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Old Apr 17th 2018, 11:24 PM   #1
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Can magnitude be subtle, intangible?

Now coming to my line of study which is Finance. We have in technical analysis the concept of ADX (Average Directional Index) which I think is a vector.

Is ADX a vector?

Here we have the magnitude which is share price and the direction which is uptrend or downtrend. Now the magnitude is subtle in this case. It is the share price. Can magnitude be subtle?
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Old Apr 17th 2018, 11:30 PM   #2
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Also clarify the following.

In case of stocks the velocity is the return of a stock during a period in time which is the velocity part. But momentum is mass x velocity. What is the mass in case of stocks?
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Old Apr 18th 2018, 01:42 AM   #3
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A Vector is a Mathematical Construct.
It has been found that this mathematical construct can be used to model certain observed physical phenomena.
Distance, Velocity, Momentum, can all be modelled using vectors (and many, many, many, more phenomena).

They can also be modelled without using vectors, but that leads to clunky awkward mathematical expressions.
Vectors were developed to provide a nice streamlined description which deals with more than one dimension of interdependent values in the same mathematical expression.

Your ADX could well be expressed as a vector,
but I would hesitate to push the analogy with velocity.
Just because two phenomena can both be described using vector notation does not mean that they have anything else in common.
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Old Apr 18th 2018, 03:40 AM   #4
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Originally Posted by avito009 View Post
Now coming to my line of study which is Finance. We have in technical analysis the concept of ADX (Average Directional Index) which I think is a vector.

Is ADX a vector?

Here we have the magnitude which is share price and the direction which is uptrend or downtrend. Now the magnitude is subtle in this case. It is the share price. Can magnitude be subtle?
Now the magnitude is subtle in this case.
What do you mean by subtle?

This is not a Philip Pullman novel.


Is ADX a vector?
I doubt it.

Can you add two of them together according to the vector addition laws?
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Old Apr 18th 2018, 04:22 AM   #5
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Addition of vectors.

I think the vectors in case of ADX can be added. But we wont use law of sines or cosines. Lets say we have two coordinates change in price and price. Suppose CMP of Studiot Inc Ltd is Rs 100 and change in price from yesterday to today i.e yesterdays price is Rs 90. So change in price is 10.

Now under vector addition law for coordinates we have.

A= (100, 10) and B= (90, 0) now adding the two we get:

A+B= (190, 10). Now is this method of vector addition correct?
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Old Apr 18th 2018, 04:47 AM   #6
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Assuming you are correct that these can be treated as vectors, then yes, that is correct "vector addition".
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Old Apr 18th 2018, 04:48 AM   #7
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Originally Posted by avito009 View Post
I think the vectors in case of ADX can be added. But we wont use law of sines or cosines. Lets say we have two coordinates change in price and price. Suppose CMP of Studiot Inc Ltd is Rs 100 and change in price from yesterday to today i.e yesterdays price is Rs 90. So change in price is 10.

Now under vector addition law for coordinates we have.

A= (100, 10) and B= (90, 0) now adding the two we get:

A+B= (190, 10). Now is this method of vector addition correct?
Vectors have additive inverses, which are also vectors. What is the additive inverse of the vector (100, 10)?

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Old Apr 18th 2018, 04:55 AM   #8
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A+B= (190, 10). Now is this method of vector addition correct?
I know very little of the maths of economics, finance or biological stuff.

Vectors mathematically obey the vector space axioms.
These are much more general than pointy arrows.

So if your finance maths supports all of these then yes, otherwise no.

A vector space V is a collection of all like vectors plus one binary operation between any two of these vectors and a second collection of objects called scalars which can be used as coefficients.

I have used bold to signify vectors.

u and v are vectors in a vector space V if

1) Denoting the binary operation as addition if u and v are in V then there is a unique vector w = (u + v) also in V. although w may also be formed from adding adifferent pair such as x and y

2) (u + v) = (v + u)

3) u + (v + w) = (u + v) + w

4)There is a vector 0 in V such that 0 + u = u + 0 = u for all u in V

5) For each u in V there is a vector -u such that u + (-u) = (-u) + u = 0

6) If k, l are any coefficients from the second collection then if u is in v then ku is in V

7) k(u + v) = ku + kv

8)(k + l)u = ku + lu

9) k(lu) = (kl) (u)

10) 1u = u

Axiom 5 refers to topsquark's additive inverse.

Last edited by studiot; Apr 18th 2018 at 05:21 AM.
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Old Apr 21st 2018, 10:52 PM   #9
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Vector Inverse.

You are right topsquark. You have a valid point.

5) For each u in V there is a vector -u such that u + (-u) = (-u) + u = 0

Lets take the vector share price which is 100. So for this vector we dont have an inverse. There is no -100 because share price cant be negative. So can we conclude that share price is not a vector.

But change in share price is a vector which in our example is + 10. Now it can be -10 also which means reduction in share price.

u + (-u) = (-u) + u = 0

10+(-10)= (-10) + 10 = 0.

Now ADX is all about velocity and velocity is change in share price so ADX is a vector.
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Old Apr 22nd 2018, 12:07 AM   #10
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Stock Momentum.

Basically Momentum = Mass * velocity

Share Momentum = Trade volume in a particular period * price change in that period.

Now Mass which is Trade volume in a particular period is a scalar but Velocity of Stock which is price change in that period is a vector.

So Momentum is scalar x vector= vector.

Since broadly ADX is a Momentum indicator it is a vector according to this logic.
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