All Heads when tossing a coin – prob^stats

A well-programmed computer simulates a biased coin. For this illustration, we can also use a physical coin. First toss is a head. You update your estimate of the true parameter (m) of the coin. Second toss is a head. You update it again. Third toss is a coin… So how exactly do you update it?

Is this statistics or probability problem? More like statistics IMO. Real data. There’s a lot of probability math in this statistics problem.

I guess you start with an estimate of m before the first toss. Safe choice is  50%. As you see more heads, you probably increase  your estimate but exactly how? Perhaps some MLE?

statistical independent =\=> no causal influence

When I was young, I ate twice as much rice as noodles; Now I still do. So the ratio of rice and noodle intake is independent of my age. This independence doesn’t imply that my age has no influence on the ratio. It only appears to have no influence.

https://bintanvictor.wordpress.com/2015/02/02/2-multivariat-normal-variables-can-be-indie/ shows that 2 RV are each controlled by a family of “driver random variables”, but the 2 RV can be independent!

Note the mathematical definition of independence is based on covariance. There must be a stream of paired data points. I would say the mathematical meaning of independence is fundamentally different from everyday English, so intuition often gets in the way.

Special context — time series. We record 2 processes X(t) and Y(t). Both could be influenced by several (perhaps shared) factors. In this context, the layman’s independence is somewhat close to the mathematical definition.
* historical data – we could analyze the paired data points and compute a covariance. We could conclude they are independent, based on the data. We aren’t sure if there’s some common factor that could later give rise to a strong covariance
* future – we are actually more interested in the future. We often rely on historical data

 

abc pairwise independence =\=> Pa*Pb*Pc

Q: If we know events A, B, C are pairwise independent, does Pa*Pb*Pc mean anything?

Q: does Pa*Pb*Pc equal P(A & B & C)?
A: no. The multivariate normal model would imply exactly that, but this is just one possibility, not guaranteed.

Jon Frye gave an excellent example to show that P(A & B & C) can have any value ranging from 0% to minimum(Pa, Pb, Pc, Pab, Pac, Pbc). Suppose Pa = Pb = Pc = 10%. Pairwise independence means Pab = 1%. However, P(abc) can be 0% or 1% (i.e. whenever A and B happen, then C also happens)

School Prom illustration – each student decides whether to go, regardless of any other individual student, but if everyone else goes, then your decision is swayed.

stable job4H1 guys#le2HenryWu

Hi Henry,

See if you can connect me to your H1 sponsor at your earliest convenience.

For most H1 immigrants, having a stable job is a top priority. We all worry about losing our job, losing the H1 status and Green card petition.

Therefore, many prefer a big, reputable employer. Some prefer a consulting firm that can help maintain our H1 status even when we change project from time to time. There are definitely risks of “gaps” between 2 jobs. In my experience, 1 to 3 months are tolerable. Beyond that, there are probably other solutions. It all depends on the last employer and the lawyer. Remember I’m not an immigration attorney.

In the Worst scenario the employer cancels the H1 right away. The USCIS regulation probably allows us (“the aliens”) to stay in the US for a few weeks looking for the next job. If we can’t find any, we should ask our lawyer when we have to leave the country. We would re-enter once we find a new employer.

The exit/reentry can (in my imagination) be a real hassle for someone with a big family, esp. if kids are in school. It might be best to avoid the exit/reentry. I guess this is one reason many H1 families are fearful of layoff and prefer a stable job even at a lower salary. (Overall, Singapore companies are less likely to layoff large number of staff.)

Therefore, if I were you I would prefer a stable job. As a risk taker, I will take a gamble that I could reduce the “gap” between jobs to 2 months, by being flexible on the salary.

serious j4 U.S. -ez2get jobs, till 60

ez2get jobs; abundance of jobs — At the risk of oversimplifying things, I would single out this one as the #1 fundamental
justification of my bold and controversial move to US.

There are many ways to /dice-n-slice/ this justification:
* It gives me confidence that I can support my kids for many years. * I don’t worry so much about aging as a techie
* I don’t worry so much about outsourcing and a shrinking job pool * when I don’t feel extremely happy on a job, I don’t need to feel trapped like I do in a Singapore job.
* I feel like an “attractive girl” not someone desperately seeking on “a men’s market”.

Look at Genn. What if I really plan to stay in one company for 10 years? I guess after 10 years I may still face problem changing job in a market like Singapore.

underlier price levels in grid^tree

grid means the PDE pricer; tree means the binomial recombinant tree pricer.

In the grid, the price levels are equidistant. In the tree, the log price levels are presumably equidistant, since the up nodes are log(X0)+log(u) -> log(X0)+2 log(u) -> log(X0)+3 log(u)

I believe the price levels in the tree are related to GBM.. but that’s another blog…