.upenn.edu/~fdiebold/p...er43/abdl4.pdf


>Do I need programing skills ( which I don't have)?

How are you going to implement a pratical solutions to a fianace problem if you do not know how to program? I'm pretty sure the reason i had so easy to find job was the fact that i had good programming skills on top of an engineering & financial education. I strongly reccomend you take a c++ course, you will master it in no time, a few months and you can solve msot problems you will ever encounter (well, maybe not efficiently, but solve never the less)

R
J

prob an easy Q. seems logical to me but not sure how to prove it.

assume two distributions, distA and distB. they are defined by their mean and standard deviations.

both have identical first, third, and fourth moments. is it possible that they have different variances?

my logical answer is, well no b/c you'd alter either the skew or kurt values if you changed the spread of the results about the mean. is that right? and if so, how would i go about proving this?

thanks,
Barron

Use symmetric distributions around 0 so the first and third moments are always zero. Now the question is if an identical fourth moment implies an identical second moment. It's pretty easy to show this isn't true for a discrete case.

Let A be a distribution of 4 points with equal weight.

A: -2, -1, 1, 2
4th moment = (2^4+1^4+1^4+2^4)/4= 8.5
2nd moment = (2^2+1^2+1^2+2^2)/4= 2.5

Let B be a distribution of 2 points with equal weight.

B: +/- 4th root of 8.5 (1.70748..) call that number Q
4th moment = (Q^4+Q^4)/2 = (8.5+8.5)/2 = 8.5
2nd moment = (Q^2+Q^2)/2 = Q^2 = sqrt(8.5) ~= 2.915.

Both have identical 1st moment (0), 3rd moment (0), 4th moment (8.5), but different 2nd moments.

Wasn't there a criteria for the distributions to "be defined by their mean and variance?" I'm not sure A and Q meet this requirement.

right. something like gaussian defined by u and s. now introduce the rest of the criteria. i THINK in that case you can't have identical 3rd/4th/1st moments w/o identical variance but not sure.

Barron

Sure, if you restrict to gaussians, then the means have to be the same (first moment), and since the fourth moment is a function of variance that doesn't allow two different variances to give the same fourth moment (this is what you would show to prove it), your condition would hold.

Does this generalise to all elliptic distributions? I guess it does in general, since the characteristic function only depends on /mu and /sigma. So if the third and fourth moments give independent equations in /mu and /sigma, the condition holds.

However, it will be possible to pick examples where this fails. e.g. choose a distribution where third moment =0; fourth moment=0 and fifth moment=/sigma. But these examples will be sparse and general elliptic distributions will have the required property.

(I believe that Elliptical distributions include normal, lognormal, Cauchy, student t and others. Reference here.)

rodekio seems to be making the simplifying assumption that there is no green on the roulette wheel. Under this assumption, Brussels Sprout is correct.

jason1990; this is a excellent thread. Thank you very much for your time.

With limited time, I have to pick my spots carefully. Since I see that you received a lot of thoughtful responses from professional statisticians in the probability forum, I will leave it to them to answer your questions. Best of luck.

repulse is correct. You should be using the Ito integration by parts formula.

My only work experience as a probabilist is in academia, so I am afraid I cannot be very specific. If you can pass yourself off as a statistician, then I believe there are a lot of great job opportunities in industry.

Again, I cannot say, specifically. I know a lot about the mathematical models used in that field, but I do not know the typical day-to-day activities of someone who works in finance.

Outside of academia, I think this is probably a must. If you decide to try to be a statistician, then you might want to learn the programming language, R.

First, you have to be more specific about the meaning of the phrase, "they are defined by their mean and standard deviations." I will suppose that it means the following. There is a family A of cumulative distribution functions such that

1. For each pair (m,s), where m is a real number and s is a nonnegative real number, there exists a unique distribution function F(x) in A with mean m and standard deviation s.
2. If F(x) is in A and a and b are real numbers with a > 0, then F(ax + b) is in A.

The question is now: if F and G are in A, and F and G have identical first, third, and fourth moments, then does it necessarily follow that F = G? The answer is yes.

To see this, let H in A be the distribution function with mean 0 and variance 1. Let X be a random variable whose distribution is H. If m is the mean of F and G, and s and t are the standard deviations of F and G, respectively, then Y = m + sX and Z = m + tX have distributions F and G, respectively.

Since F and G have the same fourth moments, we have E[(Y - m)⁴] = E[(Z - m)⁴], which gives

s⁴E[X⁴] = t⁴E[X⁴].

Note that E[X⁴] = 0 implies X = 0 almost surely, which implies X has variance 0. Therefore, since X has variance 1, we must have E[X⁴] > 0, which implies s = t.

I don't know much, either (and almost assuredly MUCH less than jason knows on the subject), but SAS would be another good program to learn in as much detail as possible, if you're interested in being a professional statistician.

Reading this gem, http://archives2.twoplustwo.com/show...st682045683150

How did he get that equation?? The gamblers risk of ruin afaik is

R(n) = p R(n-1) + (1 - p) R(n+1)

and then using the recurrence formula and finding the roots using r(0) = 1 (risk of ruin when you have no money) and r(N) = 0 (risk of ruin when you have all the money). And I get,

R(n) = (1 - r^n) / (1 - r^N)
where r = (1 - p) / p

I can list all the steps. How did he derive that equation r = q + (1 – q)*r^2 ?

I can't make sense of what he wrote. It looks like a recursive formula so is it just missing subscripts?

What does this mean? If I assume R(n) = p [ R(n+1) + R(n-1) ] where p = 1/2, then I get:

2R(n) = R(n+1) + R(n-1)
R(n+1) = 2R(n) - R(n-1)

R(0) = 1
R(N) = 0

r^2 - 2r + 1 = 0
(r - 1)^2 = 0
r = 1

R(n) = C + Dn
1 = C
0 = C + DN
D = -1/N

R(n) = 1 - n/N

So it works just following the linear recurrence equation but I want to understand WHY (what the statement means).

But that 2p2 archive article doesn't assume p = q. How does that 2p2 article get that equation? How, how, how

Thanks for your help (a very confused individual)

I'll take a shot at the first part:

His formula was:

"r" in his terminology corresponds to R(1) in your terminology. It's the risk of ruin starting with $1. Also, it appears that the two of you have your p's and q's switched - he uses p for the probability of a won flip for you, you use p for the probability of a lost flip for you. Conversely for q.

Let's take the risk-of-ruin formula you have and write down the particular expression for R(1) (I'll use q to mean probability of a lost flip):

R(1) = q * R(0) + (1 - q) * R(2)

R(0) = 1.0 and he's using r to mean R(1), so we have:

r = q + (1 - q) * R(2)

He asserts that R(2) is r * r because the probability of "[losing] a $2 bankroll... is the probability of losing $1 twice". Note, crucially, that your opponent has an infinite bankroll. Under this assumption, this statement is true. So that leads to his formula

r = q + (1 - q) * r^2

Thanks for taking the time to type out an answer, but this is the mental hurdle I have where he goes from R(2) to r^2. Isn't r (the risk of ruin) dependant on the current depth in the tree? So how is it possible to describe r in terms of itself?

Ah, I think maybe you are confusing lower case "r" with upper case "R". At the beginning of the original post, the author essentially defines "r" to be R(1) - i.e., the chance of going bankrupt given a starting bankroll of $1. The part of the derivation you are confused about should be read keeping this definition of r in mind.

Later on (after "What if our bankroll were B?") the author starts using "r" to talk about risk-of-ruin for different bankrolls so it then becomes effectively the same as your R. So I can see where a confusion would arise.

The implicit assertion (at the beginning of the original post) is just that "Given N dollars, the probability of your bankroll ever dipping as low as N-1 dollars is r". For any N.

So, in words, the risk of ruin given $2 is the probability of ever dipping as low as $1 (given that you have $2) times the probability of ever dipping as low as $0 (given that you have $1). I guess there's an assumption of independence in there. Each of those two probabilities being multiplied is r so R(2) = r^2.

Probably don't need to be a probabilists to solve it, but I still need some help with it.

Problem:
There are 16 installed units on a machine. They fail independtently, and has a average lifespan of 10 years.

What assumption do I need to make about the failure rate?

How many spare units do you need to have a 99.95% chance of having more spare parts then broken units.

Is my problem as presented a decent way to find how many spare parts you need to have if you're willing to take a certain risk (in this case you're out of spare parts 1 of 1000 times), or am I doing it wrong?

Yes I understand!!!!!!!! THANKS!!

R(1) = p R(0) + q R(1) * R(1)
because R(2) = R(1) * R(1) (losing $2 TWICE)

thanks again! /enlightenment

well for starters you state they on average last 10 years, but what is the deviation for that average (called standard deviation).

and then you don't state how long you want to use them for. Will it be 10 years? Of course you can make a graph though for 99.95% showing all the values for different years

so give this info and the answer is easy. Normal curve, and the central limit theorem (look up on Wikipedia)

X~N(µ,σ²)