The Generating New Probability Theorems (Con)

Following to previous article posted on link:  http://emfps.blogspot.com/2016/09/the-generating-new-probability-theorems.html, you can review second theorem as follows:

 Theorem (2): Rule of sixty plus (60 %+)

I start theorem (2) just like theorem (1) by using three dices but I change the function to a Transcendental function.

As we know, there are two types of functions in mathematics (Calculus). The first functions are named Algebraic function. These ones are the functions which are applied in below equation:

  0  =  (Pn (x)y^n  +...P1(x)y  + P0 (x        

                  ......... (Where P0 (x), P1(x
 Pn (x) are the polynomials of "x
  

I used Algebraic function for theorem (1).

Second functions are named Transcendental function including two group: 1) Trigonometric function 2) Exponential function. These functions are not compatible with above equation.

In theorem (2), I will apply a type of trigonometric function.

Assume we again set three dices on the table in one row from left to right. We will have:

Dice1      Dice2     Dice3

Now, consider three dices as three variables as follows:

Dice1 = x,    Dice2 = y, Dice3 = z

I am willing to define function f (x, y) below cited:

z = f (x, y) = x + COS y

We know the domain of "x" and "y" is the same the range of "z" equal to 1, 2, 3, 4, 5 and 6.

I am willing to know, what is below probability:

P (z <= x + COS y) =?

We can calculate the probability which is exactly equal to 50.00 %.

P (z <= x + COS y) = 50.00 %

Now, I will pull out another dice from basket and add forth dice on the table right side of dice 3. We will have Dice 4 and I will assign variable "w" to this dice:

Dice 4 = w

Suppose the function will be:

w = f (x, y, z) = x + COS (y + z)

What is below probability?

P (w <= x + COS (y + z)) =?

We have 6^4 = 1296   permutations with repetition.

The probability is again calculated exactly equal to 50.00 %.

P (w <= x + COS (y + z)) = 50. 00%

Then, I will pull out another dice from basket and add fifth dice on the table right side of dice 4. We will have Dice 5 and I will assign variable "t" to this dice:

Dice 5 = t

Suppose the function will be:

t = f (x, y, z, w) = x + COS (y + z + w)

It is clear, permutations with repetition is equal to 6^5 = 7776 and the probability will again be 50%.   

P (t <= x + COS (y + z + w)) = 50%

Finally, I will pull out another dice from basket and add sixth dice on the table right side of dice 5. We will have Dice 6 and I will assign variable "r" to this dice:

Dice 6 = r

Suppose the function will be:

r = f (x, y, z, w, t) = x + COS (y + z + w + t)

What is below probability?

P (r <= x + COS (y + z + w + t)) =?

Permutations with repetition is equal to 6^6 = 46656 and the probability is again 50%.

 P (w <= x + COS (y + z + w + t)) = 50%

Therefore, we can reach to a general theorem in which there are "n" dices:

If we have "n" dices with below function:

xn = f (x1, x2, x3, ……..xn-1) = x1 + COS ( x2 + x3+ ….. xn-1) 

Then, below Probability will be 50%:

 P(xn <= x1 + COS (x2 + x3+ ….. xn-1)) = 50% 

What is the application of theorem (2)?

Let me again refer to my article of "A Template for Financial Section of a Business Plan (Con)" 

posted on link:

http://emfps.blogspot.com/2016/08/a-template-for-financial-section-of_25.html

If you want to test these theorems and see real application of these theorems in the field of financial management

you should select 10, 20, 50 or more companies and apply these theorems on growth rate of sales and costs or amount of sales and costs in the sequential years, quarters and so on.

Here, I have bought an example for 20 company that I have covered the names. You can follow me as follows:

Step 1: Select 20 companies in different industries

Step 2: Go to Google finance and search name of each company

Step 3: For each company click on Financials (left side of page)

Step 4: Copy total revenue and cost of revenue for five sequential period of 13 weeks and paste on excel spreadsheet

Step 5: Calculate growth rate for total revenue and cost of revenue

Step 6: Calculate the function of theorem (2) for two sequential period of 13 weeks and compare it with third sequential period of 13 weeks

Step 7: Calculate the probability

You can see Figure (1) which is my example:

( 1) Figure

  In my example, I am predicting the periods of 13 weeks ending 2016- 03 – 26 and 13 weeks ending 2016 – 06 – 25. You can see, for function of z = x + cos y, the probability is exactly 100% for both periods (for revenue and cost). It means that the predictions are 100% true. But these predictions are not useful because they show us very high upper stream line. It is just like, I say to you that the sales will not increase more than 150 or 200% for the next period. Therefore, there are two important questions

1. Why is there the significant difference between the probability of theorem (2) and real probability which is 100% instead of 50%?

2. How can we decrease upper stream line where the predictions will be useful for us?

Answer to question (1):

As we can see, the growth rates usually are between -100% to + 100% (-1 to +1).

 Therefore, let me consider all dices contains infinity numbers which are members of Real Number between -1 to 1 as follows 

D1 = 

 {x │ x ϵ R, -1< x <1}

     =  D2 

{y │ y ϵ R, -1< y <1}

=   D3

{z │ z ϵ R, -1< z <1}

Now, if we use above conditions and apply Monte Carlo method for our analysis, we will find that the probability for theorem (2) will increase to more than 85% with Average approximately 85% by standard deviation of 0.001. Please see Figure 2

 Figure (2)

Answer to question (2):

Only way to decrease upper stream line is, to divide function of theorem (2) to some numbers. For instance, I apply the Monte Carlo analysis for below functions:

z = f (x, y) = (x + COS (y)) / 4

According to Figure (1), the probabilities are 95% and 85% for my example. In the reference with Monte Carlo analysis, we can say:

P (z <= (x + COS y) /4) > 65%   with Average > 64 % and standard deviation approximately 0.0015

z = f (x, y) = (x + COS (y)) / 6

Figure (1) shows us, the probabilities are 95% and 72.5% for my example. Regarding to Monte Carlo analysis, we can say:

P (z <= (x + COS y) /6) > 60%   with Average approximately 60 % and standard deviation 0.0018

Since theorem (2) divided to 6 is more useful for us, I can say that the Rule of sixty plus (60 %+) is as follows:

P (z <= (x + COS y) /6) > 60%

The generating New probability theorems (Con)

Following to previous article posted on link:  http://emfps.blogspot.com/2016/09/the-generating-new-probability-theorems.html, you can review second theorem as follows:

 Theorem (2): Rule of sixty plus (60 %+)

I start theorem (2) just like theorem (1) by using three dices but I change the function to a Transcendental function.

As we know, there are two types of functions in mathematics (Calculus). The first functions are named Algebraic function. These ones are the functions which are applied in below equation:

  0  =  (Pn (x)y^n  +...P1(x)y  + P0 (x        

                  ......... (Where P0 (x), P1(x
 Pn (x) are the polynomials of "x
  

I used Algebraic function for theorem (1).

Second functions are named Transcendental function including two group: 1) Trigonometric function 2) Exponential function. These functions are not compatible with above equation.

In theorem (2), I will apply a type of trigonometric function.

Assume we again set three dices on the table in one row from left to right. We will have:

Dice1      Dice2     Dice3

Now, consider three dices as three variables as follows:

Dice1 = x,    Dice2 = y, Dice3 = z

I am willing to define function f (x, y) below cited:

z = f (x, y) = x + COS y

We know the domain of "x" and "y" is the same the range of "z" equal to 1, 2, 3, 4, 5 and 6.

I am willing to know, what is below probability:

P (z <= x + COS y) =?

We can calculate the probability which is exactly equal to 50.00 %.

P (z <= x + COS y) = 50.00 %

Now, I will pull out another dice from basket and add forth dice on the table right side of dice 3. We will have Dice 4 and I will assign variable "w" to this dice:

Dice 4 = w

Suppose the function will be:

w = f (x, y, z) = x + COS (y + z)

What is below probability?

P (w <= x + COS (y + z)) =?

We have 6^4 = 1296   permutations with repetition.

The probability is again calculated exactly equal to 50.00 %.

P (w <= x + COS (y + z)) = 50. 00%

Then, I will pull out another dice from basket and add fifth dice on the table right side of dice 4. We will have Dice 5 and I will assign variable "t" to this dice:

Dice 5 = t

Suppose the function will be:

t = f (x, y, z, w) = x + COS (y + z + w)

It is clear, permutations with repetition is equal to 6^5 = 7776 and the probability will again be 50%.   

P (t <= x + COS (y + z + w)) = 50%

Finally, I will pull out another dice from basket and add sixth dice on the table right side of dice 5. We will have Dice 6 and I will assign variable "r" to this dice:

Dice 6 = r

Suppose the function will be:

r = f (x, y, z, w, t) = x + COS (y + z + w + t)

What is below probability?

P (r <= x + COS (y + z + w + t)) =?

Permutations with repetition is equal to 6^6 = 46656 and the probability is again 50%.

 P (w <= x + COS (y + z + w + t)) = 50%

Therefore, we can reach to a general theorem in which there are "n" dices:

If we have "n" dices with below function:

xn = f (x1, x2, x3, ……..xn-1) = x1 + COS ( x2 + x3+ ….. xn-1) 

Then, below Probability will be 50%:

 P(xn <= x1 + COS (x2 + x3+ ….. xn-1)) = 50% 

What is the application of theorem (2)?

Let me again refer to my article of "A Template for Financial Section of a Business Plan (Con)" 

posted on link:

http://emfps.blogspot.com/2016/08/a-template-for-financial-section-of_25.html

If you want to test these theorems and see real application of these theorems in the field of financial management

you should select 10, 20, 50 or more companies and apply these theorems on growth rate of sales and costs or amount of sales and costs in the sequential years, quarters and so on.

Here, I have bought an example for 20 company that I have covered the names. You can follow me as follows:

Step 1: Select 20 companies in different industries

Step 2: Go to Google finance and search name of each company

Step 3: For each company click on Financials (left side of page)

Step 4: Copy total revenue and cost of revenue for five sequential period of 13 weeks and paste on excel spreadsheet

Step 5: Calculate growth rate for total revenue and cost of revenue

Step 6: Calculate the function of theorem (2) for two sequential period of 13 weeks and compare it with third sequential period of 13 weeks

Step 7: Calculate the probability

You can see Figure (1) which is my example:

( 1) Figure

  In my example, I am predicting the periods of 13 weeks ending 2016- 03 – 26 and 13 weeks ending 2016 – 06 – 25. You can see, for function of z = x + cos y, the probability is exactly 100% for both periods (for revenue and cost). It means that the predictions are 100% true. But these predictions are not useful because they show us very high upper stream line. It is just like, I say to you that the sales will not increase more than 150 or 200% for the next period. Therefore, there are two important questions

1. Why is there the significant difference between the probability of theorem (2) and real probability which is 100% instead of 50%?

2. How can we decrease upper stream line where the predictions will be useful for us?

Answer to question (1):

As we can see, the growth rates usually are between -100% to + 100% (-1 to +1).

 Therefore, let me consider all dices contains infinity numbers which are members of Real Number between -1 to 1 as follows 

D1 = 

 {x │ x ϵ R, -1< x <1}

     =  D2 

{y │ y ϵ R, -1< y <1}

=   D3

{z │ z ϵ R, -1< z <1}

Now, if we use above conditions and apply Monte Carlo method for our analysis, we will find that the probability for theorem (2) will increase to more than 85% with Average approximately 85% by standard deviation of 0.001. Please see Figure 2

 Figure (2)

Answer to question (2):

Only way to decrease upper stream line is, to divide function of theorem (2) to some numbers. For instance, I apply the Monte Carlo analysis for below functions:

z = f (x, y) = (x + COS (y)) / 4

According to Figure (1), the probabilities are 95% and 85% for my example. In the reference with Monte Carlo analysis, we can say:

P (z <= (x + COS y) /4) > 65%   with Average > 64 % and standard deviation approximately 0.0015

z = f (x, y) = (x + COS (y)) / 6

Figure (1) shows us, the probabilities are 95% and 72.5% for my example. Regarding to Monte Carlo analysis, we can say:

P (z <= (x + COS y) /6) > 60%   with Average approximately 60 % and standard deviation 0.0018

Since theorem (2) divided to 6 is more useful for us, I can say that the Rule of sixty plus (60 %+) is as follows:

P (z <= (x + COS y) /6) > 60%

The Generating New Probability Theorems

The purpose of this article is to generate new theorems of probability and to find out some applications of these theorems. In this case, suppose that we have a covered basket that contains many dices. In many blind tests, we will reach in and pull out a dice and set it on the table on one row from left to right. It is clear, each dice has six events (choices) including 1, 2, 3, 4, 5, and 6.

Theorem (1): Rule of fifty plus (50 %+)

I start this theorem by using three dices. At the first, I pull out one dice from basket then second dice and finally third dice and set them on the table on one row from left to right.

 The question is: What is the probability for number of third dice less than or equal to average of numbers first and second dices?

Let us have the dices as follows:

D1 = first dice

D2 = second dice

D3 = third dice

D1    D2    D3

I am willing to know:

P (D3 ≤ ((D2 + D1) / 2)) =?

Definitely, we have 6^3 = 216   permutations with repetition.

We can calculate the probability equal to 54. 1667%.

 P (D3 ≤ ((D2 + D1) / 2)) = 54.1667%

Or, what is the probability for number of third dice more than or equal to average of numbers first and second dices? 

The answer is the same:

 P (D3 ≥ ((D2 + D1) / 2)) = 54.1667%

Now, another question is: What is the probability number of third dice less than or equal to twice number second dice minus first dice?

P (D3 ≤ ((2*D2) - D1)) =?

The calculation shows us that the probability is the same equal to 54. 1667%

P (D3 ≤ ((2*D2) - D1)) = 54.1667%

Therefore, we can say:

P (D3 ≤ ((D2 + D1) / 2)) = P (D3 ≥ ((D2 + D1) / 2)) = P (D3 ≤ ((2*D2) - D1))   

Let me expand this idea as follows:

If we assume all dices contains infinity numbers which are members of Real Number as follows:

D1 and D2 and D3 are subsets of Real Number

In this case, each dice is included a set of Real Number in which the numbers of sets will be different, then we will have:

P (D3 ≤ ((2*D2) - D1)) > 50%

I name this theorem: The rule of fifty plus (50 %+)

 What are the applications?

Here I have brought an example of financial management.

Let me again refer you to my article of "A Template for Financial Section of a Business Plan (Con)"
posted on link: 
 http://emfps.blogspot.com/2016/08/a-template-for-financial-section-of_25.html

We can apply this theorem to predict assumptions when we have the final reports of previous years 
by probability of more than fifty percent (50 %+) for instance, I utilize this theorem for growth rate of sales as follows:

If we replace the dices by years of sales (Income statement) and use from Monte Carlo method to iterate calculations, we will reach to the rule of 50 %+. (Please see below pic)

On above spreadsheet, we have sales for year 3, 4 and 5. Then I use different probability distribution and different growth rate for CUT OFF. By one way data table, you can see an average of P (x) is equal to 53.9% by standard deviation of 0.000499.

 P (x) = P (year5 ≤ ((2*year4) – year3)) = 53.9%

It means that there is a probability more than 50% in which sales in year 5 will be less than or equal to twice sales in year 4 minus sales in year 3.

You can check this theorem (50 %+) for all income statements, annual reports and so on.

Even though we have found out this theorem, there is still more than 40 % risk to use this theorem. But, how can we decrease the risk of projection for assumptions or increase the probability prediction of our assumptions?

For deducting this risks, we have to generate other theorems by increasing the number of dices or changing the functions.