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The more he talks about trust, the faster I count my silvers - George Soros

January, 2010

CrossRoads

Monday, January 18th, 2010

I have been thinking about this a long time but never managed to drum up enough enthusiasm to write it up. But there is something about today being the Martin Luther King Day that finally pushed me to write this blog.

USA is at cross roads.

As an outsider I always found USA confusing, for e.g. (Like rest of the world) things like opposition to the gun control law or willingness to continue the death penalty. What is it with these people who refuse to see the obvious and at the least ban guns?

A few days back, someone reminded me that USA is a republic. This provoked me to think further and very soon there was light.

Now I can see very clearly the basic philosophy that drives this country (or used to drive this country).

It is everyone’s right to work hard and get rewarded for it. It is everyone’s right to earn fabulous sums of money and enjoy their wealth. However, there will always be a group of people who do not want to work hard but still wish to get rewarded. In this case, it is everyone’s right to take up arms and protect themselves and if possible dispatch such people to the death row.

It is this philosophy that allowed son of a truck driver to earn the riches (read Starbucks).

It is this philosophy that enabled this country to shoot up to the nadir of wealth and power.

However, this philosophy works if and only if everyone has a chance to make it big.

This has not been the case for a long time but very few people could see this clearly.

The reality hit home after the bank bailout.

Suddenly people understood that there is now an elite class firmly entrenched in this country and however hard they might work, it is becoming difficult for other’s to join the ranks.

This is why Obama got elected; this is why he could muster support even in traditional republican alias.

This is why this country is seeing incessant demand for socialistic welfare policies.

I suppose socialistic welfare policies are an eventual certainty for any civilization that has risen to the riches.

Has this country arrived at the crossroad?

I don’t know which road we are taking.

All this discussion is not an empty theoretical debate.  What I am trying to guess is that which stocks to go long on and here is where it gets confusing for me.

I think if we see hints of de-globalization then go for small and medium companies.

If we do not see such hints then go for large corporations.

What do you think? Any comments?

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Revisiting partial derivatives (Using Stochastic article 18 of n)

Wednesday, January 13th, 2010

(Read from the beginning) OR
(If you have not read article "17 of n" already then please consider reading it before continuing here

Reinforcing the concept of partial derivatives

It seems there are two things in calculus that are vital in understanding the derivation of ITO’s lemma

One of them is Taylor series, which I have covered in my previous post.

The other one is partial derivatives. Though I had covered this topic in one of my previous posts, I was not very happy with what I had done. So I decided to spend more time to make sure that I really understand this concept.

Let us take one more theoretical example:

Find the partial derivatives for f(x,y) =   xy^3 + y^2 + y

[f(x,y) is just another way of saying function of x and y]

First let us keep x constant and find partial derivative with respect to y

Step 1: Say we need to find derivative of 3y^3 which is 9y^2. Now instead of 3y^3, we need to find derivative of xy^3. We treat x as constant (same as 3 in 3y^3) and hence the partial derivative is 3xy^2

Step 2: We already know that derivative of y^2 is 2y

Step 3: We also know that the derivative of y is 1.

Hence

\frac{{\partial f}}{{\partial y}} = 3xy^2 + 2y + 1

Now let us keep y constant and find partial derivative with respect to x.

Step 1: partial derivative of xy^3 is y^3 (derivative of x  is 1 and y^3 is a constant)

Step 2: partial derivative of y^2 is 0 (Think over this, y^2 is a constant and hence derivative of y^2 with respect to x is zero)

Step 3: Same with y, the partial derivative with respect to x is zero.

Hence

\frac{{\partial f}}{{\partial x}} = y^3

You might be initially confused about why derivative of y^2 is zero. Look at this way. If we had a function x^2 + 2x + 3 then the derivative of this function will be 2x + 2. Notice that we have calculated derivative of 3 as zero since 3 is a constant. In our example y^2 is also a constant and hence the derivative of y^2 is zero.

The trick is to be able to mentally keep one or more of the variables as constant.

Let us now take a practical example to make sure we have really understood the concept:

Consider a rectangular box with width same as height. Let us say we denote width of the box as x and the length as y.  Thus we get volume of the box as equal to x^2 y (width * height * length).

Now let us say we keep length as constant.

(a)    If the length is constant at 2 then in this case our formulae for volume is 2x^2. What is the rate of increase? Rate of increase is same as the first derivative i.e. 4x.

(b)   Now let us keep length constant at 3. In this case our formulae for volume is 3x^2. What is the rate of increase? Rate of increase is 6x

(c)    Finally let us keep length constant at 4. In this case our formulae for volume is 4x^2. What is the rate of increase? Rate of increase is 8x.

Is there any way to consolidate the above information? Yes, what we are basically doing above is to take partial derivatives i.e. keep one variable (length) constant.

Our original formulae for volume is x^2y. i.e. v = x^2 y. Now treat y as constant and take first derivative to find the rate of increase as 2xy. If you glance at the above calculations (a),(b),(c), you will realize that 2xy is just a fancy way of saying the same thing..4x, 6x, 8x etc (just substitute y with the actual values).

So now let us rewrite our fancy equation in the way mathematicians like it. It is very important to get used to this weird symbol. :)

\frac{{\partial v}}{{\partial x}} = x^2y

 (I must say there seems to be something confusing over here. I would have thought that the partial derviative would have been shown like \frac{{\partial x}}{{\partial v}} but I suppose there is some logical reason for showing it the other way i.e. \frac{{\partial v}}{{\partial x}})

 

(Continue reading: article 19 of n)
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